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Thay x = -2 vào A ta được
\(A=\frac{\left|-2+1\right|+2.\left(-2\right)}{3.\left(-2\right)^2-2\left(-2\right)-1}=\frac{1-4}{12+4-1}=\frac{-3}{15}=-\frac{1}{5}\)
Thay x = 3/4 vào A ta được :
\(A=\frac{\left|\frac{3}{4}+1\right|+\frac{2.3}{4}}{3\left(\frac{3}{4}\right)^2-\frac{2.3}{4}-1}=\frac{\frac{7}{4}+\frac{6}{4}}{\frac{3.9}{16}-\frac{6}{4}-1}=\frac{\frac{13}{4}}{-\frac{13}{16}}=-\frac{16}{4}=-4\)
\(\frac{\left(2x^3+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)
\(=\frac{2x\left(x^2+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)
\(=\frac{2\left(x^2+1\right)\left(x-2\right)}{\left(x+2\right)\left(x+1\right)}\)
Thay x=\(\frac{1}{2}\)
\(=\frac{2\left(\frac{1}{2}^2+1\right)\left(\frac{1}{2}-2\right)}{\left(\frac{1}{2}+2\right)\left(\frac{1}{2}+1\right)}\)
\(=-1\)
\(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\) ĐKXD: \(x\ne\pm2,x\ne0,x\ne3\)
\(\Leftrightarrow\left(\frac{2+x}{2-x}+\frac{4x^2}{\left(2-x\right)\left(2+x\right)}-\frac{2-x}{2+x}\right):\left(\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\right)\)
\(\Leftrightarrow\left(\frac{4+4x+x^2+4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\right):\left(\frac{x-3}{x\left(2-x\right)}\right)\)
\(\Leftrightarrow\left(\frac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}\right)\cdot\left(\frac{x\left(2-x\right)}{x-3}\right)\)
\(\Leftrightarrow\frac{4x\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}\cdot\frac{x\left(2-x\right)}{x-3}\)
\(\Leftrightarrow\frac{4x^2}{x-3}\)
b, Để A>0 thì \(\frac{4x^2}{x-3}>0\)
\(\Rightarrow4x^2>0\)
\(\Rightarrow x>0\)
c, Ta có
\(\left|x-7\right|=4\)
\(\Rightarrow\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=3\left(l\right)\end{cases}}}\)
Với \(x=11\Rightarrow\frac{4\cdot11^2}{11-3}=\frac{121}{2}\)
a) A \(=\)\(\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)\(=\)\(\frac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)
\(=\)\(\frac{2\left(x-2\right)}{x+2}\)\(=\)\(\frac{2x-4}{x+2}\)
Tại x = \(\frac{1}{2}\)thì:
A = \(\frac{2.\frac{1}{2}-4}{\frac{1}{2}+2}\)\(=\)\(\frac{-3}{\frac{5}{2}}\)\(=\)\(\frac{-6}{5}\)
\(ĐKXĐ:x\ne1;x\ne\frac{-1}{3}\)
+) Nếu \(x\ge-1\Rightarrow\left|x+1\right|=x+1\)
\(\Rightarrow A=\frac{x+1+2x}{3x^2-2x-1}=\frac{3x+1}{\left(x-1\right)\left(3x+1\right)}=\frac{1}{x-1}\)
Với x = -2 thì \(A=\frac{-1}{3}\)
Với \(x=\frac{3}{4}\)thì \(A=-4\)
+) Nếu \(x< -1\Rightarrow\left|x+1\right|=-x-1\)
\(\Rightarrow A=\frac{-x-1+2x}{3x^2-2x-1}=\frac{x-1}{\left(x-1\right)\left(3x+1\right)}=\frac{1}{3x+1}\)
Với x = -2 thì \(A=\frac{-1}{5}\)
Với \(x=\frac{3}{4}\)thì \(A=\frac{4}{13}\)