1) phân tích thành phân tử:
a) x^3y^3 + 1/125
b) (x+5)^3-(x-5)^3
c) (6-x)^3+(6+x)^3
d) 27x^3 -135x^2+225x-125
2) tìm x:
a)4x^2-25x^4=0
b) x^2-8x+16=0
c)x^3-3x^2+3x-1=0
Phân tích các đa thức sau đây thành nhân tử
a, 36x^2 - ( 3x -2 ) ^2
b, 16(4x+5)^5 - 25 (2x+2)^2
c, ( x - y + 4 )^2
d, (x+1)^4 - (x-1)^4
e, 16x^2 - 24xy + 9y^2
f, -x^4/4 + 2x^2y^3 - 4y^6
g , 64x^3 +1
h, x^3y^6z^9 - 125
k, 27x^6 - 8x^3
I , x^6 - y^6
m, 27x^3 - 54x^2y + 36xy^2 - 8y^3
n, y^9 - 9x^2y^6 + 27x^4y^3 - 27x^6
làm ơn giải chi tiết giúp mik vs ạ , cảm ơn
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)
Bài 1 : Phân tích đa thức thành nhân tử
a) 5x^2y-20xy^2
b) 1-8x+16x^2-y^2
c) 4x-4-x^2
d) x^3-2x^2+x-xy^2
e)27-3x^2
f) 2x^2+4x+2-2y^2
Bài 2: tìm x, biết
a) x^2(x-2023)-2023+x=0
b) -x(x-4)+(2x^3-4x^2-9x):x=0
c) x^2+2x-3x-6=0
d) 3x(x-10)-2x+20=0
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: \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x+1+y\right)\left(x+1-y\right)\)
Bài 2
a) x²(x - 2023) - 2023 + x = 0
x²(x - 2023) - (x - 2023) = 0
(x - 2023)(x² - 1) = 0
x - 2023 = 0 hoặc x² - 1 = 0
*) x - 2023 = 0
x = 2023
*) x² - 1 = 0
x² = 1
x = 1 hoặc x = -1
Vậy x = -1; x = 1; x = 2023
b) -x(x - 4) + (2x³ - 4x² - 9x) : x = 0
-x² + 4x + 2x² - 4x - 9 = 0
x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
Vậy x = 3; x = -3
c) 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 = -2; x = 3
d) 3x(x - 10) - 2x + 20 = 0
3x(x - 10) - (2x - 20) = 0
3x(x - 10) - 2(x - 10) = 0
(x - 10)(3x - 2) = 0
x - 10 = 0 hoặc 3x - 2 = 0
*) x - 10 = 0
x = 10
*) 3x - 2 = 0
3x = 2
x = 2/3
Vậy x = 2/3; x = 10
Tìm x:
a) (2x - 1) (x^2 - x + 1) = 2x^3 - 3x^2 + 2
b) (x + 1) (x^2 + 2x + 4) - x^3 - 3x^2 + 16 = 0
c) (x + 1) (x + 2) (x + 5) - x^3 - 8x^2 = 27
a) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Leftrightarrow6x=-20\)
hay \(x=-\dfrac{10}{3}\)
c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)
\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)
\(\Leftrightarrow17x=17\)
hay x=1
Tìm x
a) 3x(4x - 3) - 2x(5 - 6x) = 0
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
d) 3x (x + 1) - 5x(3 - x) + 6(x^2 + 2x + 3) = 0
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
b) 5(2x-3)+4x(x-2)+2x(3-2x)=0
\(\Leftrightarrow\)10x-15+4x2-8x+6x-4x2=0
\(\Leftrightarrow8x-15=0\)
\(\Leftrightarrow8x=15\)
\(\Leftrightarrow x=\dfrac{15}{8}\)
vậy x=\(\dfrac{15}{8}\)
c)3x(2-x)+2x(x-1)=5x(x+3)
\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)
\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)
Phân tích các đa thức thành nhân tử
1) -25x^6-y^8+10x^3y^4
2) 2x(3x-5)+10-6x
3) x^2-9-x^2(x^2-9)
4) 4x^2-9-(3x+1).(2x-3)
5) 8x^3-y^3-4x+2y
1) -25x6 - y8 + 10x3y4 = -( 25x6 - 10x3y4 + y8 ) = -[ ( 5x3 )2 - 2.5x3.y4 + ( y4 ) ] = -( 5x3 - y4 )2
2) 2x( 3x - 5 ) + 10 - 6x = 2x( 3x - 5 ) - 2( 3x - 5 ) = ( 3x - 5 )( 2x - 2 ) = 2( 3x - 5 )( x - 1 )
3) x2 - 9 - x2( x2 - 9 ) = ( x2 - 9 ) - x2( x2 - 9 ) = ( x2 - 9 )( 1 - x2 ) = ( x - 3 )( x + 3 )( 1 - x )( 1 + x )
4) 4x2 - 9 - ( 3x + 1 )( 2x - 3 ) = ( 2x - 3 )( 2x + 3 ) - ( 3x + 1 )( 2x - 3 )
= ( 2x - 3 )[ ( 2x + 3 ) - ( 3x + 1 ) ]
= ( 2x - 3 )( 2x + 3 - 3x - 1 )
= ( 2x - 3 )( 2 - x )
5) 8x3 - y3 - 4x + 2y = ( 8x3 - y3 ) - ( 4x - 2y )
= [ ( 2x )3 - y3 ) - 2( 2x - y )
= ( 2x - y )( 4x2 + 2xy + y2 ) - 2( 2x - y )
= ( 2x - y )( 4x2 + 2xy + y2 - 2 )
Phân tích đa thức thành nhân tử:
a) x^3 - x^2 + 8x - 8
b) 8x^3 - 8x^2y + 2xy^2
c) (x^2 + y^2 - z^2)^2 - 4x^2y^2
d) (x^2 - y^2 - 5)^2 - 4(x^2y^2 + 4xy + 4)
e) x^3 - y^3 - 3x^2 + 3x - 1
a) (x3-x2)+(8x-8)=x(x-1)+8(x-1)=(x2+8)(x-1)
b) 8x3-8x2y+2xy2=2x(4x2-4xy+y2)
c) (x2+y2-z2)2 - 4x2y2=(x2+y2-z2)2 - (2xy)2=(x2+y2-z2-2xy)(x2+y2-z2+2xy)
Tìm x biết:
a, 16x² – 9(x + 1)²= 0
b, x2 (x – 1) – 4x2 + 8x – 4 = 0
c, x(2x – 3) – 2(3 – 2x) = 0
d, (x – 3)(x² + 3x + 9) – x(x + 2)(x – 2) = 1
e, 4x² + 4x – 6 = 2
f, 2x² + 7x + 3 = 0
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Tìm x:
a)(x+2)^2-2(x+2)(x-5)=0
b)2x^2+3x-5=0
c)x+2√2x^2+2x^3=0
d)(3x-1)^2-4(x+5)^2=0
a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)
1 phân tích đa thức thành nhân tử
a,\(3x^2-6xy+3y^2\)
b,\(\left(x-y\right)^2-4x^2\)
2.tìm x biết
a,2x(x-3)-x+3=0
b,\(x^2+5x+6=0\)
`1)`
`a)3x^2-6xy+3y^2=3(x^2-2xy+y^2)=3(x-y)^2`
`b)(x-y)^2-4x^2=(x-y-2x)(x-y+2x)=(-x-y)(3x-y)`
`2)`
`a)2x(x-3)-x+3=0`
`<=>2x(x-3)-(x-3)=0`
`<=>(x-3)(2x-1)=0`
`<=>[(x=3),(x=1/2):}`
`b)x^2+5x+6=0`
`<=>x^2+2x+3x+6=0`
`<=>(x+2)(x+3)=0`
`<=>[(x=-2),(x=-3):}`