x+1 phần x-1 - x-1 phần x+1 = -16 phần 1-x mũ 2
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13 tháng 4 2023
\(M=\left(\dfrac{x-x^2}{\left(x-1\right)^2}+\dfrac{1}{1+x}-\dfrac{x}{x-1}\right)\cdot\left(\dfrac{3x-1}{x}+\dfrac{1}{x+1}-1\right)\)
\(=\left(\dfrac{-x}{x-1}-\dfrac{x}{x-1}+\dfrac{1}{x+1}\right)\cdot\dfrac{\left(3x-1\right)\left(x+1\right)+x-x\left(x+1\right)}{x\left(x+1\right)}\)
\(=\dfrac{-2x\left(x+1\right)+x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{3x^2+2x-1+x-x^2-x}{x\left(x+1\right)}\)
\(=\dfrac{-2x^2-2x+x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2x^2+2x-1}{x\left(x+1\right)}\)
\(=\dfrac{-2x^2-x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2x^2+2x-1}{x\left(x+1\right)}\)
\(=\dfrac{\left(-2x^2-x-1\right)\left(2x^2+2x-1\right)}{x\left(x+1\right)^2\cdot\left(x-1\right)}\)
MD
14 tháng 4 2023
\(\left(\dfrac{3x-1}{x+1}-1\right)\)bạn sửa lại đề bào thế này
S
1
10 tháng 9 2019
\(P:\frac{4x-2-16}{2x+1}=\frac{4x^2+4x+1}{x-2}\)
\(\Rightarrow P=\frac{4x^2+4x+1}{x-2}.\frac{4x^2-16}{2x+1}\)
= \(\frac{\left(2x+1\right)^2}{x-2}.\frac{4.\left(x-2\right)\left(x+2\right)}{2x+1}\)
\(\Rightarrow P=4.\left(2x+1\right).\left(x+2\right)\)
\(=4.\left(2x^2+x+4x+2\right)\)
= \(8x^2+40x+8\)
Chúc bạn học tốt !!!
HM
1
17 tháng 12 2015
a)\(\frac{1}{4}.x=-\frac{1}{3}\)
\(x=-\frac{1}{3}:\frac{1}{4}\)
\(x=-\frac{4}{3}\)
b)\(-\frac{3}{7}+x=\frac{5}{8}\)
\(\text{ }x=\frac{5}{8}-\left(-\frac{3}{7}\right)\)
\(x=\frac{59}{56}\)
c)\(\frac{16}{2^x}=2\)
\(2^x=\frac{16}{2}\)
\(2^x=8\)
\(\Rightarrow2^x=2^3\)
vậy x=3
18 tháng 3 2020
\(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0^2\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy x = 1/2
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2=1^2\)
\(\Leftrightarrow x-2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy x = 3 hoặc x = 1
\(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\)
<=> 2x = -1
<=> x = -0,5
Vậy x = -0,5
18 tháng 3 2020
\(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1+2\\x=-1+2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
Vậy\(x\in\left\{3;1\right\}\)
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=-1\times2\)
\(x=-2\)
\(x\left(\frac{1}{2}\right)^2=\frac{1}{16}\)
\(x\left(\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x\frac{1}{2}=\frac{1}{4}\\x\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}:\frac{1}{2}\\x=-\frac{1}{4}:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)
MD
2
12 tháng 4 2023
\(M=\dfrac{x-x^2}{1-x^2}+\dfrac{1}{1+x}\cdot\dfrac{3x-1}{x+1}-1\)
\(=\dfrac{x\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}+\dfrac{3x-1}{\left(x+1\right)^2}-1\)
\(=\dfrac{x^2+x+3x-1-x^2-2x-1}{\left(x+1\right)^2}\)
\(=\dfrac{2x-2}{\left(x+1\right)^2}\)
\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{-16}{1-x^2}\left(x\ne\pm1\right)\)
\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{16}{1-x^2}=0\)
\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{16}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{16}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{x^2+2x+1-x^2+2x-1-16}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{4x-16}{\left(x-1\right)\left(x+1\right)}=0\)
=> 4x-16=0
<=> 4x=16
<=> x=4 (tmđk)
Vậy x=4