\(\frac{x-2}{7}+\frac{x-7}{2}=\frac{7}{x-2}-\frac{2}{x-7}\)

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3 tháng 8 2020

\(\frac{x-2}{7}+\frac{x-7}{2}=\frac{7}{x-2}-\frac{2}{x-7}\)ĐKXĐ: \(x\ne2\)và  \(x\ne7\)

\(\Leftrightarrow\frac{2\left(x-2\right)}{14}+\frac{7\left(x-7\right)}{14}=\frac{\left(x-7\right)7}{\left(x-7\right)\left(x-2\right)}-\frac{\left(x-2\right)2}{\left(x-7\right)\left(x-2\right)}\)

\(\Leftrightarrow2\left(x-2\right)+7\left(x-7\right)=\left(x-7\right)7-\left(x-2\right)2\)

\(\Leftrightarrow2x-4+7x-49=7x-49-2x-4\)

\(\Leftrightarrow\left(2x-4+7x-49\right)-\left(7x-49-2x-4\right)=0\)

\(\Leftrightarrow2x-4+7x-49+7x+49+2x+4=0\)

\(\Leftrightarrow4x-49x-8-98=0\)

\(\Leftrightarrow x\left(4-49\right)-106=0\)

\(\Leftrightarrow x\left(-45\right)=106\)

\(\Leftrightarrow x=\frac{-106}{45}\)

<=> \(\frac{2x-3}{35}\)\(\frac{x^2-2x}{7}\)\(\frac{5x^2}{35}\)\(\frac{7\left(2x-5\right)}{35}\)

<=> \(\frac{2x-3}{35}\)\(\frac{5\left(x^2-2x\right)}{35}\)\(\frac{5x^2}{35}\)\(\frac{7\left(2x-5\right)}{35}\)

<=> 2x - 3 + 5( x2 - 2x ) < 5x2 - 7( 2x - 5 )

<=> 2x - 3 + 5x2 - 10x < 5x2 - 14x + 35

<=> 2x + 5x2 - 10x - 5x2 + 14x < 35 + 3

<=> 6x < 38

<=> x < \(\frac{19}{3}\)

có sai thì thông cảm :)

20 tháng 1 2020

\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)

<=> \(\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]-24=0\)

<=> \(\left(x^2+x\right)\left(x^2+2x-x-2\right)-24=0\)

<=> \(\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)

Đặt t = x2 + x 

<=> t(t - 2) - 24 = 0

<=> t2 - 2t - 24 = 0

<=> t2 - 6t + 4t - 24 = 0

<=> (t + 4)(t - 6) = 0

<=> \(\orbr{\begin{cases}x^2+x+4=0\\x^2+x-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x^2+x+\frac{1}{4}\right)+\frac{15}{4}=0\\x^2+3x-2x-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\left(ktm\right)\\\left(x-2\right)\left(x+3\right)=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

Vậy S = {2; -3}

(lưu ý: thay "ktm" thành vô lý và giải thích thêm)

\(\left(x+3\right)^4+\left(x+5\right)^4=2\)

<=> (x + 4 - 1)4 + (x + 4 + 1)4 - 2 = 0

Đặt y = x + 4

<=> (y - 1)4 + (y + 1)4 - 2 = 0

<=> y4 - 4y3 + 6y2 - 4y + 1 + y4 + 4y3 + 6y2 + 4y + 1 - 2 = 0

<=> 2y4 + 12y2 = 0

<=> 2y2(y2 + 6) = 0

<=> \(\orbr{\begin{cases}y^2=0\\y^2+6=0\left(ktm\right)\end{cases}}\)

<=> y = 0

<=> x + 4 = 0

<=> x = -4

Vậy S = {-4}

20 tháng 1 2020

\(\frac{x^2+x+4}{2}+\frac{x^2+x+7}{3}=\frac{x^2+x+13}{5}+\frac{x^2+x+16}{6}\)

<=> \(\frac{x^2+x+4}{2}-3+\frac{x^2+x+7}{3}-3=\frac{x^2+x+13}{5}-3+\frac{x^2+x+16}{6}-3\)

<=> \(\frac{x^2+x+4-6}{2}+\frac{x^2+x+7-9}{3}=\frac{x^2+x+13-15}{5}+\frac{x^2+x+16-18}{6}\)

<=> \(\frac{x^2+x-2}{2}+\frac{x^2+x-2}{3}=\frac{x^2+x-2}{5}+\frac{x^2+x-2}{6}\)

<=> \(\left(x^2+2x-x-2\right)\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}-\frac{1}{6}\right)=0\)

<=> (x + 2)(x - 1) = 0 (do \(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}-\frac{1}{6}\ne0\))

<=> \(\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)

Vậy S = {-2; 1}

câu cuối: + 3 vào sau các phân số của pt như trên

Bài 1:

ĐKXĐ: x≠1

Ta có: \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)

\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4\left(x-1\right)}{\left(x^2+x-1\right)\left(x-1\right)}=0\)

\(\Leftrightarrow x^2+x+1+2x^2-5-4\left(x-1\right)=0\)

\(\Leftrightarrow x^2+x+1+2x^2-5-4x+4=0\)

\(\Leftrightarrow3x^2-3x=0\)

\(\Leftrightarrow3x\left(x-1\right)=0\)

Vì 3≠0

nên \(\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\)

Vậy: x=0

Bài 2:

ĐKXĐ: x≠2; x≠3; \(x\ne\frac{1}{2}\)

Ta có: \(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)

\(\Leftrightarrow\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1-\left(2x+5\right)}{\left(x-3\right)\left(2x-1\right)}=0\)

\(\Leftrightarrow\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1-2x-5}{\left(x-3\right)\left(2x-1\right)}=0\)

\(\Leftrightarrow\frac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\frac{\left(-x-4\right)\left(x-2\right)}{\left(x-3\right)\left(2x-1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x^2+x-12-x^2-2x+8=0\)

\(\Leftrightarrow-x-4=0\)

\(\Leftrightarrow-x=4\)

hay x=-4(tm)

Vậy: x=-4

Bài 3:

ĐKXĐ: x≠1; x≠-1

Ta có: \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=3x\left(1-\frac{x-1}{x+1}\right)\)

\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}=3x-\frac{3x\left(x-1\right)}{x+1}\)

\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}-3x+\frac{3x\left(x-1\right)}{x+1}=0\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{3x\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)-\left(x^2-2x+1\right)-3x\left(x^2-1\right)+3x\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow x^2+2x+1-x^2+2x-1-3x^3+3x+3x^3-6x^2+3x=0\)

\(\Leftrightarrow-6x^2+10x=0\)

\(\Leftrightarrow2x\left(-3x+5\right)=0\)

Vì 2≠0

nên \(\left[{}\begin{matrix}x=0\\-3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{5}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\frac{5}{3}\right\}\)

Bài 4:

ĐKXĐ: x≠1; x≠-3

Ta có: \(\frac{2x}{x-1}+\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}\)

\(\Leftrightarrow\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x-5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}=0\)

\(\Leftrightarrow2x^2+6x+4-\left(2x^2-7x+5\right)=0\)

\(\Leftrightarrow2x^2+6x+4-2x^2+7x-5=0\)

\(\Leftrightarrow13x-1=0\)

\(\Leftrightarrow13x=1\)

hay \(x=\frac{1}{13}\)(tm)

Vậy: \(x=\frac{1}{13}\)

Bài 5:

ĐKXĐ: x≠1; x≠-2

Ta có: \(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{x^2+x-2}\)

\(\Leftrightarrow\frac{x+2}{\left(x-1\right)\left(x+2\right)}-\frac{7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}-\frac{3}{\left(x+2\right)\left(x-1\right)}=0\)

\(\Leftrightarrow x+2-7\left(x-1\right)-3=0\)

\(\Leftrightarrow x+2-7x+7-3=0\)

\(\Leftrightarrow-6x+6=0\)

\(\Leftrightarrow-6\left(x-1\right)=0\)

Vì -6≠0

nên x-1=0

hay x=1(ktm)

Vậy: x∈∅

Bài 6:

ĐKXĐ: x≠4; x≠2

Ta có: \(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{6x-8-x^2}=0\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{-\left(x^2-6x+8\right)}=0\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}+\frac{2}{\left(x-4\right)\left(x-2\right)}=0\)

\(\Leftrightarrow\frac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}+\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\frac{2}{\left(x-4\right)\left(x-2\right)}=0\)

\(\Leftrightarrow x^2+x-6+x^2-5x+4+2=0\)

\(\Leftrightarrow2x^2-4x=0\)

\(\Leftrightarrow2x\left(x-2\right)=0\)

Vì 2≠0

nên \(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\)

Vậy: x=0

Bài 7:

ĐKXĐ: x≠1; x≠-2; x≠-1

Ta có: \(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{1-x^2}\)

\(\Leftrightarrow\frac{1}{x-1}-\frac{7}{x+2}+\frac{3}{x^2-1}=0\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}-\frac{7\left(x-1\right)\left(x+1\right)}{\left(x+2\right)\left(x-1\right)\left(x+1\right)}+\frac{3\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}=0\)

\(\Leftrightarrow x^2+3x+2-7\left(x^2-1\right)+3x+6=0\)

\(\Leftrightarrow x^2+3x+2-7x^2+7x+3x+6=0\)

\(\Leftrightarrow-6x^2+13x+8=0\)
\(\Leftrightarrow-6x^2+16x-3x+8=0\)

\(\Leftrightarrow2x\left(-3x+8\right)+\left(-3x+8\right)=0\)

\(\Leftrightarrow\left(-3x+8\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x+8=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-8\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=\frac{-1}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{8}{3};\frac{-1}{2}\right\}\)

25 tháng 3 2020

\( 1)\dfrac{1}{{x - 1}} + \dfrac{{2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{4}{{{x^2} + x + 1}}\\ DK:x \ne 1\\ \Leftrightarrow \dfrac{{{x^2} + x + 1 + 2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{{4\left( {x - 1} \right)}}{{{x^3} - 1}}\\ \Leftrightarrow {x^2} + x + 1 + 2{x^2} - 5 = 4x - 4\\ \Leftrightarrow 3{x^2} - 3x = 0\\ \Leftrightarrow 3x\left( {x - 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\left( {tm} \right)\\ x = 1\left( {ktm} \right) \end{array} \right.\\ 2)\dfrac{{x + 4}}{{2{x^2} - 5x + 2}} + \dfrac{{x + 1}}{{2{x^2} - 7x + 3}} = \dfrac{{2x + 5}}{{2{x^2} - 7x + 3}}\\ + DK:x \ne \dfrac{1}{2};x \ne 2;x \ne 3\\ \Leftrightarrow \dfrac{{x + 4}}{{\left( {2x - 1} \right)\left( {x - 2} \right)}} + \dfrac{{x + 1}}{{\left( {x - 3} \right)\left( {2x - 1} \right)}} = \dfrac{{2x + 5}}{{\left( {x - 3} \right)\left( {2x - 1} \right)}}\\ \Leftrightarrow \left( {x + 4} \right)\left( {x - 3} \right) + \left( {x + 1} \right)\left( {x - 2} \right) = \left( {2x + 5} \right)\left( {x - 2} \right)\\ \Leftrightarrow {x^2} + x - 12 + {x^2} - x - 2 = 2{x^2} + x - 10\\ \Leftrightarrow x = - 4\left( {tm} \right)\\ 3)\dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}} = 3x\left( {1 - \dfrac{{x - 1}}{{x + 1}}} \right)\\ DK:x \ne \pm 1\\ \Leftrightarrow {\left( {x + 1} \right)^2} - {\left( {x - 1} \right)^2} = 3x\left( {x - 1} \right)\left( {x + 1 - x + 1} \right)\\ \Leftrightarrow {x^2} + 2x + 1 - {x^2} + 2x - 1 = 6x\left( {x - 1} \right)\\ \Leftrightarrow 4x = 6{x^2} - 6x\\ \Leftrightarrow 2x\left( {3x - 5} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \dfrac{5}{3} \end{array} \right.\left( {tm} \right) \)

Còn lại tương tự mà làm nhé!

10 tháng 2 2018

a) \(\left(3x^2+10x-8\right)^2=\left(5x^2-2x+10\right)^2\)

\(3x^2+10x-8=5x^2-2x+10\)

\(3x^2-5x^2+10x+2x-8-10=0\)

\(-2x^2+12x-18=0\)

\(x^2-6x+9=0\)

\(\left(x-3\right)^2=0\)

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)

b) \(\frac{x^2-x-6}{x-3}=0\)

\(\Rightarrow x^2-x-6=0\)

\(\Rightarrow x^2-2x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}-6=0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2-\frac{25}{4}=0\)

\(\Rightarrow\left(x-\frac{1}{2}-\frac{5}{2}\right)\left(x-\frac{1}{2}+\frac{5}{2}\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

10 tháng 2 2018

Gin hotaru  

8 tháng 2 2020

a, \(\frac{x-3}{5}\) = 6 - \(\frac{1-2x}{3}\)

3(x - 3) = 90 - 5(1 - 2x)

⇔ 3x - 9 = 90 - 5 + 10x

⇔ 3x - 10x = 90 - 5 + 9

⇔ -7x = 94

⇔ x = \(\frac{-94}{7}\)

S = { \(\frac{-94}{7}\) }

b, \(\frac{3x-2}{6}\) - 5 = \(\frac{3-2\left(x+7\right)}{4}\)

⇔ 2(3x - 2) - 60 = 9 - 6(x + 7)

⇔ 6x - 4 - 60 = 9 - 6x - 42

⇔ 6x + 6x = 9 - 42 + 60 + 4

⇔ 12x = 31

⇔ x = \(\frac{31}{12}\)

S = { \(\frac{31}{12}\) }

c, \(\frac{x+8}{6}\) - \(\frac{2x-5}{5}\) = \(\frac{x+1}{3}\) - x + 7

⇔ 5(x+ 8) - 6(2x - 5) = 10(x+1) - 30x+210

⇔ 5x+ 40 - 12x+ 30 = 10x+ 10 - 30x+210

⇔ 5x - 12x - 10x+ 30x = 10+ 210 - 30- 40

⇔ 13x = 150

⇔ x = \(\frac{150}{13}\)

S = { \(\frac{150}{13}\) }

d, \(\frac{7x}{8}\) - 5(x - 9) = \(\frac{2x+1,5}{6}\)

⇔ 21x - 120(x - 9) = 4(2x + 1,5)

⇔ 21x - 120x + 1080 = 8x + 6

⇔ 21x - 120x - 8x = 6 - 1080

⇔ -107x = -1074

⇔ x = \(\frac{1074}{107}\)

S = { \(\frac{1074}{107}\) }

e, \(\frac{5\left(x-1\right)+2}{6}\) - \(\frac{7x-1}{4}\) = \(\frac{2\left(2x+1\right)}{7}\) - 5

⇔ 140(x-1)+56 - 42(7x-1) = 48(2x+1)-840

⇔ 140x -140+56 -294x+42= 96x+48 -840

⇔ 140x -294x -96x = 48 -840 -42 -56+140

⇔ -250x = -750

⇔ x = 3

S = { 3 }

f, \(\frac{x+1}{3}\) + \(\frac{3\left(2x+1\right)}{4}\) = \(\frac{2x+3\left(x+1\right)}{6}\) + \(\frac{7+12x}{12}\)

⇔ 4(x+1)+9(2x+1) = 4x+6(x+1)+7+12x

⇔ 4x+4+18x+9 = 4x+6x+6+7+12x

⇔ 4x+18x - 4x - 6x - 12x = 6+7- 9 - 4

⇔ 0x = 0

S = R

Chúc bạn học tốt !

22 tháng 4 2020

Bạn ơi giải giúp mình 2 bài này với ạ : https://hoc24.vn/hoi-dap/question/969683.html

Mình cảm ơn trước nhaa

ĐKXĐ: \(x\ne2;x\ne1\)

Ta có: \(\frac{4x}{x-2}-\frac{1}{x-1}=\frac{8x^2-7}{3x-6}\)

\(\Leftrightarrow\frac{4x}{x-2}-\frac{1}{x-1}-\frac{8x^2-7}{3x-6}=0\)

\(\Leftrightarrow\frac{4x\left(x-1\right)\cdot3}{\left(x-2\right)\left(x-1\right)\cdot3}-\frac{1\left(x-2\right)\cdot3}{\left(x-1\right)\left(x-2\right)\cdot3}-\frac{\left(8x^2-7\right)\left(x-1\right)}{3\left(x-2\right)\left(x-1\right)}=0\)

\(\Leftrightarrow12x\left(x-1\right)-3\left(x-2\right)-\left(8x^2-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow12x^2-12x-\left(3x-6\right)-\left(8x^3-8x^2-7x+7\right)=0\)

\(\Leftrightarrow12x^2-12x-3x+6-8x^3+8x^2+7x-7=0\)

\(\Leftrightarrow-8x^3+20x^2-8x-1=0\)

20 tháng 2 2020

x=??

19 tháng 2 2020

\(\frac{12x\left(x-1\right)-3x+6}{3\left(x-2\right)\left(x-1\right)}=\frac{\left(8x^2-7\right)\left(x-1\right)}{3\left(x-2\right)\left(x-1\right)}\)

tiếp theo nhân vào và khử mẫu nha bạn!

20 tháng 2 2020

mong mn giải ra kq lun xem

23 tháng 2 2019

pt đầu \(\Leftrightarrow x+1+\frac{1}{x+1}+x+7+\frac{7}{x+7}=x+3+\frac{3}{x+3}+x+5+\frac{5}{x+5}\)

\(\Rightarrow\frac{1}{x+1}+\frac{7}{x+7}=\frac{3}{x+3}+\frac{5}{x+5}\\ \Rightarrow\frac{8x+14}{x^2+8x+7}=\frac{8x+30}{x^2+8x+15}\)

\(\Leftrightarrow\left(4x+7\right)\left(x^2+8x+15\right)=\left(4x+15\right)\left(x^2+8x+7\right)\)

Đặt a=4x+7

      b=x2 +8x+7

như vậy ta được pt mới có dạng \(a\left(b+8\right)=b\left(a+8\right)\Leftrightarrow ab+8a=ab+8b\Rightarrow a=b\)

hay\(4x+7=x^2+8x+7\Rightarrow x^2+4x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)