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Bài 1 : Cộng thêm 1 vào 3 phân thức đầu, trừ cho 3 ở phân thức thứ 4, có nhân tử chung là (x+2020)
Bài 2 : Trừ mỗi phân thức cho 1, chuyển vế và có nhân tử chung là (x-2021)
Bài 3 : Phân thức thứ nhất trừ đi 1, phân thức hai trù đi 2, phân thức ba trừ đi 3, phân thức bốn trừ cho 4, phân thức 5 trừ cho 5. Có nhân tử chung là (x-100)
bài 3
\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15.\)
=>\(\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=0\)
=>\(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
=>\(\left(x-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
=>(x-100)=0 do \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
=> x=100
a. \(\frac{5x-2}{3}=\frac{5x-3x}{2}\)
\(\Leftrightarrow2.\left(5x-2\right)=3.\left(5x-3x\right)
\)
\(\Leftrightarrow10x-4=15x-9x\)
\(\Leftrightarrow4x=4\)
\(\Leftrightarrow x=1\)
Vậy...
b. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\left(1\right)\)
MC = 36.
pt (1) <=>
\(\frac{3\left(10x+3\right)}{36}=\frac{36}{36}+\frac{4\left(6+8x\right)}{36}\)
=> 3.(10x+3) = 36 + 4(6+8x)
<=> 30x+9 = 36+24+32x
<=> -2x = 51
<=> x = \(\frac{-51}{2}\)
Vậy...
c. \(\frac{7x-1}{6}+2=\frac{16-x}{5}\left(2\right)\)
MC = 30.
pt (2) <=>
\(\frac{5\left(7x-1\right)}{30}+\frac{60x}{30}=\frac{6\left(16-x\right)}{30}\)
=> 5(7x-1) + 60x = 6(16-x)
<=> 35x-5 + 60x = 96-6x
<=> 101x = 101
<=> x = 1
Vậy...
d. \(\frac{3x+2}{2}-\frac{3x+1}{6}=5\) (3)
MC = 12.
pt (3)<=>
\(\frac{6\left(3x+2\right)}{12}-\frac{2\left(3x+1\right)}{12}=\frac{60}{12}\)
=> 6(3x+2) - 2(3x+1) = 60
<=> 18x+12 - 6x-2 = 60
<=> 12x = 50
<=> x = \(\frac{25}{6}\)
Vậy...
e. \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\) (4)
MC = 30.
pt (4) <=>
\(\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)
=> 6(x+4) - 30x + 120 = 10x - 15(x-2)
<=> 6x+24 - 30x + 120 = 10x - 15x+30
<=> -19x = -114
<=> x = \(\frac{114}{19}=6\)
Vậy...
Bài làm
j) \(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\) ĐKXĐ: \(x\ne\pm5\)
\(\Leftrightarrow\frac{\left(x+5\right)^2}{x^2-25}-\frac{\left(x-5\right)^2}{x^2-25}=\frac{20}{x^2-25}\)
\(\Rightarrow x^2+10x+25-x^2+10x-25=20\)
\(\Leftrightarrow20x=20\)
\(\Leftrightarrow x=1\)
Vậy x = 1 là nghiệm phương trình.
k) \(\frac{3}{x-4}+\frac{5x-2}{x^2-16}=\frac{4}{x+4}\)
\(\Leftrightarrow\frac{3\left(x+4\right)}{x^2-16}+\frac{5x-2}{x^2-16}=\frac{4\left(x-4\right)}{x^2-16}\)
\(\Rightarrow3x+12+5x-2=4x-16\)
\(\Leftrightarrow4x=-26\)
<=> \(x=-\frac{13}{2}\)
Vậy x = -13/2 là nghiệm phương trình.
l) \(\frac{2x-1}{3}-\frac{5x+2}{4}=2x\)
\(\Leftrightarrow4x-4-15x-6=24x\)
\(\Leftrightarrow-35x=10\)
\(\Leftrightarrow x=-\frac{2}{7}\)
Vậy x = -2/7 là nghiệm phương trình.
Bài làm
2 - x = 3x + 1
<=> - x - 3x = -2 + 1
<=> -4x = -1
<=> x = 1/4
Vậy x = 1/4 là nghiệm phương trình.
4x + 7( x - 2 ) = -9x + 5
<=> 4x + 7x - 14 = -9x + 5
<=> 4x + 7x + 9x = 14 + 5
<=> 20x = 19
<=> x = 19/20
Vậy x = 19/20 là nghiệm phương trình.
5x - 2( 3x - 5 ) = 7x + 11
<=> 5x - 6x + 10 = 7x + 11
<=> 5x - 6x - 7x = 11 - 10
<=> -8x = -21
<=> x = 21/8
Vậy x = 21/8 là nghiệm phương trình.
( 5x + 2 )( x - 7 ) = 0
<=> \(\left[{}\begin{matrix}5x+2=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2}{5}\\x=7\end{matrix}\right.\)
Vậy tập nghiệm phương trình S = { -2/5; 7 }
2x( x - 5 ) + 3( x - 5 ) = 0
<=> ( 2x + 3 )( x - 5 ) = 0
<=> \(\left[{}\begin{matrix}2x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=5\end{matrix}\right.\)
Vậy tập nghiệm phương trìh S = { -3/2; 5 }
\(\frac{5x-3}{6}=\frac{-2x+5}{9}\)
\(\Rightarrow6\left(-2x+5\right)=9\left(5x-3\right)\)
\(\Leftrightarrow-12x+30=45x-27\)
\(\Leftrightarrow-57x=-57\)
\(\Leftrightarrow x=1\)
Vậy x = 1 là nghiệm phương trình.
\(\frac{x}{3}-\frac{2x+1}{2}=\frac{5x}{6}\)
\(\Leftrightarrow2x-3\left(2x+1\right)=5x\)
\(\Leftrightarrow2x-6x-3=5x\)
\(\Leftrightarrow-9x=3\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy x = -1/3 là nghiệm phương trình.
\(\frac{x}{3}-\frac{2x+1}{2}=\frac{x}{6}-x\)
\(\Leftrightarrow2x-3\left(2x+1\right)=x-6x\)
\(\Leftrightarrow2x-6x-3=x-6x\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy x = 3/2 là nghiệm phương trình.
\(\frac{3}{x+1}=\frac{5}{2x+2}\) ĐKXĐ: x khác 1
<=> \(\frac{6}{2x+2}=\frac{5}{2x+2}\)( vô lí )
Vậy phương trình trên vô nghiệm.
# Học tốt #
ĐKXĐ: \(x\ne\pm\frac{3}{2}\)
\(\frac{1}{\left(2x-3\right)^2}+\frac{3}{\left(2x-3\right)\left(2x+3\right)}-\frac{4}{\left(2x+3\right)^2}=0\)
\(\Leftrightarrow\frac{1}{\left(2x-3\right)^2}-\frac{1}{\left(2x-3\right)\left(2x+3\right)}+\frac{4}{\left(2x-3\right)\left(2x+3\right)}-\frac{4}{\left(2x-3\right)^2}=0\)
\(\Leftrightarrow\frac{1}{2x-3}\left(\frac{1}{2x-3}-\frac{1}{2x+3}\right)-\frac{4}{2x-3}\left(\frac{1}{2x-3}-\frac{1}{2x+3}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{2x-3}-\frac{4}{2x+3}\right)\left(\frac{1}{2x-3}-\frac{1}{2x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2x-3\left(vn\right)\\2x+3=4\left(2x-3\right)\Rightarrow x=\frac{5}{2}\end{matrix}\right.\)
1/ \(\frac{3\left(x+3\right)}{4}+\frac{1}{2}=\frac{5x+9}{3}-\frac{7x-9}{4}\)
=> \(\frac{9\left(x+3\right)}{12}+\frac{6}{12}=\frac{4\left(5x+9\right)}{12}-\frac{3\left(7x-9\right)}{12}\)
=> \(9\left(x+3\right)+6=4\left(5x+9\right)-3\left(7x-9\right)\)
=> \(9x+27+6=20x+36-21x+27\)
=> \(9x-20x+21x=27-27-6+36\)
=> \(10x=30\)
=> \(x=3\)
Vậy phương trình có tập nghiệm là \(S=\left\{3\right\}\)
2.Ta có : \(\frac{2x-3}{3}-\frac{x-3}{6}=\frac{4x+3}{5}-17\)
=> \(\frac{10\left(2x-3\right)}{30}-\frac{5\left(x-3\right)}{30}=\frac{6\left(4x+3\right)}{30}-\frac{510}{30}\)
=> \(10\left(2x-3\right)-5\left(x-3\right)=6\left(4x+3\right)-510\)
=> \(20x-30-5x+15=24x+18-510\)
=> \(20x-5x-24x=18-510+30-15\)
=> \(-9x=-477\)
=> \(x=53\)
Vậy phương trình có tập nghiệm là \(S=\left\{53\right\}\)
3/ Ta có : \(\frac{5x-1}{6}+\frac{2\left(x+4\right)}{9}=\frac{7x-5}{15}+x-1\)
=> \(\frac{30\left(5x-1\right)}{180}+\frac{40\left(x+4\right)}{180}=\frac{12\left(7x-5\right)}{180}+\frac{180x}{180}-\frac{180}{180}\)
=> \(30\left(5x-1\right)+40\left(x+4\right)=12\left(7x-5\right)+180x-180\)
=> \(150x-30+40x+160=84x-60+180x-180\)
=> \(150x+40x-180x-84x=-60-180-160+30\)
=> \(-74x=-370\)
=> \(x=5\)
Vậy phương trình có tập nghiệm là \(S=\left\{5\right\}\)
\(a.\frac{7x-3}{x-1}=\frac{2}{3}\\\Leftrightarrow \frac{3\left(7x-3\right)}{3\left(x-1\right)}= \frac{2\left(x-1\right)}{3\left(x-1\right)}\\ \Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\\\Leftrightarrow 3\left(7x-3\right)-2\left(x-1\right)=0\\ \Leftrightarrow21x-9-2x+2=0\\ \Leftrightarrow19x-7=0\\ \Leftrightarrow19x=7\\ \Leftrightarrow x=\frac{7}{19}\)
\(b.\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}\\ \Leftrightarrow\frac{4\left(3-7x\right)}{2\left(1+x\right)}=\frac{1\left(1+x\right)}{2\left(1+x\right)}\\\Leftrightarrow 4\left(3-7x\right)=1\left(1+x\right)\\ \Leftrightarrow4\left(3-7x\right)-1\left(1+x\right)=0\\ \Leftrightarrow12-28x-1-x=0\\ \Leftrightarrow11-29x=0\\ \Leftrightarrow-29x=-11\\ \Leftrightarrow x=\frac{-11}{-29}=\frac{11}{29}\)
\(c.\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\\ \Leftrightarrow\frac{\left(5x-1\right)\left(3x-1\right)}{\left(3x+2\right)\left(3x-1\right)}=\frac{\left(5x-7\right)\left(3x+2\right)}{\left(3x+2\right)\left(3x-1\right)}\\ \Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\\ \Leftrightarrow\left(5x-1\right)\left(3x-1\right)-\left(5x-7\right)\left(3x+2\right)=0\\ \Leftrightarrow15x^2-5x-3x+1-15x^2-10x+21x+14=0\\ \Leftrightarrow3x+15=0\\\Leftrightarrow 3x=-15\\\Leftrightarrow x=-5\)
\(d.\frac{4x+7}{x-1}=\frac{12x+5}{3x+4}\\\Leftrightarrow \frac{\left(4x+7\right)\left(3x+4\right)}{\left(x-1\right)\left(3x+4\right)}=\frac{\left(12x+5\right)\left(x-1\right)}{\left(3x+4\right)\left(x-1\right)}\\\Leftrightarrow \left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\\\Leftrightarrow \left(4x+7\right)\left(3x+4\right)-\left(12x+5\right)\left(x-1\right)=0\\ \Leftrightarrow12x^2+16x+21x+28-12x^2-12x+5x-5=0\\ \Leftrightarrow30x+23=0\\ \Leftrightarrow30x=-23\\ \Leftrightarrow x=\frac{-23}{30}\)
\(e.\frac{1}{x-2}+3=\frac{3-x}{x-2}\\ \Leftrightarrow\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}=\frac{3-x}{x-2}\\ \Leftrightarrow1+3\left(x-2\right)=3-x\\\Leftrightarrow 1+3x-6=3-x\\\Leftrightarrow 1+3x-6-3+x=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow4x=8\\ \Leftrightarrow x=2\)
\(f.\frac{8-x}{x-7}-8=\frac{1}{x-7}\\ \Leftrightarrow\frac{8-x}{x-7}-\frac{8\left(x-7\right)}{x-7}=\frac{1}{x-7}\\ \Leftrightarrow8-x-8\left(x-7\right)=1\\ \Leftrightarrow8-x-8\left(x-7\right)-1=0\\\Leftrightarrow 8-x-8x+56-1=0\\\Leftrightarrow 63-9x=0\\\Leftrightarrow -9x=-63\\ \Leftrightarrow x=\frac{-63}{-9}=7\)
\(g.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\\ \Leftrightarrow\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{\left(x-5\right)\left(x+5\right)}\\\Leftrightarrow \frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{20}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=20\\\Leftrightarrow \left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)-20=0\\ \Leftrightarrow x^2+5x+5x+25-x^2+5x+5x-25-20=0\\ \Leftrightarrow20x-20=0\\ \Leftrightarrow20x=20\\ \Leftrightarrow x=1\)
\(j.\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\\\Leftrightarrow \frac{x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{2.2x}{2\left(x+1\right)\left(x-3\right)}\\ \Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\\\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)-4x=0\\\Leftrightarrow x^2+x+x^2-3x-4x=0\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\left(x-3\right)=0\\\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right. \)
a) \(-7x^2+10x-2016=-7\left(x^2-\frac{10x}{7}\right)-2016=-7\left(x^2-2.x.\frac{5}{7}+\frac{25}{49}\right)+\frac{25}{49}.7-2016=-7\left(x-\frac{5}{7}\right)^2-\frac{14087}{7}\le-\frac{14087}{7}\)Vậy Max = \(-\frac{14087}{7}\Leftrightarrow x=\frac{5}{7}\)
b) \(\frac{x+5}{11}+\frac{x+2010}{6}\ge\frac{x-1}{2017}+\frac{x+6}{2010}\)
\(\Leftrightarrow\frac{x}{2011}+\frac{x}{6}+\frac{5}{2011}+335\ge\frac{x}{2017}+\frac{x}{2010}-\frac{1}{2017}+\frac{1}{335}\)
\(\Leftrightarrow x\left(\frac{1}{2011}+\frac{1}{6}-\frac{1}{2017}-\frac{1}{2010}\right)\ge\frac{1}{335}-\frac{1}{2017}-\frac{5}{2011}-335\)
\(\Leftrightarrow\frac{677389259}{4076467935}x\ge\frac{-455205582048}{1358822645}\) \(\Leftrightarrow x\ge-2016\)
Câu b) còn cách khác nữa bạn nhé. Mình làm cách này "xù" quá ^^
có bị sai đề ko
nhầm 100%