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✫¸.•°*”˜˜”*°•✫ Ṱђầภ Ḉђết ✫•°*”˜˜”*°... · Accepted Answer

Bài làmBài 3a) $A=\left(x^2+\frac{4x^2}{x^2-4}\right)\cdot\left(\frac{x+2}{2x-4}+\frac{2-3x}{x^3-4x}\cdot\frac{x^2-4}{x-2}\right)$$A=\left(x^2+\frac{4x^2}{x^2-4}\right)\cdot\left(\frac{x+2}{2\left(x-2\right)}+\frac{2-3x}{x\left(x^2-4\right)}\cdot\frac{x^2-4}{x-2}\right)$$A=\left(x^2+\frac{4x^2}{x^2-4}\right)\cdot\left(\frac{x+2}{2\left(x-2\right)}+\frac{2-3x}{x\left(x-2\right)}\right)$$A=\left(x^2+\frac{4x^2}{x^2-4}\right)\cdot\left(\frac{x\left(x+2\right)}{2x\left(x-2\right)}+\frac{2\left(2-3x\right)}{2x\left(x-2\right)}\right)$$A=\left(x^2+\frac{4x^2}{x^2-4}\right)\cdot\frac{x^2+2x+4-6x}{2x\left(x-2\right)}$$A=\left(x^2+\frac{4x^2}{x^2-4}\right)\cdot\frac{x^2-4x+4}{2x\left(x-2\right)}$$A=\left(\frac{x^4-4x^2}{x^2-4}+\frac{4x^2}{x^2-4}\right)\cdot\frac{\left(x-2\right)^2}{2x\left(x-2\right)}$$A=\frac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{2x}$$A=\frac{x^3}{2\left(x+2\right)}$Vậy $A=\frac{x^3}{2\left(x+2\right)}$b) Ta có: $\left|2x-1\right|=3$<=> $\orbr{\begin{cases}2x-1=3\2x-1=-3\end{cases}}$<=> $\orbr{\begin{cases}2x=4\2x=-2\end{cases}}$<=> $\orbr{\begin{cases}x=2\x=-1\end{cases}}$*) Với x = 2 thì ta thay x = 2 vào A ta được:$A=\frac{2^3}{2\left(2+2\right)}=\frac{8}{8}=1$Vậy với x = 2 thì A = 1*) Với x = -1 thì ta thay x = -2 vào ta, ta được:$A=\frac{\left(-1\right)^3}{2\left(-1+2\right)}=\frac{-1}{2}$Vậy với x = -1 thì x = -1/2Bài 2:a) $A=\left(\frac{x+2}{x^2+2x+1}-\frac{x-2}{x^2-1}\right):\frac{2x^2+x}{x^3+x^2-x-1}$$A=\left(\frac{x+2}{\left(x+1\right)\left(x+1\right)}-\frac{x-2}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x^2+x}{x^2\left(x+1\right)-\left(x+1\right)}$$A=\left(\frac{\left(x+2\right)\left(x-1\right)}{\left(x+1\right)\left(x+1\right)\left(x-1\right)}-\frac{\left(x-2\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)\left(x+1\right)}\right):\frac{2x^2+x}{\left(x+1\right)\left(x+1\right)\left(x-1\right)}$$A=\left(\frac{x^2+2x-x-2}{\left(x+1\right)^2\left(x-1\right)}-\frac{x^2+x-2x-2}{\left(x+1\right)^2\left(x-1\right)}\right)\cdot\frac{\left(x+1\right)^2\left(x-1\right)}{x\left(2x+1\right)}$$A=\frac{x^2+2x-x-2-x^2-x+2x+2}{\left(x+1\right)^2\left(x-1\right)}\cdot\frac{\left(x+1\right)^2\left(x-1\right)}{x\left(2x+1\right)}$$A=4x-2x\cdot\frac{1}{x\left(2x+1\right)}$$A=\frac{2x\left(2-1\right)}{x\left(2x+1\right)}$$A=\frac{4-2}{2x+1}$$A=\frac{2}{2x+1}$Để A xác định<=> 2x + 1 khác 0<=> 2x khác -1<=> x khác -1/2Vậy x khác -1/2 thì A xác định.b) Thay x = -3 vào A ta được:$A=\frac{2}{2\left(-3\right)+1}=\frac{2}{-6+1}=\frac{2}{-5}$Vậy x = -3 thì A = 2/-5Thay x = 1/4 vào A ta được$A=\frac{2}{2\cdot\frac{1}{4}+1}=\frac{2}{\frac{1}{2}+1}=\frac{2}{\frac{3}{2}}=2:\frac{3}{2}=2\cdot\frac{2}{3}=\frac{4}{3}$Vậy x = 1/4 thì A = 4/3Vì x = -1/2 (Không thỏa mãn điều kiện)Do đó với x = -1/2 thì A không xác định.c) Để |A| = 3<=> $\left|\frac{2}{2x+1}\right|=3$<=> $\orbr{\begin{cases}\frac{2}{2x+1}=3\\frac{2}{2x+1}=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}6x+3=2\-6x-3=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}6x=-1\-6x=5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{6}\x=-\frac{5}{6}\end{cases}}}$Vậy x = -1/6 hoặc x = -5/6 thì |A| = 3Câu trả lời: Bài làmBài 3a) $A=\left(x^2+\frac{4x^2}{x^2-4}\right)\cdot\left(\frac{x+2}{2x-4}+\frac{2-3x}{x^3-4x}\cdot\frac{x^2-4}{x-2}\right)$$A=\left(x^2+\frac{4x^2}{x^2-4}\right)\cdot\left(\frac{x+2}{2\left(x-2\right)}+\frac{2-3x}{x\left(x^2-4\right)}\cdot\frac{x^2-4}{x-2}\right)$$A=\left(x^2+\frac{4x^2}{x^2-4}\right)\cdot\left(\frac{x+2}{2\left(x-2\right)}+\frac{2-3x}{x\left(x-2\right)}\right)$$A=\left(x^2+\frac{4x^2}{x^2-4}\right)\cdot\left(\frac{x\left(x+2\right)}{2x\left(x-2\right)}+\frac{2\left(2-3x\right)}{2x\left(x-2\right)}\right)$$A=\left(x^2+\frac{4x^2}{x^2-4}\right)\cdot\frac{x^2+2x+4-6x}{2x\left(x-2\right)}$$A=\left(x^2+\frac{4x^2}{x^2-4}\right)\cdot\frac{x^2-4x+4}{2x\left(x-2\right)}$$A=\left(\frac{x^4-4x^2}{x^2-4}+\frac{4x^2}{x^2-4}\right)\cdot\frac{\left(x-2\right)^2}{2x\left(x-2\right)}$$A=\frac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{2x}$$A=\frac{x^3}{2\left(x+2\right)}$Vậy $A=\frac{x^3}{2\left(x+2\right)}$b) Ta có: $\left|2x-1\right|=3$<=> $\orbr{\begin{cases}2x-1=3\2x-1=-3\end{cases}}$<=> $\orbr{\begin{cases}2x=4\2x=-2\end{cases}}$<=> $\orbr{\begin{cases}x=2\x=-1\end{cases}}$*) Với x = 2 thì ta thay x = 2 vào A ta được:$A=\frac{2^3}{2\left(2+2\right)}=\frac{8}{8}=1$Vậy với x = 2 thì A = 1*) Với x = -1 thì ta thay x = -2 vào ta, ta được:$A=\frac{\left(-1\right)^3}{2\left(-1+2\right)}=\frac{-1}{2}$Vậy với x = -1 thì x = -1/2Bài 2:a) $A=\left(\frac{x+2}{x^2+2x+1}-\frac{x-2}{x^2-1}\right):\frac{2x^2+x}{x^3+x^2-x-1}$$A=\left(\frac{x+2}{\left(x+1\right)\left(x+1\right)}-\frac{x-2}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x^2+x}{x^2\left(x+1\right)-\left(x+1\right)}$$A=\left(\frac{\left(x+2\right)\left(x-1\right)}{\left(x+1\right)\left(x+1\right)\left(x-1\right)}-\frac{\left(x-2\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)\left(x+1\right)}\right):\frac{2x^2+x}{\left(x+1\right)\left(x+1\right)\left(x-1\right)}$$A=\left(\frac{x^2+2x-x-2}{\left(x+1\right)^2\left(x-1\right)}-\frac{x^2+x-2x-2}{\left(x+1\right)^2\left(x-1\right)}\right)\cdot\frac{\left(x+1\right)^2\left(x-1\right)}{x\left(2x+1\right)}$$A=\frac{x^2+2x-x-2-x^2-x+2x+2}{\left(x+1\right)^2\left(x-1\right)}\cdot\frac{\left(x+1\right)^2\left(x-1\right)}{x\left(2x+1\right)}$$A=4x-2x\cdot\frac{1}{x\left(2x+1\right)}$$A=\frac{2x\left(2-1\right)}{x\left(2x+1\right)}$$A=\frac{4-2}{2x+1}$$A=\frac{2}{2x+1}$Để A xác định<=> 2x + 1 khác 0<=> 2x khác -1<=> x khác -1/2Vậy x khác -1/2 thì A xác định.b) Thay x = -3 vào A ta được:$A=\frac{2}{2\left(-3\right)+1}=\frac{2}{-6+1}=\frac{2}{-5}$Vậy x = -3 thì A = 2/-5Thay x = 1/4 vào A ta được$A=\frac{2}{2\cdot\frac{1}{4}+1}=\frac{2}{\frac{1}{2}+1}=\frac{2}{\frac{3}{2}}=2:\frac{3}{2}=2\cdot\frac{2}{3}=\frac{4}{3}$Vậy x = 1/4 thì A = 4/3Vì x = -1/2 (Không thỏa mãn điều kiện)Do đó với x = -1/2 thì A không xác định.c) Để |A| = 3<=> $\left|\frac{2}{2x+1}\right|=3$<=> $\orbr{\begin{cases}\frac{2}{2x+1}=3\\frac{2}{2x+1}=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}6x+3=2\-6x-3=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}6x=-1\-6x=5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{6}\x=-\frac{5}{6}\end{cases}}}$Vậy x = -1/6 hoặc x = -5/6 thì |A| = 3

Ngô Chi Lan · Answer

$A=\left(x^2+\frac{4x^2}{x^2-4}\right)\left(\frac{x+2}{2x-4}+\frac{2-3x}{x^3-4}.\frac{x^2-4}{x-2}\right)$$=\left(\frac{x^2\left(x^2-4\right)}{x^2-4}+\frac{4x^2}{x^2-4}\right)\left(\frac{x+2}{2\left(x-2\right)}+\frac{\left(2-3x\right)\left(x^2-4\right)}{x\left(x^2-4\right)\left(x-2\right)}\right)$$=\left(\frac{x^2\left(x^2-4\right)+4x^2}{x^2-4}\right)\left(\frac{x+2}{2\left(x-2\right)}+\frac{2-3x}{x\left(x-2\right)}\right)$$=\left(\frac{x^4-4x^2+4x^2}{x^2-4}\right)\left(\frac{x\left(x+2\right)}{2x\left(x-2\right)}+\frac{2\left(2-3x\right)}{2x\left(x-2\right)}\right)$$=\left(\frac{x^4}{x^2-4}\right)\left(\frac{x^2+2x+4-6x}{2x\left(x-2\right)}\right)=\left(\frac{x^4}{x^2-4}\right)\left(\frac{x^2-4x+4}{2x\left(x-2\right)}\right)$$=\left(\frac{x^4}{x^2-4}\right)\left(\frac{\left(x-2\right)^2}{2x\left(x-2\right)}\right)=\left(\frac{x^4}{x^2-4}\right)\left(\frac{x-2}{2x}\right)=\frac{x^4.\left(x-2\right)}{\left(x^2-4\right)2x}$$=\frac{x^4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)2x}=\frac{x^3}{2\left(x+2\right)}$b,Ta có:$\left|2x-1\right|=3$$\Leftrightarrow2x-1=\pm3$$\Leftrightarrow\orbr{\begin{cases}2x-1=3\2x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\x=-1\end{cases}}}$Với x=2 thì giá trị của A là:$\frac{2^3}{2\left(2+2\right)}=1$Với x=-1 thì giá trị biểu thức là:$\frac{\left(-1\right)^3}{-2\left(-1+2\right)}=-\frac{1}{2}$Câu trả lời: $A=\left(x^2+\frac{4x^2}{x^2-4}\right)\left(\frac{x+2}{2x-4}+\frac{2-3x}{x^3-4}.\frac{x^2-4}{x-2}\right)$$=\left(\frac{x^2\left(x^2-4\right)}{x^2-4}+\frac{4x^2}{x^2-4}\right)\left(\frac{x+2}{2\left(x-2\right)}+\frac{\left(2-3x\right)\left(x^2-4\right)}{x\left(x^2-4\right)\left(x-2\right)}\right)$$=\left(\frac{x^2\left(x^2-4\right)+4x^2}{x^2-4}\right)\left(\frac{x+2}{2\left(x-2\right)}+\frac{2-3x}{x\left(x-2\right)}\right)$$=\left(\frac{x^4-4x^2+4x^2}{x^2-4}\right)\left(\frac{x\left(x+2\right)}{2x\left(x-2\right)}+\frac{2\left(2-3x\right)}{2x\left(x-2\right)}\right)$$=\left(\frac{x^4}{x^2-4}\right)\left(\frac{x^2+2x+4-6x}{2x\left(x-2\right)}\right)=\left(\frac{x^4}{x^2-4}\right)\left(\frac{x^2-4x+4}{2x\left(x-2\right)}\right)$$=\left(\frac{x^4}{x^2-4}\right)\left(\frac{\left(x-2\right)^2}{2x\left(x-2\right)}\right)=\left(\frac{x^4}{x^2-4}\right)\left(\frac{x-2}{2x}\right)=\frac{x^4.\left(x-2\right)}{\left(x^2-4\right)2x}$$=\frac{x^4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)2x}=\frac{x^3}{2\left(x+2\right)}$b,Ta có:$\left|2x-1\right|=3$$\Leftrightarrow2x-1=\pm3$$\Leftrightarrow\orbr{\begin{cases}2x-1=3\2x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\x=-1\end{cases}}}$Với x=2 thì giá trị của A là:$\frac{2^3}{2\left(2+2\right)}=1$Với x=-1 thì giá trị biểu thức là:$\frac{\left(-1\right)^3}{-2\left(-1+2\right)}=-\frac{1}{2}$

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