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Vì \(x^2-4x+5=x^2-4x+4+1=\left(x-2\right)^2+1\ge1>0\) với mọi giá trị của \(x\) nên giá trị của biểu thức luôn luôn âm với mọi giá trị khác 0 và khác -3 của \(x\)
\(a,\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)
\(=\frac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)
\(=\frac{2\left(x-2\right)}{x+2}\)
Với \(x=\frac{1}{2}\)
\(\Rightarrow\frac{2\left(x-2\right)}{x+2}=\frac{2\left(\frac{1}{2}-2\right)}{\frac{1}{2}+2}=\frac{2.-\frac{3}{2}}{\frac{5}{2}}=-3.\frac{2}{5}=\frac{-6}{5}\)
b,Do x = -5; y = 10=> y = -2x
Thay y = -2x vào biểu thức ta được
\(\frac{x^3-x^2\left(-2x\right)+x\left(-2x\right)^2}{x^3+\left(-2x\right)^3}\)
\(=\frac{x^3+2x^3+2x^2}{x^3-8x^3}\)
\(=\frac{3x^3+2x^2}{-7x^3}=\frac{3}{-7}+\frac{2}{-7x}\)
Thay x = -5 là đc
Giải:
a) \(8\left(3x-2\right)-13x=5\left(12-3x\right)+7x\)
\(\Leftrightarrow24x-16-13x=60-15x+7x\)
\(\Leftrightarrow24x-13x+15x-7x=60+16\)
\(\Leftrightarrow19x=76\)
\(\Leftrightarrow x=\dfrac{76}{19}=4\)
Vậy ...
b) \(\dfrac{5x}{x+2}-\dfrac{3}{x-2}+\dfrac{3x^2+6}{\left(x-2\right)\left(x+2\right)}=0\) (1)
ĐKXĐ: \(x\ne\pm2\)
\(\left(1\right)\Leftrightarrow\dfrac{5x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{3x^2+6}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow5x\left(x-2\right)-3\left(x+2\right)+3x^2+6=0\)
\(\Leftrightarrow5x^2-10x-3x-6+3x^2+6=0\)
\(\Leftrightarrow8x^2-13x=0\)
\(\Leftrightarrow x\left(8x-13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\8x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=\dfrac{13}{8}\left(TM\right)\end{matrix}\right.\)
Vậy ...
c) \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\) (2)
ĐKXĐ: \(x\ne-1;x\ne3\)
\(\left(2\right)\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{4x}{2\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)
\(\Leftrightarrow x\left(x+1+x-3\right)=4x\)
\(\Leftrightarrow x\left(2x-2\right)=4x\)
\(\Leftrightarrow2x-2=4\)
\(\Leftrightarrow x=3\)
Vậy ...
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
\(\left(\dfrac{5x-x-12}{x\left(x+3\right)}\right):\dfrac{x-3}{x+3}=\dfrac{4\left(x-3\right)}{x\left(x+3\right)}.\dfrac{x+3}{x-3}=\dfrac{4}{x}\)