a,(2x-1)^2+2(x-1)(x+1)+(x+1)^2
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21 tháng 5 2022
a: \(=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{2x^2-x^3}{x^2-3x}\)
\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(=\dfrac{-4x^2-8x}{x+2}\cdot\dfrac{-x}{x-3}\)
\(=\dfrac{-4x\left(x+2\right)}{x+2}\cdot\dfrac{-x}{x-3}=\dfrac{4x^2}{x-3}\)
b: \(=\dfrac{2x-1}{2x+1}:\left(2x-1+\dfrac{2-4x}{2x+1}\right)\)
\(=\dfrac{2x-1}{2x+1}:\dfrac{4x^2-1+2-4x}{2x+1}\)
\(=\dfrac{2x-1}{4x^2-4x+1}=\dfrac{1}{2x-1}\)
c: \(=\left(\dfrac{1}{1-x}-1\right):\left(x+1-\dfrac{2x-1}{x-1}\right)\)
\(=\dfrac{1-1+x}{1-x}:\dfrac{x^2-1-2x+1}{x-1}\)
\(=\dfrac{-x}{x-1}\cdot\dfrac{x-1}{x\left(x-2\right)}=\dfrac{-1}{x-2}\)
NN
21 tháng 11 2017
1)
\(\dfrac{7x-1}{2x^2+6x}=\dfrac{7x-12}{x\left(x+3\right)}\)
\(\dfrac{3-2x}{x^2-9}=\dfrac{3-2x}{\left(x-3\right)\left(x+3\right)}\)
MTC: \(x\left(x-3\right)\left(x+3\right)\)
\(\dfrac{7x-1}{2x^2+6x}=\dfrac{7x-12}{x\left(x+3\right)}=\dfrac{\left(x-3\right)\left(7x-12\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{7x^2-12x-21x+36}{x\left(x-3\right)\left(x+3\right)}=\dfrac{7x^2-33x+36}{x\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{3-2x}{x^2-9}=\dfrac{3-2x}{\left(x-3\right)\left(x+3\right)}=\dfrac{ x\left(3-2x\right)}{x\left(x-3\right)\left(x+3\right)}\dfrac{3x-2x^2}{x\left(x-3\right)\left(x+3\right)}\)
2)
\(\dfrac{2x-1}{x-x^2}=\dfrac{2x-1}{x\left(1-x\right)}\)
\(\dfrac{x+1}{2-4x+2x^2}=\dfrac{x+1}{2\left(1-2x+x^2\right)}=\dfrac{x+1}{2\left(1-x\right)^2}\)
MTC: \(2x\left(1-x\right)^2\)
\(\dfrac{2x-1}{x-x^2}=\dfrac{2x-1}{x\left(1-x\right)}=\dfrac{2\left(1-x\right)\left(2x-1\right)}{2x\left(1-x\right)^2}=\dfrac{\left(2-2x\right)\left(2x-1\right)}{2x\left(1-x\right)^2}=\dfrac{4x-2-4x^2+2x}{2x\left(1-x\right)^2}=\dfrac{6x-2-4x^2}{2x\left(1-x\right)^2}\)
\(\dfrac{x+1}{2-4x+2x^2}=\dfrac{x+1}{2\left(1-2x+x^2\right)}=\dfrac{x+1}{2\left(1-x\right)^2}=\dfrac{ x\left(x+1\right)}{2x\left(1-x\right)^2}=\dfrac{x^2+x}{2x\left(1-x\right)^2}\)
PT
21 tháng 11 2017
Phần còn lại nhé :v
3.
\(x^3+1=\left(x+1\right)\left(x^2-x+1\right)\)
\(x^2-x+1=x^2-x+1\)
\(x+1=x+1\)
MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)
\(\dfrac{x-1}{x^3+1}=\dfrac{x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\dfrac{2x}{x^2-x+1}=\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\dfrac{2}{x+1}=\dfrac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
4.
\(5x\)
\(x-2y=x-2y=-\left(2y-x\right)\)
\(8y^2-2x^2=2\left(4y^2-x^2\right)=2\left(2y-x\right)\left(2y+x\right)\)
MTC: \(-10x\left(2y-x\right)\left(2y+x\right)\)
\(\dfrac{7}{5x}=\dfrac{7\left(2y-x\right)\left(2y+x\right)-2}{5x\left(2y-x\right)\left(2y+x\right)-2}=\dfrac{-14\left(2y-x\right)\left(2y+x\right)}{-10x\left(2y-x\right)\left(2y+x\right)}\)
\(\dfrac{4}{x-2y}=\dfrac{4\left(2y-x\right)\left(2y+x\right)10x}{-\left(2y-x\right)\left(2y+x\right)10x}=\dfrac{40x\left(2y-x\right)\left(2y+x\right)}{-10x\left(2y-x\right)\left(2y+x\right)}\)
\(\dfrac{x-y}{8y^2-2x^2}=\dfrac{\left(x-y\right)-5x}{2\left(2y-x\right)\left(2y+x\right)-5x}=\dfrac{-5x\left(x-y\right)}{-10x\left(2y-x\right)\left(2y+x\right)}\)
5.
\(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
\(x^2-x=x\left(x-1\right)\)
\(x^2+x+1\)
MTC: \(x\left(x-1\right)\left(x^2+x+1\right)\)
\(\dfrac{x}{x^3-1}=\dfrac{x.x}{\left(x-1\right)\left(x^2+x+1\right)x}=\dfrac{x^2}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{x+1}{x^2-x}=\dfrac{\left(x+1\right)\left(x^2+x+1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{x-1}{x^2+x+1}=\dfrac{x\left(x-1\right)\left(x-1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x\left(x-1\right)^2}{x\left(x-1\right)\left(x^2+x+1\right)}\)
6.
\(x^2-2ax+a^2=\left(x-a\right)^2\)
\(x^2-ax=x\left(x-a\right)\)
MTC: \(x\left(x-a\right)^2\)
\(\dfrac{x}{x^2-2ax+a^2}=\dfrac{x.x}{\left(x-a\right)^2x}=\dfrac{x^2}{x\left(x-a\right)^2}\)
\(\dfrac{x+a}{x^2-ax}=\dfrac{\left(x+a\right)\left(x-a\right)}{x\left(x-a\right)\left(x-a\right)}=\dfrac{x^2-a^2}{x\left(x-a\right)^2}\)
KK
17 tháng 2 2021
sửa lại chút: a) (2x+1)^2-2x-1=2 b) (x^2-3x)^2+5(x^2-3x)+6=0 c) (x^2-x-1)(x^2-x)-2=0 d) (5-2x)^2+4x-10=8 e) (x^2+2x+3)(x^2+2x+1)=3 f) x(x-1)(x^2-x+1)-6=0
17 tháng 2 2021
a) Ta có: \(\left(2x+1\right)^2-2x-1=2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)+\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1-2\right)+\left(2x+1-2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{1}{2};-1\right\}\)
VD
24 tháng 3 2022
\(a,3\left(x+4\right)=x+2\\ \Leftrightarrow3x+12-x-2=0\\ \Leftrightarrow2x+10=0\\ \Leftrightarrow x=-5\\ b,\dfrac{3x+2}{3}=\dfrac{x}{2}-\dfrac{1}{6}\\ \Leftrightarrow\dfrac{2\left(3x+2\right)}{6}-\dfrac{3x}{6}-\dfrac{1}{6}=0\\ \Leftrightarrow6x+4-3x-1=0\\ \Leftrightarrow3x+3=0\\ \Leftrightarrow x=-1\\ c,\left(2x-1\right)^2=\left(2-x\right)\left(2x-1\right)\\ \Leftrightarrow\left(2x-1\right)^2-\left(2-x\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-2+x\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(3x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
\(d,ĐKXĐ:x\ne0,x\ne-1\\ \dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}+\dfrac{x}{x\left(x+1\right)}-\dfrac{2x-1}{x\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-1+x-2x+1}{x\left(x+1\right)}=0\\ \Rightarrow x^2-x=0\\ \Leftrightarrow x\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
17 tháng 5 2020
a, \(\frac{1+2x-5}{6}=\frac{3-x}{4}\)
\(\frac{4+8x-20}{24}=\frac{18-6x}{24}\)
\(-16-8x=18-6x\)
\(-16-8x-18+6x=0\)
\(-34-2x=0\)
\(2x=-34\Leftrightarrow x=-17\)
17 tháng 5 2020
b, \(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)ĐKXĐ : x \(\ne\)-1 ; 0
\(\frac{x^2+3x}{x^2+x}+\frac{x^2-x-2}{x^2+x}=\frac{2x^2+2x}{x^2+x}\)
\(x^2+3x+x^2-x-2=2x^2+2x\)
\(2x^2+2x-2=2x^2+2x\)
\(2x^2+2x-2x^2-2x-2=0\)
\(-2\ne0\) Nên phuwong trình vô nghiệm. (xem lại hộ)
\(\left(2x-1\right)^2+2\left(x-1\right)\left(x+1\right)+\left(x+1\right)^2\)
\(=4x^2-4x+1+2\left(x^2-1\right)+x^2+2x+1\)
\(=4x^2-4x+1+2x^2-2+x^2+2x+1\)
\(=7x^2-2x\)
\(=x\left(7x-2\right)\)
#H