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1, a,= (x+2)^2/3.(x+2) = x+2/3
b, = 3x.(x+4)/2x.(x+4) = 3/2
k mk nha
a: \(=2x^2+2xy-xy-y^2\)
\(=2X\left(x+y\right)-y\left(x+y\right)=\left(x+y\right)\left(2x-y\right)\)
c: \(x^4-3x^2+1\)
\(=x^4-2x^2+1-x^2\)
\(=\left(x^2-1\right)^2-x^2\)
\(=\left(x^2-x-1\right)\left(x^2+x-1\right)\)
d: \(=\left(x-2\right)\left(x-3\right)\)
e: \(=x^2+10x+25-y^2-6y-9\)
\(=\left(x+5\right)^2-\left(y+3\right)^2\)
\(=\left(x+5-y-3\right)\left(x+5+y+3\right)\)
\(=\left(x-y+2\right)\left(x+y+8\right)\)
a) (x3 + 8y3) : (2y + x)
= (x + 2y)(x2 - 2xy + 4y2) : (2y + x)
= x2 - 2xy + 4y2
b) (x3 + 3x2y + 3xy2 + y3) : (2x + 2y)
= (x + y)3 : 2(x + y)
= \(\dfrac{\left(x+y\right)^2}{2}\)
c) (6x5y2 - 9x4y3 + 15x3y4) : 3x3y2
= 3x3y2(2x2 - 3xy + 5y2) : 3x3y2
= 2x2 - 3xy + 5y2
Bài 1:
a) \(\left(x+3\right)^2+x\left(x-2\right)=2x^2\)
\(x^2+6x+9+x^2-2x-2x^2=0\)
\(4x+9=0\)
\(x=\frac{-9}{4}\)
b) \(5x\left(x-4\right)-x+4=0\)
\(5x\left(x-4\right)-\left(x-4\right)=0\)
\(\left(x-4\right)\left(5x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\5x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{1}{5}\end{cases}}}\)
Bài 2:
a) \(x^2-4x=x\left(x-4\right)\)
b) \(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=\left(x+5\right)^2\)
c) \(x^2-y^2+2y-1\)
\(=x^2-\left(y^2-2y+1\right)\)
\(=x^2-\left(y-1\right)^2\)
\(=\left(x-y+1\right)\left(x+y-1\right)\)
d) \(x^2-11x+18\)
\(=x^2-2x-9x+18\)
\(=x\left(x-2\right)-9\left(x-2\right)\)
\(=\left(x-2\right)\left(x-9\right)\)
(x + 3)2 + x(x - 2) = 2x2
x2 + 6x + 9 + x2 - 2x = 2x2
<=> 2x2 + 4x + 9 = 2x2
<=> 4x = -9
<=> x = -9/4
\(a.\left(2x-3\right)\left(4x^2+6x+9\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\\ =\left(2x\right)^3-3^3-\left[\left(2x\right)^3+3^3\right]\\ =8x^3-9-\left(8x^3+9\right)\\ =8x^3-9-8x^3-9=-18\)
\(b.\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\\ =x^3+1-\left(x^3-1\right)\\ =x^3+1-x^3+1=2\)
\(c.\left(3x-1\right)\left(3x+1\right)-\left(3x-2\right)^2\\ =9x^2-1-\left(9x^2-12x+4\right)\\ =9x^2-1-9x^2+12x-4\\ =12x-5\)
\(d.\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\\ =\left(2x-3\right)\cdot\left[\left(2x-3\right)-\left(2x+3\right)\right]\\ =\left(2x-3\right)\cdot\left(2x-3-2x-3\right)\\ =\left(2x-3\right)\cdot\left(-6\right)\\ =-12x\cdot18\)
\(e.\left(3x-4\right)^2-\left(2x+4\right)^2\\ =9x^2-24x+16-\left(4x^2+16x+16\right)\\ =9x^2-24x+16-4x^2-16x-16\\ =5x^2-40x\)
\(f.\left(3x-5\right)^3-\left(3x+5\right)^3\\ =27x^3-135x^2+225x-125-\left(27x^3+135x^2+225x+125\right)\\ =27x^3-135x^2+225x-125-27x^3-135x^2-225x-125\\ =-270x^2-250\)
\(g.\left(2x-1\right)^2-\left(3x-1\right)^2\\ =4x^2-4x+1-\left(9x^2-6x+1\right)\\ =4x^2-4x+1-9x^2+6x-1\\ =-5x^2+2x\)
\(h.\left(x-2y\right)\left(x^2+2xy+4y^2\right)+\left(x^3-6y^3\right)\\ =x^3-8y^3+x^3-6y^3\\ =2x^3-14y^3\)
1: \(\Leftrightarrow5x^2+4x-1-2x^2+12x-18=3x^2+5x-2-x^2-8x-16+x^2-x\)
\(\Leftrightarrow3x^2+16x-19=3x^2-4x-18\)
=>20x=1
hay x=1/20
2: \(\Leftrightarrow5x^2-20x-41=x^2-10x+25+4x^2+4x+1-\left(x^2-2x\right)+\left(x-1\right)^2\)
\(\Leftrightarrow5x^2-20x-41=4x^2-4x+26+x^2-2x+1\)
\(\Leftrightarrow-20x-41=-6x+27\)
=>-14x=68
hay x=-34/7