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a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)
b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)
\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)
a) ĐKXĐ : \(x\ne0\);\(x\ne2;-2\)
A=\(\left(\frac{1}{x-2}-\frac{2x}{4-x^2}+\frac{1}{2+x}\right).\left(\frac{2}{x}-1\right)\)
=\(\left(\frac{1}{x-2}+\frac{2x}{x^2-4}+\frac{1}{x+2}\right).\left(\frac{2}{x}-\frac{x}{x}\right)\)
=\(\frac{x+2+2x+x-2}{\left(x+2\right)\left(x-2\right)}.\frac{2-x}{x}\)
=\(\frac{4x}{\left(x+2\right)\left(x-2\right)}.\frac{-\left(x-2\right)}{x}\)
= \(\frac{-4}{x+2}\)
b) Ta có : \(2x^2+x=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=\frac{-1}{2}\end{cases}}\left(tm\right)\)
Để A = -1/2 thì
\(\Leftrightarrow\frac{-4}{x+2}=\frac{-1}{2}\)
\(\Leftrightarrow-\left(x+2\right)=-8\)
\(\Leftrightarrow x+2=8\)
\(\Leftrightarrow x=6\)
c) Để A =0,5 thì
\(\frac{-4}{x+2}=0,5\)
\(\Leftrightarrow-8=x+2\)
\(\Leftrightarrow x=-10\)
d) Để A \(\inℤ\)thì
\(-4⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(-4\right)\)
\(\Leftrightarrow x+2\in\left\{1;2;4;-1;-2;-4\right\}\)
Lập bảng giá trị
| x+2 | -1 | 1 | -2 | 2 | -4 | 4 |
| x | -3 | -1 | -4 | 0 | -6 | 2 |
Mà \(x\ne0\)và \(x\ne2;-2\)
\(\Rightarrow x\in\left\{-1;-3;-4;-6\right\}\)
a) Ta có :A = \(\left(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}+\frac{1}{x-1}\right):\frac{x^2+x}{x^3+x}\)
ĐK: \(\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)
A = \(\left(\frac{\left(x-1\right)^2}{x^2+x+1}-\frac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\right):\frac{x\left(x+1\right)}{x\left(x^2+1\right)}\)
= \(\frac{\left(x-1\right)^3-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)
= \(\frac{x^3-3x^2+3x-1+3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)
= \(\frac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}=1.\frac{x^2+1}{x+1}=\frac{x^2+1}{x+1}\)
b) Để A > - 1 <=> \(\frac{x^2+1}{x+1}>-1\)
<=> \(\frac{x^2+1}{x+1}+1>0\)
<=> \(\frac{x^2+x+2}{x+1}>0\)
Vì x2 + x + 2 >0 \(\forall x\)
=> A > 0 <=> x + 1 > 0 <=> x > -1
b1:
ĐKXĐ: \(x\ne0;x\ne\pm2\)
Ta có : \(A=\left(\frac{4x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{8x^2}{x^2-4}\right)\left(\frac{x-1}{x\left(x-2\right)}-\frac{2\left(x-2\right)}{x\left(x-2\right)}\right)\)
\(=\left(\frac{4x^2-8x-8x^2}{\left(x-2\right)\left(x+2\right)}\right)\left(\frac{x-1-2x+4}{x\left(x-2\right)}\right)\)
\(=\left(\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right)\left(\frac{3-3x}{x\left(x-2\right)}\right)\)
\(=\frac{12\left(x-1\right)}{x-2}\)
Vậy ....
Ta có : \(A< 0\Rightarrow\frac{12\left(x-1\right)}{x-2}< 0\)
Đến đây xét 2 TH 12(x-1)<0 & (x-2)>0 hoặc 12(x-1)>0 & (x-2)<0
\(A=\left(\frac{x^3-1}{x^2-x}+\frac{x^2-4}{x^2-2x}-\frac{2-x}{x}\right)\div\frac{x+1}{x}\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne2\end{cases}}\)
\(=\left(\frac{x^2+x+1}{x}+\frac{x+2}{x}-\frac{2-x}{x}\right)\times\frac{x}{x+1}\)
\(=\left(\frac{x^2+x+1+x+2-2+x}{x}\right)\times\frac{x}{x+1}\)
\(=\frac{x^2+3x+1}{x}\times\frac{x}{x+1}=\frac{x^2+3x+1}{x+1}\)
b) x3 - 4x2 + 3x = 0
<=> x( x2 - 4x + 3 ) = 0
<=> x( x - 1 )( x - 3 ) = 0
<=> x = 0 (ktm) hoặc x = 1(tm) hoặc x = 3(tm)
Bạn tự thế các giá trị tm nhé ;)
b) Ta có: \(x^3-4x^2+3x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-3\right)=0\)
<=> x=0 ( loại) hoặc x=1 (loại) hoặc x=3 ( thỏa mãn)
Thay x=3 vào A ta có:
\(A=\frac{3^2+3.3+1}{3+1}=\frac{19}{4}\)