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a: =(6x)^2-(3x-2)^2
=(6x-3x+2)(6x+3x-2)
=(9x-2)(3x+2)
d: \(=\left[\left(x+1\right)^2-\left(x-1\right)^2\right]\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\)
\(=4x\cdot\left[x^2+2x+1+x^2-2x+1\right]\)
=8x(x^2+1)
e: =(4x)^2-2*4x*3y+(3y)^2
=(4x-3y)^2
f: \(=-\left(\dfrac{1}{4}x^4-2\cdot\dfrac{1}{2}x^2\cdot2y^3+4y^6\right)\)
\(=-\left(\dfrac{1}{2}x^2-2y^3\right)^2\)
g: =(4x)^3+1^3
=(4x+1)(16x^2-4x+1)
k: =x^3(27x^3-8)
=x^3(3x-2)(9x^2+6x+4)
l: =(x^3-y^3)(x^3+y^3)
=(x-y)(x+y)(x^2-xy+y^2)(x^2+xy+y^2)
a) \(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
b) \(x^2-3y^2\)
\(=x^2-\left(y\sqrt{3}\right)^2\)
\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)
\(=\left(5x-5y\right)\left(x+y\right)\)
\(=5\left(x-y\right)\left(x+y\right)\)
d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)
\(=\left[3\left(x-y\right)+2\left(x+y\right)\right]\left[3\left(x-y\right)-2\left(x+y\right)\right]\)
\(=\left(3x-3y+2x+2y\right)\left(3x-3y-2x-2y\right)\)
\(=\left(5x-y\right)\left(x-5y\right)\)
e) \(\left(4x^2-4x+1\right)-\left(x+1\right)^2\)
\(=\left(2x-1\right)^2-\left(x+1\right)\)
\(=\left(2x-1+x+1\right)\left(2x-1-x-1\right)\)
\(=3x\left(x-2\right)\)
f) \(x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
g) \(27x^3-0,001\)
\(=\left(3x\right)^3-\left(0,1\right)^3\)
\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)
h) \(125x^3-1\)
\(=\left(5x\right)^3-1^3\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
c: \(=\left(5x-y\right)\left(5x+y\right)\)
e: \(=\left(x-2\right)\left(x-3\right)\)
a) x(4y-10x)
b)3(x+2y)+(x+1)
c)(5x-y)(5x+y)
d)5x(y-z)2
e)(x-3)(x-2)
f)(2x+y)3
Ta có : x3 - 7x + 6
= x3 - x - 6x + 6
= x(x2 - 1) - 6(x - 1)
= x(x + 1)(x - 1) - 6(x - 1)
= (x - 1) [x(x + 1) - 6]
= (x - 1) (x2 + x - 6) .
CÁC Ý SAU TƯƠNG TỰ
1
x3-7x+6
=x3+0x2-7x +6
= x3-x2+x2-x-6x+6
=(x3-x2)+(x2-x)-(6x-6)
=x2(x-1)+x(x-1)-6(x-1)
=(x-1)(x2+x-6)
=(x-1)(x2+3x-2x-6)
=(x-1)[x(x+3)-2(x+3)]
=(x-1)(x-2)(x+3)
7) (x+2)(x+3)(x+4)(x+5)-24
=(x+2)(x+5) (x+3)(x+4)-24
=[x(x+5)+2(x+5)][x(x+4)+3(x+4)]-24
=[x2+5x+2x+10][x2+4x+3x+12]-24
=[x2+7x+10][x2+7x+12]-24
đặt a=x2+7x+10
=>x2+7x+12=a+2
=a(a+2)-24
=a2+2a-24
=a2+6a-4a-24
=(a2+6a)-(4a+24)
=a(a+6)-4(a+6)
=(a+6)(a-4)
thay a= x2+7x+10 vào ta được
(x2+7x+10+6)(x2+7x+10-4)
=(x2+7x+16)(x2+7x+6)
Bài 1
a) 5x²y - 20xy²
= 5xy(x - 4y)
b) 1 - 8x + 16x² - y²
= (1 - 8x + 16x²) - y²
= (1 - 4x)² - y²
= (1 - 4x - y)(1 - 4x + y)
c) 4x - 4 - x²
= -(x² - 4x + 4)
= -(x - 2)²
d) x³ - 2x² + x - xy²
= x(x² - 2x + 1 - y²)
= x[(x² - 2x+ 1) - y²]
= x[(x - 1)² - y²]
= x(x - 1 - y)(x - 1 + y)
= x(x - y - 1)(x + y - 1)
e) 27 - 3x²
= 3(9 - x²)
= 3(3 - x)(3 + x)
f) 2x² + 4x + 2 - 2y²
= 2(x² + 2x + 1 - y²)
= 2[(x² + 2x + 1) - y²]
= 2[(x + 1)² - y²]
= 2(x + 1 - y)(x + 1 + y)
= 2(x - y + 1)(x + y + 1)
Bài 2:
a: \(x^2\left(x-2023\right)+x-2023=0\)
=>\(\left(x-2023\right)\left(x^2+1\right)=0\)
mà \(x^2+1>=1>0\forall x\)
nên x-2023=0
=>x=2023
b:
ĐKXĐ: x<>0
\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)
=>\(-x\left(x-4\right)+2x^2-4x-9=0\)
=>\(-x^2+4x+2x^2-4x-9=0\)
=>\(x^2-9=0\)
=>(x-3)(x+3)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
c: \(x^2+2x-3x-6=0\)
=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)
=>\(x\left(x+2\right)-3\left(x+2\right)=0\)
=>(x+2)(x-3)=0
=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
d: 3x(x-10)-2x+20=0
=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)
=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)
=>\(\left(x-10\right)\left(3x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)
Câu 1:
a: \(5x^2y-20xy^2\)
\(=5xy\cdot x-5xy\cdot4y\)
\(=5xy\left(x-4y\right)\)
b: \(1-8x+16x^2-y^2\)
\(=\left(16x^2-8x+1\right)-y^2\)
\(=\left(4x-1\right)^2-y^2\)
\(=\left(4x-1-y\right)\left(4x-1+y\right)\)
c: \(4x-4-x^2\)
\(=-\left(x^2-4x+4\right)\)
\(=-\left(x-2\right)^2\)
d: \(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1-y\right)\left(x-1+y\right)\)
e: \(27-3x^2\)
\(=3\left(9-x^2\right)\)
\(=3\left(3-x\right)\left(3+x\right)\)
f: \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x+1+y\right)\left(x+1-y\right)\)
