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Bài 1:
a. $3x^3-12x^2+12x=3x(x^2-4x+4)=3x(x-2)^2$
b. $x^2-25+4xy+4y^2=(x^2+4xy+4y^2)-25=(x+2y)^2-5^2=(x+2y-5)(x+2y+5)$
c. $4x^3-x=x(4x^2-1)=x[(2x)^2-1^2]=x(2x-1)(2x+1)$
d. $x^2-x+2y-4y^2=(x^2-4y^2)-(x-2y)=(x-2y)(x+2y)-(x-2y)=(x-2y)(x+2y+1)$
Bài 2:
a. $3x(x-1)+x-1=0$
$\Leftrightarrow (x-1)(3x+1)=0$
$\Leftrightarrow x-1=0$ hoặc $3x+1=0$
$\Leftrightarrow x=1$ hoặc $x=\frac{-1}{3}$
b. $x(2x+1)-4x^2+1=0$
$\Leftrightarrow x(2x+1)-(4x^2-1)=0$
$\Leftrightarrow x(2x+1)-(2x-1)(2x+1)=0$
$\Leftrightarrow (2x+1)[x-(2x-1)]=0$
$\Leftrightarrow (2x+1)(-x+1)=0$
$\Leftrightarrow 2x+1=0$ hoặc $-x+1=0$
$\Leftrightarrow x=\frac{-1}{2}$ hoặc $x=1$
a) \(x^3-4x=0\)
\(x\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}}\)
b) \(5x\left(3x-2\right)=4-9x^2\)
\(5x\left(3x-2\right)-\left(4-9x^2\right)=0\)
\(5x\left(3x-2\right)-\left(2-3x\right)\left(2+3x\right)=0\)
\(5x\left(3x-2\right)+\left(3x-2\right)\left(2+3x\right)=0\)
\(\left(3x-2\right)\left(5x+3x+2\right)=0\)
\(\left(3x-2\right)\left(8x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-2=0\\8x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{-1}{4}\end{cases}}}\)
c) \(x^2+7x=8\)
\(x^2+7x-8=0\)
\(x^2+8x-x-8=0\)
\(x\left(x+8\right)-\left(x+8\right)=0\)
\(\left(x+8\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+8=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-8\\x=1\end{cases}}}\)
d) \(2x^2+4y^2+10x+4xy=-25\)
\(x^2+x^2+4y^2+10x+4xy+25=0\)
\(\left(4y^2+4xy+x^2\right)+\left(x^2+10x+25\right)=0\)
\(\left(2y+x\right)^2+\left(x+5\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}2y+x=0\\x+5=0\end{cases}\Rightarrow\hept{\begin{cases}y=\frac{5}{2}\\x=-5\end{cases}}}\)
a) \(2x^2-2y^2\)
\(=2\left(x^2-y^2\right)\)
\(=2\left(x-y\right)\left(x+y\right)\)
b) \(x^2-4x+4\)
\(=x^2-2\cdot x\cdot2+2^2\)
\(=\left(x-2\right)^2\)
c) \(x^2+2x+1-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x-y+1\right)\left(x+y+1\right)\)
d) \(x^2-4x\)
\(=x\left(x-4\right)\)
e) \(x^2+10x+25\)
\(=x^2+2\cdot x\cdot5+5^2\)
\(=\left(x+5\right)^2\)
g) \(x^2-2xy+y^2-9\)
\(=\left(x-y\right)^2-3^2\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
h) \(2x^2-2\)
\(=2\left(x^2-1\right)\)
\(=2\left(x-1\right)\left(x+1\right)\)
i) \(5x^2-5xy+9x-9y\)
\(=5x\left(x-y\right)+9\left(x-y\right)\)
\(=\left(x-y\right)\left(5x+9\right)\)
k) \(y^2-4y+4-x^2\)
\(=\left(y-2\right)^2-x^2\)
\(=\left(y-x-2\right)\left(y+x-2\right)\)
l) \(x^2-16\)
\(=x^2-4^2\)
\(=\left(x-4\right)\left(x+4\right)\)
m) \(3x^2-3xy+2x-2y\)
\(=3x\left(x-y\right)+2\left(x-y\right)\)
\(=\left(x-y\right)\left(3x+2\right)\)
o) \(3x^4-6x^3+3x^2\)
\(=3x^2\left(x^2-2x+1\right)\)
\(=3x^2\left(x-1\right)^2\)
a) 2x2 - 2y2
= (2x - 2y)(2x + 2y)
= 4(x - y)(x + y)
b) x2 - 4x + 4
= (x - 2)2
c) x2 + 2x + 1 - y2
= (x + 1)2 - y2
= (x + 1 - y)(x + 1 + y)
d) x2 - 4x
= x(x - 4)
e) x2 +10x + 25
= (x + 5)2
g) x2 - 2xy + y2 - 9
= (x - y)2 - 32
= (x - y - 3)(x - y + 3)
h) 2x2 - 2
= 2(x2 - 1)
= 2(x - 1)(x + 1)
i) 5x2 - 5xy + 9x - 9y
= 5x(x - y) + 9(x- y)
= (5x + 9)(x - y)
k) y2 - 4y + 4 - x2
= (y - 2)2 - x2
= (y - 2 - x)(y - 2 + x)
l) x2 - 16
= x2 - 42
= (x - 4)(x + 4)
m) 3x2 - 3xy + 2x -2y
= 3x(x - y) +2(x-y)
= (3x + 2)(x - y)
o) 3x4 - 6x3 + 3x2
= 3x4 - 3x3 - 3x3 + 3x2
= 3x3(x - 1) - 3x2(x - 1)
= (3x3 - 3x2)(x - 1)
= 3x2(x - 1)(x - 1)
= 3x2.(x - 1)2
a: 2x^2y-50xy=2xy(x-25)
b: 5x^2-10x=5x(x-2)
c: 5x^3-5x=5x(x^2-1)=5x(x-1)(x+1)
d: \(x^2-xy+x=x\left(x-y+1\right)\)
e: x(x-y)-2(y-x)
=x(x-y)+2(x-y)
=(x-y)(x+2)
f: 4x^2-4xy-8y^2
=4(x^2-xy-2y^2)
=4(x^2-2xy+xy-2y^2)
=4[x(x-2y)+y(x-2y)]
=4(x-2y)(x+y)
f1: x^2ỹ-y^2+y
=(x-y)(x+y)+(x+y)
=(x+y)(x-y+1)
\(1.\)
\(x^2-2x+1-xy-y=\left(x-1\right)^2-y\left(x-1\right)=\left(x-1\right)\left(x-1-y\right)\)
\(2.\)
\(x^3-4x^2+4x-2x+2=x\left(x^2-4x+4\right)-2\left(x-1\right)=x\left(x-2\right)^2-2\left(x-1\right)\)
\(3.\)
\(10x-25-x^2+4y^2=4y^2-\left(x^2-10x+25\right)=4y^2-\left(x-5\right)^2=\left(2y+x-5\right)\left(2y-x+5\right)\)
\(4.\)
\(4x^2-2x+2xy-y=2x\left(2x-1\right)+y\left(2x-1\right)=\left(2x-1\right)\left(2x+y\right)\)
\(5.\)
\(4x\left(x-3\right)^2-3x^2+9x=4x\left(x-3\right)^2-3x\left(x-3\right)=\left(x-3\right)\left(4x^2-12x-3x\right)\)
a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)
\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)
\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)
c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)
\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)
làm ơn giúp em với
Bài 2:
a: \(\Leftrightarrow12x-8x^2+8x^2-12x-8+12x=4\)
\(\Leftrightarrow12x=12\)
hay x=1
b: \(\Leftrightarrow\left(x-4\right)\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)