Đại số lớp 8

\(A=\frac{\left|x-1\right|+\left|x\right|-x}{3x^2+4x+1}=\frac{1-x-x-x}{3x^2+3x+x+1}=\frac{1-3x}{\left(x+1\right)\left(3x+1\right)}\)

\(B=\frac{\left|2x-1\right|+x}{3x^2-22x+7}=\frac{1-2x+x}{3x^2-21x-x+7}=\frac{1-x}{\left(x-7\right)\left(3x-1\right)}\)

a. 2x

b.\({3x}\over x^2-1\)

1

a . X.(X-2)-Y.(X-2)=(X-Y).(X-2)

b .(X² +1+2X).(X² +1-2X)

2

3X² +2X+X² +2X+1-4X² -10X+10X+5=(-12)

4X+6= -12

X=9/2

1. a, x²-2x+2y-xy = x(x-2)+y(2-y) = x(x-2)-y(x-2) = (x-y)(x-2)

b, (x²+1)²-4x² = (x²+1-2x)(x²+1+2x) = (x-1)²(x+1)²

2. x(3x+2)+(x+1)²-(2x-5)(2x+5) = -12

=> (3x²+2x+x²+2x+1)-(2x)²-5² = -12

_{=> 3x²+2x+x²+2x+1-4x²-25 = -12}

_{=> 4x-24 = -12 => 4x = 12 => x = 3}

2, \(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=-12\)

<=> \(3x^2+2x+x^2+2x+1-4x^2-10x+10x+25=-12\)

<=> \(4x+26=-12\)

<=> \(4x=-12-26=-38\)

<=> \(x=-38:4=\frac{-19}{2}\)

vậy x=\(\frac{-19}{2}\)

Ta có:

\(C_1:\left(3x+1\right)^2-4\left(x-2\right)^2=\left(9x^2+6x+1\right)-4\left(x^2-4x+4\right)\)

\(=9x^2+6x+1-4x^2+16x-16=5x^2+22x-15=5x\left(x+5\right)-3\left(x+5\right)=\left(5x-3\right)\left(x+5\right)\)

\(C_2:\left[\left(3x+1\right)-2\left(x-2\right)\right]\left[\left(3x+1\right)+2\left(x-2\right)\right]=\left(x+5\right)\left(5x-3\right)\)

Ta có:

\(\frac{2x^2-3x+1}{x-1}=\frac{\left(2x^2-2x\right)-\left(x-1\right)}{x-1}=\frac{2x\left(x-1\right)-\left(x-1\right)}{x-1}=\frac{\left(2x-1\right)\left(x-1\right)}{x-1}=2x-1\)

=> Với mọi số nguyên \(x\) thì \(2x-1\) là số nguyên

x=2007,5

Ta có :

\(1+a+a^2+....+a^{63}\)

\(=\left(1+a\right)+a^2\left(1+a\right)+....+a^{62}\left(1+a\right)\)

\(=\left(1+a\right)\left(1+a^2+a^4+....+a^{62}\right)\)

\(=\left(1+a\right)\left[\left(1+a^2\right)+a^4\left(1+a^2\right)+.....+a^{60}\left(1+a^2\right)\right]\)

\(=\left(1+a\right)\left(1+a^2\right)\left(1+a^4+....+a^{60}\right)\)

.....

\(=\left(1+a\right)\left(1+a^2\right).....\left(1+a^{32}\right)\)

\(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)

\(=\frac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{x\left(x-2y\right)}{\left(x+2y\right)\left(x-2y\right)}-\frac{4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2x}{x+2y}\)

ngu

\({2x^2-4xy}\over x^2-4y^2\)

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