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KV
14 tháng 12 2023
a) \(\left(x+2\right)^2=4\left(2x-1\right)^2\)
\(\left(x+2\right)^2-4\left(2x-1\right)^2=0\)
\(\left(x+2\right)^2-\left[2\left(2x-1\right)\right]^2=0\)
\(\left(x+2\right)^2-\left(4x-2\right)^2=0\)
\(\left(x+2-4x+2\right)\left(x+2+4x-2\right)=0\)
\(6x\left(-3x+4\right)=0\)
\(\Rightarrow6x=0\) hoặc \(-3x+4=0\)
*) \(6x=0\)
\(x=0\)
*) \(-3x+4=0\)
\(3x=4\)
\(x=\dfrac{4}{3}\)
Vậy \(x=0;x=\dfrac{4}{3}\)
b) \(4x\left(x-2019\right)-x+2019=0\)
\(4x\left(x-2019\right)-\left(x-2019\right)=0\)
\(\left(x-2019\right)\left(4x-1\right)=0\)
\(\Rightarrow x-2019=0\) hoặc \(4x-1=0\)
*) \(x-2019=0\)
\(x=2019\)
*) \(4x-1=0\)
\(4x=1\)
\(x=\dfrac{1}{4}\)
Vậy \(x=\dfrac{1}{4};x=2019\)
H
28 tháng 7 2023
\(\left(x+2\right)^2=\left(2x-1\right)^2\\ \Leftrightarrow\left(x+2\right)^2-\left(2x-1\right)^2=0\\\Leftrightarrow\left[x+2-\left(2x-1\right)\right]\left[x+2+2x-1\right]=0\\ \Leftrightarrow\left(x+2-2x+1\right)\left(x+2+2x-1\right)=0\\ \Leftrightarrow\left(-x+3\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x+3=0\\3x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=-3\\3x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{3}\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
28 tháng 7 2023
\(\left(x+2\right)^2=\left(2x-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=2x-1\\x+2=-\left(2x-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2x=-1-2\\x+2=-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=-3\\x+2x=1-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\3x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{3}\end{matrix}\right.\)
21 tháng 10 2017
a) \(49-\left(3x-1\right)^2=0\)
\(\Leftrightarrow7^2-\left(3x-1\right)^2=0\)
\(\Leftrightarrow\left(7-3x+1\right)\left(7+3x-1\right)=0\)
\(\Leftrightarrow\left(8-3x\right)\left(6+3x\right)=0\)
\(\hept{\begin{cases}8-3x=0\\6+3x=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{8}{3}\\x=-2\end{cases}}\)
Vậy \(x=\frac{8}{3};x=-2\)
b) \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)
\(\Leftrightarrow\left(x-1\right)^3-\left(x^3+2^3\right)-3\left(1-x^2\right)=0\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8-3+3x^2=0\)
\(\Leftrightarrow3x-12=0\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
17 tháng 10 2016
A= 2006 X 2008 - 20072
A = 2006 . 2008 - 2007 . 2007
A = 2006 . ( 2007 + 1 ) - 2007 . ( 2006 + 1 )
A = 2006 . 2007 + 2006 - 2007 . 2006 + 2007
A = -1
B= 2016 X 2018 - 20172
B= 2016 . 2018 - 2017 . 2017
B = 2016 . ( 2017 + 1 ) - 2017 . ( 2016 + 1 )
B = 2016 . 2017 + 2016 - 2017 . 2016 + 2017
B = -1
7 tháng 2 2019
Bài 1 :
Mình nghĩ phải sửa đề ntn :
\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)
\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right]\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\7x+23=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{-23}{7}\end{cases}}}\)
Vậy....
b) \(A=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(q=x^2+x+1\)ta có :
\(A=q\left(q+1\right)-12\)
\(A=q^2+q-12\)
\(A=q^2+4q-3q-12\)
\(A=q\left(q+4\right)-3\left(q+4\right)\)
\(A=\left(q+4\right)\left(q-3\right)\)
Thay \(q=x^2+x+1\)ta có :
\(A=\left(x^2+x+1+4\right)\left(x^2+x+1-3\right)\)
\(A=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(A=\left(x^2+x+5\right)\left(x^2+2x-x-2\right)\)
\(A=\left(x^2+x+5\right)\left[x\left(x+2\right)-\left(x+2\right)\right]\)
\(A=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)
(x-1)(x+1)(x+2)=(x^2-1)(x+2)=x^3+2x^2-x-2