\(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}-\frac{1}{x-1}\)
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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
Trước tiên ta đi rút gọn biểu thức trên :
Đặt \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
ĐKXĐ : \(x\ne\pm2,x\ne0\)
Ta có : \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left(\frac{x^2}{x\left(x^2-4\right)}+\frac{6}{3\left(2-x\right)}+\frac{1}{x+2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\right)\)
\(=\left(\frac{x\cdot3-6\cdot\left(x+2\right)+3\cdot\left(x-2\right)}{3\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)\)
\(=\frac{-18}{3\left(x-2\right)\left(x+2\right)}:\left(-\frac{6}{x+2}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{\left(-6\right)}=\frac{1}{x-2}\)
Để \(A\) nhận giá trị nguyên
\(\Leftrightarrow\frac{1}{x-2}\inℤ\) \(\Leftrightarrow1⋮x-2\) \(\Leftrightarrow x-2\inƯ\left(1\right)\)
\(\Leftrightarrow x-2\in\left\{-1,1\right\}\)
\(\Leftrightarrow x\in\left\{1,3\right\}\) ( Thỏa mãn ĐKXĐ )
Vậy : \(x\in\left\{1,3\right\}\) thì A nhận giá trị nguyên.
ĐKXĐ: \(x\notin\left\{0;1;-1\right\}\)
a: \(A=\left(\dfrac{\left(x-1\right)^2}{x^2+x+1}-\dfrac{-2x^2+4x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right)\cdot\dfrac{x\left(x^2+1\right)}{x\left(x+1\right)}\)
\(=\dfrac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{\left(x^2+1\right)}{x+1}\)
\(=\dfrac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{x+1}=\dfrac{x^2+1}{x+1}\)
Để R=0 thì \(x^2+1=0\)(vô lý)
b: Ta có: |x|=1
=>x=1(loại) hoặc x=-1(loại)
1) \(\left(x-2\right)\left(\frac{x+1}{3}-x+1\right)=0\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{3}-x^2+x-\frac{2\left(x+1\right)}{3}+2x-2=0\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{3}-x^2+3x-\frac{2\left(x+1\right)}{3}-2=0\)
\(\Leftrightarrow x\left(x+1\right)-3x^2+9x-2\left(x+1\right)-6=0\)
\(\Leftrightarrow x^2+x-3x^2+9x-2x-2-6=0\)
\(\Leftrightarrow-2x^2+8x-8=0\)
\(\Leftrightarrow-2\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow-2.\left(x^2-2.x.2+2^2\right)=0\)
\(\Leftrightarrow-2\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy nghiệm của phương trình là: {2}
2) \(\left(3x+4x\right)\left(\frac{x}{2}-x-\frac{3x}{5}+1\right)=0\)
\(\Leftrightarrow7x\left(\frac{x}{2}-x-\frac{3x}{5}+1\right)=0\)
\(\Leftrightarrow7x\left(-\frac{11x}{10}+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}7x=0\\-\frac{11x}{10}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{11}{10}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{10}{11}\end{cases}}\)
Vậy: nghiệm của phương trình là: \(\left\{0;\frac{10}{11}\right\}\)
3) \(\left|x-1\right|=x^2-x\)
\(\Leftrightarrow x-1=x^2-x\)
\(\Leftrightarrow1=x^2-x-x\)
\(\Leftrightarrow1=x^2\)
\(\Leftrightarrow x^2=1\)
\(\Rightarrow x=\pm1\)
Vậy nghiệm phương trình là: {1; -1}
4) \(\left|x^2-3x+1\right|=2x-3\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-3x+1=2x-3\\x^2-3x+1=-\left(2x-3\right)\end{cases}}\)
Xét trường hợp này rồi làm tiếp, dễ rồi :))
\(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}-\frac{1}{x-1}\)
\(=\frac{\left(x-1\right)^2}{x^2-2x+1+3x}-\frac{1-2x^2+4x}{x^3-1}-\frac{1}{x-1}\)
\(=\frac{\left(x-1\right)^2}{x^2+x+1}-\frac{1-2x^2+4x}{x^3-1}-\frac{1}{x-1}\)
a)\(ĐKXĐ:x\ne1\)
\(MTC:\left(x-1\right)^3=\left(x-1\right)\left(x^2+x+1\right)\)
b)\(\frac{\left(x-1\right)^3}{x^3-1}-\frac{1-2x^2+4x}{x^3-1}-\frac{x^2+x+1}{x^3-1}=0\)
\(\Rightarrow\left(x-1\right)^3-\left(1-2x^2+4x\right)-\left(x^2+x+1\right)=0\)
\(\Leftrightarrow x^3-3x^2+27x-1-1+2x^2-4x-x^2-x-1=0\)
\(\Leftrightarrow x^3-2x^2+22x-3=0\)
ĐẾN ĐÂY THÌ BÍ RỒI T_T