TH
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
8 tháng 1 2020
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
20 tháng 2 2020
a) \(\frac{4x-8}{2x^2+1}=0\)
\(\Rightarrow4x-8=0\left(2x^2+1\ne0\right)\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
Vậy x=2
b)
\(\frac{x^2-x-6}{x-3}=0\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x+2\right)}{x-3}=0\)
\(\Rightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy x=-2
29 tháng 4 2020
Bài 1:
a) Ta có: \(2,3x-2\left(0,7+2x\right)=3,6-1,7x\)
\(\Leftrightarrow2,3x-1,4-4x-3,6+1,7x=0\)
\(\Leftrightarrow-5=0\)(vl)
Vậy: \(x\in\varnothing\)
b) Ta có: \(\frac{4}{3}x-\frac{5}{6}=\frac{1}{2}\)
\(\Leftrightarrow\frac{4}{3}x=\frac{1}{2}+\frac{5}{6}=\frac{8}{6}=\frac{4}{3}\)
hay x=1
Vậy: x=1
c) Ta có: \(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(\Leftrightarrow\frac{9x}{90}-\frac{3x}{90}-\frac{4x}{90}-\frac{72}{90}=0\)
\(\Leftrightarrow2x-72=0\)
\(\Leftrightarrow2\left(x-36\right)=0\)
mà 2>0
nên x-36=0
hay x=36
Vậy: x=36
d) Ta có: \(\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\Leftrightarrow12\left(10x+3\right)=8\left(7-8x\right)\)
\(\Leftrightarrow120x+36=56-64x\)
\(\Leftrightarrow120x+36-56+64x=0\)
\(\Leftrightarrow184x-20=0\)
\(\Leftrightarrow184x=20\)
hay \(x=\frac{5}{46}\)
Vậy: \(x=\frac{5}{46}\)
e) Ta có: \(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}-\frac{12-x}{9}\)
\(\Leftrightarrow\frac{2\left(10x-5\right)}{36}+\frac{3\left(x+3\right)}{36}-\frac{6\left(7x+3\right)}{36}+\frac{4\left(12-x\right)}{36}=0\)
\(\Leftrightarrow2\left(10x-5\right)+3\left(x+3\right)-6\left(7x+3\right)+4\left(12-x\right)=0\)
\(\Leftrightarrow20x-10+3x+9-42x-18+48-4x=0\)
\(\Leftrightarrow-23x+29=0\)
\(\Leftrightarrow-23x=-29\)
hay \(x=\frac{29}{23}\)
Vậy: \(x=\frac{29}{23}\)
f) Ta có: \(\frac{x+4}{5}-x-5=\frac{x+3}{2}-\frac{x-2}{2}\)
\(\Leftrightarrow\frac{2\left(x+4\right)}{10}-\frac{10x}{10}-\frac{50}{10}=\frac{25}{10}\)
\(\Leftrightarrow2x+8-10x-50-25=0\)
\(\Leftrightarrow-8x-67=0\)
\(\Leftrightarrow-8x=67\)
hay \(x=\frac{-67}{8}\)
Vậy: \(x=\frac{-67}{8}\)
g) Ta có: \(\frac{2-x}{4}=\frac{2\left(x+1\right)}{5}-\frac{3\left(2x-5\right)}{10}\)
\(\Leftrightarrow5\left(2-x\right)-8\left(x+1\right)+6\left(2x-5\right)=0\)
\(\Leftrightarrow10-5x-8x-8+12x-30=0\)
\(\Leftrightarrow-x-28=0\)
\(\Leftrightarrow-x=28\)
hay x=-28
Vậy: x=-28
h) Ta có: \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)
\(\Leftrightarrow\frac{4\left(x+2\right)}{12}+\frac{9\left(2x-1\right)}{12}-\frac{2\left(5x-3\right)}{12}-\frac{12x}{12}-\frac{5}{12}=0\)
\(\Leftrightarrow4x+8+18x-9-10x+6-12x-5=0\)
\(\Leftrightarrow0x=0\)
Vậy: \(x\in R\)
Bài 2:
a) Ta có: \(5\left(x-1\right)\left(2x-1\right)=3\left(x+8\right)\left(x-1\right)\)
\(\Leftrightarrow5\left(x-1\right)\left(2x-1\right)-3\left(x-1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[5\left(2x-1\right)-3\left(x+8\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(10x-5-3x-24\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x-29\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\7x-29=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\7x=29\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{29}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{29}{7}\right\}\)
b) Ta có: \(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+5\ge5\ne0\forall x\)(2)
Từ (1) và (2) suy ra:
\(\left[{}\begin{matrix}3x-2=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-6\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{\frac{2}{3};-6\right\}\)
c) Ta có: \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)-x+4=0\)
\(\Leftrightarrow27x^3-8-27x^3+1-x+4=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\)
hay x=-3
Vậy: Tập nghiệm S={-3}
d) Ta có: \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)
\(\Leftrightarrow x^2-x-\left(x^2+x-12\right)-5x=0\)
\(\Leftrightarrow x^2-x-x^2-x+12-5x=0\)
\(\Leftrightarrow12-7x=0\)
\(\Leftrightarrow7x=12\)
hay \(x=\frac{12}{7}\)
Vậy: Tập nghiệm \(S=\left\{\frac{12}{7}\right\}\)
e) Ta có: (2x+1)(2x-1)=4x(x-7)-3x
\(\Leftrightarrow4x^2-1-4x^2+28x+3x=0\)
\(\Leftrightarrow31x-1=0\)
\(\Leftrightarrow31x=1\)
hay \(x=\frac{1}{31}\)
Vậy: Tập nghiệm \(S=\left\{\frac{1}{31}\right\}\)
LB
7 tháng 5 2017
bài 1
\(ĐKXĐ:1+x\ne0\Rightarrow x\ne-1\)
\(\frac{3-7x}{1+x}=\frac{1}{2}\Rightarrow2\left(3-7x\right)=1+x\)
\(\Leftrightarrow6-14x=1+x\\
\Leftrightarrow-14x-x=1-6\\
\Leftrightarrow-15x=-5\\
\Leftrightarrow x=\frac{1}{3}\left(N\right)\)
XU
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
AH
Akai Haruma
Giáo viên
29 tháng 4 2019
a)
ĐKXĐ: \(x\neq 0; x\neq -10\)
\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)
\(\Leftrightarrow \frac{x+10+x}{x(x+10)}=\frac{1}{12}\)
\(\Leftrightarrow \frac{2x+10}{x(x+10)}=\frac{1}{12}\)
\(\Rightarrow 12(2x+10)=x(x+10)\)
\(\Leftrightarrow x^2-14x-120=0\)
\(\Leftrightarrow (x+6)(x-20)=0\Rightarrow \left[\begin{matrix} x=-6\\ x=20\end{matrix}\right.\) (đều thỏa mãn)
b)
ĐKXĐ: \(x\neq 0; x\neq 3\)
PT\(\Leftrightarrow \frac{(x+3).x-(x-3)}{x(x-3)}=\frac{3}{x(x-3)}\)
\(\Leftrightarrow \frac{x^2+2x+3}{x(x-3)}=\frac{3}{x(x-3)}\)
\(\Rightarrow x^2+2x+3=3\)
\(\Leftrightarrow x^2+2x=0\Leftrightarrow x(x+2)=0\Rightarrow \left[\begin{matrix} x=0\\ x=-2\end{matrix}\right.\) . Kết hợp với đkxđ suy ra $x=-2$
AH
Akai Haruma
Giáo viên
29 tháng 4 2019
c)
ĐKXĐ: \(x\neq \pm 2\)
\(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{3(x-2)-2(x+2)}{(x+2)(x-2)}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{x-10}{x^2-4}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{x-2}{x^2-4}=0\Leftrightarrow \frac{1}{x+2}=0\) (vô lý)
Vậy pt vô nghiệm.
d)
ĐKXĐ: \(x\neq -2; x\neq 3\)
PT \(\Leftrightarrow \frac{3(x-3)-2(x+2)}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)
\(\Leftrightarrow \frac{x-13}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)
\(\Rightarrow x-13=8\Rightarrow x=21\) (thỏa mãn)
Vậy..........
19 tháng 2 2020
Bài 3 :
Ta có : \(A=x^2+x+2012\)
=> \(A=x^2+x+\left(\frac{1}{2}\right)^2+\frac{8047}{4}\)
=> \(A=\left(x+\frac{1}{2}\right)^2+\frac{8047}{4}\)
- Ta thấy : \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
=> \(\left(x+\frac{1}{2}\right)^2+\frac{8047}{4}\ge\frac{8047}{4}\forall x\)
- Dấu "=" xảy ra <=> \(x+\frac{1}{2}=0\)
<=> \(x=-\frac{1}{2}\)
Vậy MinA = \(\frac{8047}{4}\) <=> x = \(-\frac{1}{2}\) .
Bài 1 :
a, Ta có : \(\left(3x-2\right)\left(4+5x\right)=0\)
=> \(\left[{}\begin{matrix}3x-2=0\\4+5x=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}3x=2\\5x=-4\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{2}{3}\\x=-\frac{4}{5}\end{matrix}\right.\)
Vậy phương trình có nghiệm là x = \(\frac{2}{3}\), x = \(-\frac{4}{5}\) .
b,- ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
=> \(x\ne\pm1\)
Ta có : \(\frac{x+1}{x-1}-\frac{4}{x+1}=\frac{3-x^2}{1-x^2}\)
=> \(\frac{\left(x+1\right)^2}{x^2-1}-\frac{4\left(x-1\right)}{x^2-1}=\frac{x^2-3}{x^2-1}\)
=> \(\left(x+1\right)^2-4\left(x-1\right)=x^2-3\)
=> \(x^2+2x+1-4x+4=x^2-3\)
=> \(-2x=-3-5\)
=> \(x=4\left(TM\right)\)
Vậy phương trình có nghiệm là x = 4 .
c, Ta có : \(\frac{10x+3}{2009}+\frac{10x-1}{2013}=\frac{10x+1}{2011}-\frac{2-10x}{2014}\)
=> \(\frac{10x+3}{2009}+\frac{10x-1}{2013}=\frac{10x+1}{2011}+\frac{10x-2}{2014}\)
=> \(\frac{10x+3}{2009}+1+\frac{10x-1}{2013}+1=\frac{10x+1}{2011}+1+\frac{10x-2}{2014}+1\)
=> \(\frac{10x+3}{2009}+\frac{2009}{2009}+\frac{10x-1}{2013}+\frac{2013}{2013}=\frac{10x+1}{2011}+\frac{2011}{2011}+\frac{10x-2}{2014}+\frac{2014}{2014}\)
=> \(\frac{10x+2012}{2009}+\frac{10x+2012}{2013}=\frac{10x+2012}{2011}+\frac{10x+2012}{2014}\)
=> \(\frac{10x+2012}{2009}+\frac{10x+2012}{2013}-\frac{10x+2012}{2011}-\frac{10x+2012}{2014}=0\)
=> \(\left(10x+2012\right)\left(\frac{1}{2009}+\frac{1}{2013}-\frac{1}{2011}-\frac{1}{2014}\right)=0\)
=> \(10x+2012=0\)
=> \(x=-\frac{2012}{10}\)
Vậy phương trình có nghiệm là x = \(-\frac{2012}{10}\) .
JS
19 tháng 2 2020
Bài 3:
Giải:
Ta có : A = x2 + x + 2012
= x2 + 2.\(\frac{1}{2}\).x + \(\frac{1}{4}\) + \(\frac{8047}{4}\)
= (x + \(\frac{1}{2}\))2 + \(\frac{8047}{4}\) ≥ \(\frac{8047}{4}\)
⇒ Amin = \(\frac{8047}{4}\) ⇔ (x + \(\frac{1}{2}\))2 = 0 ⇔ x = \(-\frac{1}{2}\)
Vậy Amin = \(\frac{8047}{4}\) tại x = \(-\frac{1}{2}\)
Chúc bạn học tốt@@
\(\frac{x-1}{6}-\frac{x-3}{12}=\frac{x-2}{8}\)
\(\frac{8x-8}{48}-\frac{4x-12}{48}=\frac{6x-12}{48}\)
\(8x-8-4x+12=6x-12\)
\(8x-4x-6x=8-12-12\)
\(-2x=-16\Rightarrow x=8\)
\(\frac{x-1}{6}-\frac{x-3}{12}=\frac{x-2}{8}\)
\(\frac{8x-8}{48}-\frac{4x-12}{48}=\frac{6x-12}{48}\)
\(8x-8-4x+12=6x-12\)
\(8x-4x-6x=8-12-12\)
\(-2x=-16\Rightarrow x=8\)