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a) M = ( 2x + 3)(2x - 3) - 2(x + 5)2 - 2(x - 1)(x + 2)
= 4x2 - 9 - 2(x2 + 10x + 25) - 2(x2 + x - 2)
= 4x2 - 9 - 2x2 - 20x - 50 - 2x2 - 2x + 4
= -22x - 55 = -11(2x + 5)
b) M = -11(2x + 5) = - 11(2.\(\frac{-7}{3}\)+ 5) = \(\frac{-11}{3}\)
b) M = -11(2x + 5) = 0
\(\Rightarrow\)2x + 5 = 0
\(\Rightarrow\)x = \(\frac{-5}{2}\)
Ta có: M = (2x+3)(2x-3) - 2(x+5)2 - 2(x-1)(x+2) \(=\left(2x\right)^2-3^2-2\left(x^2+10x+25\right)-\) \(2\left(x^2+x-2\right)\)
\(=4x^2-9-2x^2-20x-50-2x^2-2x+4\) =\(\left(4x^2-2x^2-2x^2\right)-\left(20x+2x\right)-\left(50+9-4\right)\) \(=-22x-55\)
b, Với x = \(-2\frac{1}{3}=\frac{-7}{3}\)
\(\Rightarrow M=-22.\frac{-7}{3}-55=\frac{154}{3}-55=\frac{-11}{3}\)
c, Để M = 0 => -22x - 55 = 0 \(\Rightarrow-22x=55\Rightarrow x=\frac{-55}{22}=\frac{-5}{2}\)
Vậy \(x=\frac{-5}{2}\)
\(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne0\end{cases}}\)
a) \(P=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(\Leftrightarrow P=\left(\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right):\frac{x^2-4+10-x^2}{x-2}\)
\(\Leftrightarrow P=\frac{x^2-2x\left(x+2\right)+x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}:\frac{6}{x-2}\)
\(\Leftrightarrow P=\frac{x^2-2x^2-4x+x^2-2x}{x\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{6}\)
\(\Leftrightarrow P=\frac{-6x}{6x\left(x+2\right)}\)
\(\Leftrightarrow P=\frac{-1}{x+2}\)
b) Khi \(\left|x\right|=\frac{3}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=-\frac{3}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}P=-\frac{1}{\frac{3}{4}+2}=-\frac{4}{11}\\P=-\frac{1}{-\frac{3}{4}+2}=-\frac{4}{5}\end{cases}}\)
c) Để P = 7
\(\Leftrightarrow-\frac{1}{x+2}=7\)
\(\Leftrightarrow7\left(x+2\right)=-1\)
\(\Leftrightarrow7x+14=-1\)
\(\Leftrightarrow7x=-15\)
\(\Leftrightarrow x=-\frac{15}{7}\)
Vậy để \(P=7\Leftrightarrow x=-\frac{15}{7}\)
d) Để \(P\inℤ\)
\(\Leftrightarrow1⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{-3;-1\right\}\)
Vậy để \(P\inℤ\Leftrightarrow x\in\left\{-3;-1\right\}\)
Ta có:
a) M = \(\left(\frac{6x}{x^2-9}-\frac{1}{x+3}+\frac{5}{3-x}\right):\frac{4}{x^2-3x}\)
M = \(\left(\frac{6x}{\left(x-3\right)\left(x+3\right)}-\frac{x-3}{\left(x+3\right)\left(x-3\right)}-\frac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\cdot\frac{x^2-3x}{4}\)
M = \(\left(\frac{6x-x+3-5x-15}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x\left(x-3\right)}{4}\)
M = \(\frac{-12.x\left(x-3\right)}{\left(x-3\right)\left(x+3\right).4}\)
M = \(-\frac{3x}{x+3}\)
b) Với x = 2 => M = \(-\frac{3.2}{3+2}=-\frac{6}{5}\)
\(ĐK:x\ne\pm1;x\ne0;x\ne3\)
Với \(x\ne\pm1;x\ne0;x\ne3\)thì\(M=\frac{x^3+2x^2-x-2}{x^3-2x^2-3x}\left[\frac{\left(x+2\right)^2-x^2}{4x^2-4}-\frac{3}{x^2-x}\right]=\frac{x^2\left(x+2\right)-\left(x+2\right)}{\left(x^3-x\right)-\left(2x^2+2x\right)}\left[\frac{x^2+4x+4-x^2}{4x^2-4}-\frac{3}{x\left(x-1\right)}\right]\)\(=\frac{\left(x-1\right)\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)\left(x-1\right)-2x\left(x+1\right)}\left[\frac{4\left(x+1\right)}{4\left(x+1\right)\left(x-1\right)}-\frac{3}{x\left(x-1\right)}\right]=\frac{\left(x-1\right)\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-3x\right)}\left[\frac{1}{x-1}-\frac{3}{x\left(x-1\right)}\right]\)\(=\frac{\left(x-1\right)\left(x+2\right)}{x\left(x-3\right)}.\frac{x-3}{x\left(x-1\right)}=\frac{x+2}{x^2}\)
M = 3 \(\Leftrightarrow\frac{x+2}{x^2}=3\Leftrightarrow3x^2-x-2=0\Leftrightarrow\left(x-1\right)\left(3x+2\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-2}{3}\end{cases}}\)
Mà \(x\ne1\)(theo điều kiện) nên x =-2/3