rút gọn m=(x+1)^x-2*(x^2+x-2)+2 a tính giá trị biểu thức tại x=-3 b tìm x khi m=0
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MTC = (x - 2)(x + 2)
\(M=\dfrac{1}{x-2}-\dfrac{1}{x+2}+\dfrac{x^2+4x}{x^2-4}\)
\(=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x+2-x+2+x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x+2}{x-2}\)
\(M=\dfrac{x^2}{x^2-3x}\left(x\ne0;x\ne3\right)\\ M=\dfrac{x^2}{x\left(x-3\right)}\\ M=\dfrac{x}{x-3}\)
\(N=\dfrac{x}{x+1}+\dfrac{3x+1}{x^2-1}\left(x\ne\pm1\right)\\ N=\dfrac{x-1+3x+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{4x}{\left(x+1\right)\left(x-1\right)}\)
Còn tiếp:
\(\dfrac{4x}{\left(x+1\right)\left(x-1\right)}=\dfrac{4x}{x^2-1}\)
a: \(M=\dfrac{x^2\left(x-2\right)}{x-2}+\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{x^2-x+1}=x^2+x+1\)
b: Để M=7 thì (x+3)(x-2)=0
=>x=-3(nhận) hoặc x=2(loại)
Vậy: x=-3
Bài 1:
a.\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2\left(x+y\right)\)
b.\(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=4x^2\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne-1\end{cases}}\)
\(M=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x-x^2+1}{3x}\)
\(=\left[\frac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\frac{6x}{3x\left(x+1\right)}-\frac{9x\left(x+1\right)}{3x\left(x+1\right)}\right].\frac{x+1}{2-4x}+\frac{x^2-3x-1}{3x}\)
\(=\left[\frac{x^2+3x+2}{3x\left(x+1\right)}+\frac{6x}{3x\left(x+1\right)}-\frac{9x^2+9x}{3x\left(x+1\right)}\right].\frac{x+1}{2-4x}+\frac{x^2-3x-1}{3x}\)
\(=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}.\frac{x+1}{2-4x}+\frac{x^2-3x-1}{3x}\)
\(=\frac{2-8x^2}{3x}.\frac{1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{2\left(1-4x^2\right)}{3x}.\frac{1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{2\left(1-2x\right)\left(1+2x\right)}{3x}.\frac{1}{2\left(1-2x\right)}+\frac{x^2-3x-1}{3x}\)
\(=\frac{1+2x}{3x}+\frac{x^2-3x-1}{3x}\)
\(=\frac{1+2x+x^2-3x-1}{3x}=\frac{x^2-x}{3x}=\frac{x\left(x-1\right)}{3x}=\frac{x-1}{3}\)
b) Với \(x=6013\)( thỏa mãn ĐKXĐ )
Thay \(x=6013\)vào biểu thức ta được:
\(M=\frac{6013-1}{3}=\frac{6012}{3}=2004\)
a) Ta có: \(M=\dfrac{x-2}{x+2\sqrt{x}}-\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-2-\sqrt{x}-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b) Ta có: M-1
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{-2}{\sqrt{x}}< 0\forall x\) thỏa mãn ĐKXĐ
hay M<1
a: \(M=\dfrac{1-x}{1+x}:\dfrac{x^2-9-x^2+4+x+2}{\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{1-x}{1+x}\cdot\dfrac{\left(x-3\right)\left(x-2\right)}{x-3}=\dfrac{\left(1-x\right)\left(x-2\right)}{\left(1+x\right)}\)
b: M<0
=>(x-1)(x-2)/(x+1)>0
=>-1<x<1 hoặc x>2
c: M nguyên
=>(x-1)(x-2) chia hết cho x+1
=>x^2-3x+2 chia hết cho x+1
=>x^2+x-4x-4+6 chia hết cho x+1
=>x+1 thuộc {1;-1;2;-2;3;-3;6;-6}
=>x thuộc {0;-2;1;-3;-4;7;-5}
\(m=\left(x+1\right)^x-2\left(x^2+x-2\right)+2\)
a, Thay x = -3 ta được :
\(=\left(-3+1\right)^{-3}-2\left[\left(-3\right)^2-3-2\right]+2\)
\(=-\frac{1}{8}-8+2=-\frac{1}{8}-\frac{64}{8}+\frac{16}{8}=\frac{-49}{8}\)
b, Ta có : \(m=0\)hay \(\left(x+1\right)^x-2\left(x^2+x-2\right)+2=0\)
... =))?