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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 \(x=\frac{-5}{2}\)
a) A = x2(m + 5) - x(m + 5)(x + 3/2) + (x - m)
A = mx2 + 5x2 - mx2 - 3/2mx - 5x2 - 15/2x + x - m
A = -3/2mx - m - 13/2x
b) Khi m = -1, ta có:
(-3/2).(-1).x - (-1) - 13/2x = 0
<=> 3/2x - 13/2x + 1 = 0
<=> 3/2x - 13/2x = 0 - 1
<=> 3/2x - 13/2x = -1
<=> 3x - 13x = -2
<=> -10x = -2
<=> x = -2/-10 = 1/5
ĐKXĐ \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
\(M=\frac{\left(x-3\right)^2}{2x\left(x-3\right)}\left(1-\frac{6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\)
\(=\frac{x-3}{2x}\left(1-\frac{6}{x-3}\right)\)
\(=\frac{x-3}{2x}.\frac{x-9}{x-3}=\frac{x-9}{2x}\)
\(M=\frac{\left(x-3\right)^2}{2x^2-6x}\left(1-\frac{6x+18}{x^2-9}\right)\left(x\ne\pm3;x\ne0\right)\)
\(\Leftrightarrow M=\frac{\left(x-3\right)^2}{2x\left(x-3\right)}\left(1-\frac{6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\)
\(\Leftrightarrow M=\frac{x-3}{2x}\cdot\left(1-\frac{6}{x-3}\right)\)
\(\Leftrightarrow M=\frac{x-3}{2x}\cdot\frac{x-9}{x-3}\)
\(\Leftrightarrow M=\frac{x-9}{2x}\)
Vậy với \(x\ne\pm3;x\ne0\)thì \(M=\frac{x-9}{2x}\)
\(M=2\left(a^3+b^3\right)-3\left(a^2+b^2\right)\)
\(\Leftrightarrow M=2\left[\left(a+b\right)\left(a^2-ab+b^2\right)\right]-3\left(a^2+b^2\right)\)
\(\Leftrightarrow M=2\left[\left(a^2-ab+b^2\right)\right]-3\left(a^2+b^2\right)\)
\(\Leftrightarrow M=2a^2-2ab+2b^2-3a^2-3b^2\)
\(\Leftrightarrow M=-a^2-2ab-b^2\)
\(\Leftrightarrow M=-\left(a+b\right)^2\)
Ta có:
\(\left(m^3-m+1\right)^2+\left(m^2-3\right)^2-2\left(m^2-3\right)\left(m^3-m+1\right)\)\(=\left(m^3-m+1-m^2+3\right)^2=\left(m^3-m^2-m+4\right)^2\)