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10 tháng 5 2023
a: Khi x>0 thì A=3x-3x+2=2
Khi x<0 thì A=-3x-3x+2=-6x+2
b: B=4-x-x+5=9-2x
c: TH1: 5/4<x<5/2
A=5-2x-3x+7=12-5x
TH2: x>=5/2
A=2x-5-3x+7=-x+2
d: D=3-5x+|5x-3|
TH1: x>=3/5
D=3-5x+5x-3=0
TH2: x<3/5
D=3-5x+3-5x=6-10x
HV
1
29 tháng 9 2016
\(6x\left(1-3x\right)+9x\left(2x-7\right)+171=0\)
\(\Leftrightarrow6x-18x^2+18x^2-63x+171=0\)
\(\Leftrightarrow-57x=-171\)
\(\Leftrightarrow x=3\)
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\left(\frac{x+1}{2015}+1\right)+\left(\frac{x+2}{2014}+1\right)-\left(\frac{x+3}{2013}+1\right)-\left(\frac{x+4}{2012}+1\right)=0\)
\(\Leftrightarrow\)\(\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}+\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
\(\Leftrightarrow x+2016=0\) ( vì \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\) )
\(\Leftrightarrow x=-2016\)
10 tháng 10 2021
b: Ta có: \(\left(x-3\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+9\left(x+2\right)^2\)
\(=x^3-9x^2+27x-27-x^3-8+9x^2+36x+36\)
\(=53x+1\)
C
1
3 tháng 7 2021
a, \(\left(5x-4\right)\left(5x+4\right)-\left(5x-4\right)^2=\left(25x^2-16\right)-\left(25x^2-40x+16\right)=40x-32\)
b,\(\left(5x+3\right)^2-\left(4x-1\right)^2-\left(9x^2+8\right)=\left(x+4\right)\left(9x-2\right)-\left(9x^2+8\right)\)
\(=9x^2+34x-8-\left(9x^2+8\right)=34x\)
c,\(2\left(x-5y\right)\left(x+5y\right)+\left(x+5y\right)^2+\left(x-5y\right)^2=\left(2x\right)^2=4x^2\)
TN
2 tháng 8 2016
Bài 1: (x-7)(x-8)-(x-5)(x-2)
=x^2 - 15x +56 -( x^2 -7x +10)
=46-8x.Thay x=-1/5 vào bt ta có:
A=46-8*(-1)/5=47,6
Bài 2:(x - 3)^2 - 2(x - 3)(x + 2)+ (x+2)^2
=(x - 3)[x - 3 - 2(x+2)] +(x+2)^2
=(x-3)[-x-7] + x^2+4x+4
=-x^2 -4x +21 +x^2+4x+4
=25
Bài 3:
a)2x^2 - 6x=0
<=>2x(x-3)=0
<=>2x=0 hoặc x-3=0
<=>x=0 hoặc x=3
b)x^2-6x+9=0 <-- chắc đề thế này
<=>(x-3)^2=0 dùng HĐT
<=>x-3=0 =>x=3
23 tháng 12 2021
a: \(=x^2+2x-8-x^2-2x-1=-9\)
b: \(=\dfrac{x^2+6x+9+3x-9+2x^2-18x}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x^2-9x}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
JP
3
LD
11 tháng 1 2022
\(A=\dfrac{x}{x-2}-\dfrac{x^2+x-2}{x^2-4}=\dfrac{x^2+2x-x^2-x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\)
11 tháng 1 2022
\(A=\dfrac{x}{x-2}+\dfrac{x^2+x-2}{4-x^2}\left(x\ne\pm2\right).\)
\(A=\dfrac{x}{x-2}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x-2}-\dfrac{x-1}{x-2}=\dfrac{x-x+1}{x-2}=\dfrac{1}{x-2.}\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
Bài 2:
Ta có: \(x\left(x-4\right)-x^2+8=0\)
\(\Leftrightarrow x^2-4x-x^2+8=0\)
\(\Leftrightarrow-4x=-8\)
hay x=2
1)=x2-49-x2
=-49
2)=>x2-4x-x2+8=0
=>-4x+8=0
=>-4x=-8
=>x=2