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=>0,2x+0,4-0,5x=0,25-0,5x+0,25
=>0,2x+0,4=0,5
=>0,2x=0,1
=>x=1/2
\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
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Bài 1.
a) ( x - 3 )( x + 7 ) = 0
<=> x - 3 = 0 hoặc x + 7 = 0
<=> x = 3 hoặc x = -7
Vậy S = { 3 ; -7 }
b) ( x - 2 )2 + ( x - 2 )( x - 3 ) = 0
<=> ( x - 2 )( x - 2 + x - 3 ) = 0
<=> ( x - 2 )( 2x - 5 ) = 0
<=> x - 2 = 0 hoặc 2x - 5 = 0
<=> x = 2 hoặc x = 5/2
Vậy S = { 2 ; 5/2 }
c) x2 - 5x + 6 = 0
<=> x2 - 2x - 3x + 6 = 0
<=> x( x - 2 ) - 3( x - 2 ) = 0
<=> ( x - 2 )( x - 3 ) = 0
<=> x - 2 = 0 hoặc x - 3 = 0
<=> x = 2 hoặc x = 3
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
\(\Leftrightarrow\frac{6x^2+3}{24}-\frac{10x-4}{24}=\frac{6x^2-6}{24}-\frac{4x-12}{24}\)
\(\Leftrightarrow\frac{6x^2+3-10x+4}{24}=\frac{6x^2-6-4x+12}{24}\)
\(\Leftrightarrow6x^2-10x+7=6x^2-4x+6\)
\(\Leftrightarrow-6x+1=0\)
\(\Rightarrow-6x=-1\)
\(\Leftrightarrow x=\frac{1}{6}\)
Vậy ...
\(a,x+\frac{4}{5}-x+4=\frac{x}{3}-x-1\)
\(x+\frac{24}{5}-x=\frac{x}{3}-x-1\)
\(x+\frac{24}{5}-x-\frac{x}{3}+x+1=0\)
\(x+\frac{29}{5}-\frac{x}{3}=0\)
\(x-\frac{1}{3}x=-\frac{29}{5}\)
\(\frac{2}{3}x=-\frac{29}{5}\)
\(x=-\frac{87}{10}\)
a, \(x^2-8x+16=81\Leftrightarrow x^2-8x-65=0\)
\(\Leftrightarrow\left(x-13\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=13\)
Vậy tập nghiệm của pt là S = { -5 ; 13 }
b, \(\frac{2x+2}{5}+\frac{3}{10}< \frac{3x-2}{4}\)
\(\Leftrightarrow\frac{8x+8+6}{20}< \frac{15x-10}{20}\Leftrightarrow8x+14< 15x-10\)
\(\Leftrightarrow-7x< -24\Leftrightarrow x>\frac{24}{7}\)
Vậy tập nghiệm của BFT là S = { x | x > 24/7 }
c, \(\frac{2}{x-2}+\frac{3}{x-3}=\frac{3x-20}{x^2}\)ĐK : \(x\ne0;2;3\)
\(\Leftrightarrow\frac{2x^2\left(x-3\right)+3x^2\left(x-2\right)}{x^2\left(x-2\right)\left(x-3\right)}=\frac{\left(3x-20\right)\left(x-2\right)\left(x-3\right)}{x^2\left(x-2\right)\left(x-3\right)}\)
tự khử mẫu, làm tiếp nhé, mình bị lười :>
d, \(3\left(x-11\right)-2\left(x+11\right)=1964\)
\(\Leftrightarrow3x-33-2x-22=1964\Leftrightarrow x-55=1964\Leftrightarrow x=2019\)
Vâỵ tập nghiệm của pt là S = { 2019 }
e, \(\left|2x-3\right|=5\)
Với \(x\ge\frac{3}{2}\)pt có dạng : \(2x-3=5\Leftrightarrow x=4\)( tm )
Với \(x< \frac{3}{2}\)pt có dạng : \(-2x+3=5\Leftrightarrow-2x=2\Leftrightarrow x=-1\)( tm )
Vậy tập nghiệm của pt là S = { -1; 4 }
g, \(\frac{-2x+14}{x-5}+\frac{5x-3}{2x}=\frac{8}{x\left(x-5\right)}\)ĐK : \(x\ne0;5\)
\(\Leftrightarrow\frac{2x\left(-2x+14\right)+\left(5x-3\right)\left(x-5\right)}{2x\left(x-5\right)}=\frac{16}{2x\left(x-5\right)}\)
Tự khử mẫu tự giải nhá :>
\(2x^2-7x-15=22.\)
\(\Leftrightarrow2x^2-7x-37=0.\)
\(\Leftrightarrow x^2-\frac{7}{2}x-\frac{37}{2}=0.\)
\(\Leftrightarrow x^2-2.\frac{7}{4}x+\frac{49}{16}-\frac{49}{16}-\frac{37}{2}=0.\)
\(\Leftrightarrow\left(x-\frac{7}{4}\right)^2=\frac{345}{16}.\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{7}{4}=\frac{\sqrt{345}}{4}\\x-\frac{7}{4}=-\frac{\sqrt{345}}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{345}+7}{4}\\x=\frac{-\sqrt{345}+7}{4}\end{cases}}\)
Học tốt
(2x+3)(x-5)=42+6
2x2-10x+3x-15x=16+6
2x2-7x-15=22
2x2-7x-15-22=0
2x2-7x-37=0
\( {7 \pm\sqrt{345} \over 2}\)\(+{7 - \sqrt{345} \over 2}\)=0
x1=-2,89354
x2=6,39354
neu co sai cho to xin loi
chuc ban hoc tot