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a) \(8x\left(x-3\right)+x-3=0\)
\(\Rightarrow8x\left(x-3\right)+\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{8}\end{matrix}\right.\)
b) \(x^2+36=12x\)
\(\Rightarrow x^2-12x+36=0\)
\(\Rightarrow\left(x-6\right)^2=0\)
\(\Rightarrow x=6\)
\(x^2-8x-9\)
\(=x^2-9x+x-9\)
\(=x\left(x-9\right)+\left(x-9\right)\)
\(=\left(x-9\right)\left(x+1\right)\)
Bài 3
a) x² + 10x + 25
= x² + 2.x.5 + 5²
= (x + 5)²
b) 8x - 16 - x²
= -(x² - 8x + 16)
= -(x² - 2.x.4 + 4²)
= -(x - 4)²
c) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
d) (x + y)² - 9x²
= (x + y)² - (3x)²
= (x + y - 3x)(x + y + 3x)
= (y - 2x)(4x + y)
e) (x + 5)² - (2x - 1)²
= (x + 5 - 2x + 1)(x + 5 + 2x - 1)
= (6 - x)(3x + 4)
Bài 4
a) x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
b) (x - 4)² - 36 = 0
(x - 4 - 6)(x - 4 + 6) = 0
(x - 10)(x + 2) = 0
x - 10 = 0 hoặc x + 2 = 0
*) x - 10 = 0
x = 10
*) x + 2 = 0
x = -2
Vậy x = -2; x = 10
c) x² - 10x = -25
x² - 10x + 25 = 0
(x - 5)² = 0
x - 5 = 0
x = 5
d) x² + 5x + 6 = 0
x² + 2x + 3x + 6 = 0
(x² + 2x) + (3x + 6) = 0
x(x + 2) + 3(x + 2) = 0
(x + 2)(x + 3) = 0
x + 2 = 0 hoặc x + 3 = 0
*) x + 2 = 0
x = -2
*) x + 3 = 0
x = -3
Vậy x = -3; x = -2
= x^2(X-1) - 4(x^2-2x+1)
=x^2(x-1)-4(x-1)^2
=(x-1)(x^2-4x+4)
=(x-1)(x-2)^2
a) Ta có: \(8x+4x^2-12xy\)
\(=4x\left(2+x-3y\right)\)
b) Ta có: \(5x^3-10x^2+5x\)
\(=5x\left(x^2-2x+1\right)\)
\(=5x\left(x-1\right)^2\)
c) Ta có: \(x^3+x^2y-xy^2-y^3\)
\(=x^2\left(x+y\right)-y^2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)^2\)
d) Ta có: \(x^2-8x-9\)
\(=x^2-9x+x-9\)
\(=\left(x-9\right)\left(x+1\right)\)
a. `8x+4x^2-12xy=4x(2+x-3y)`
b) `5x^3-10x^2+5x=5x(x^2-2x+1)`
c) `x^3+x^2y-xy^2-y^3=x^2(x+y)-y^2(x+y)=(x+y)(x^2-y^2)=(x+y)^2 (x-y)`
d) `x^2-8x-9=(x^2-2.x.4+4^2)-25=(x-4)^2-5^2=(x+1)(x-9)`
( x2 + 8x + 7 ) ( x2 + 8x + 15 ) + 15
Đặt x2 + 8x + 7 = y ta có:
y ( y + 8 ) + 15
= y2 + 8y + 15
= ( y + 3 ) ( y + 5 )
= ( x2 + 8x + 10 ) ( x2 + 8x + 12 )
= ( x2 + 8x + 10 ) ( x + 2 ) ( x + 6 )
Đặt x2 + 8x + 7 = y ta có:
y ( y + 8 ) + 15
= y2 + 8y + 15
= ( y + 3 ) ( y + 5 )
= ( x2 + 8x + 10 ) ( x2 + 8x + 12 )
= ( x2 + 8x + 10 ) ( x + 2 ) ( x + 6 )
a: 2x+4=2(x+2)
b: \(x^2+2xy+y^2-9=\left(x+y-3\right)\left(x+y+3\right)\)
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
\(8x\left(x^2-9\right)=0\Rightarrow8x\left(x-3\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x-3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)