Bài 2 : Phân tích đa thức sau thành nhân tử
a) 5x^2 + 30y
b) x^3 - 2x^2 - 4xy^2 + x
Bài 3 : Tìm x , biết
a) 2x(x - 3 ) - x + 3 = 0
b) ( 3x - 1 ) ( 2x + 1 ) - (x + 1)^2 = 5x^2
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KV
29 tháng 12 2023
Bài 3
a) 2x(x - 3) - x + 3 = 0
2x(x - 3) - (x - 3) = 0
(x - 3)(2x - 1) = 0
x - 3 = 0 hoặc 2x - 1 = 0
*) x - 3 = 0
x = 3
*) 2x - 1 = 0
2x = 1
x = 1/2
Vậy x = 1/2; x = 3
b) (3x - 1)(2x + 1) - (x + 1)² = 5x²
6x² + 3x - 2x - 1 - x² - 2x - 1 - 5x² = 0
(6x² - x² - 5x²) + (3x - 2x - 2x) = 0 + 1 + 1
-x = 2
x = -2
KV
29 tháng 12 2023
Bài 2
a) 5x² + 30y
= 5(x² + 6y)
b) x³ - 2x² - 4xy² + x
= x(x² - 2x - 4y² + 1)
= x[(x² - 2x + 1) - 4y²]
= x[(x - 1)² - (2y)²]
= x(x - 1 - 2y)(x - 1 + 2y)
MH
1 tháng 11 2021
1.
a) \(2x^4-4x^3+2x^2\)
\(=2x^2\left(x^2-2x+1\right)\)
\(=2x^2\left(x-1\right)^2\)
b) \(2x^2-2xy+5x-5y\)
\(=\left(2x^2-2xy\right)+\left(5x-5y\right)\)
\(=2x\left(x-y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left(2x+5\right)\)
KJ
1 tháng 11 2021
2 .
a,
\(4x\left(x-3\right)-x+3=0\)
⇒\(4x\left(x-3\right)-\left(x-3\right)=0\)
⇒\(\left(x-3\right)\left(4x-1\right)=0\)
⇒\(\left[{}\begin{matrix}x-3=0\\4x-1=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\4x=1\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)
vậy \(x\in\left\{3;\dfrac{1}{4}\right\}\)
b,
\(\)\(\left(2x-3\right)^2-\left(x+1\right)^2=0\)
⇒\(\left(2x-3-x-1\right)\left(2x-3+x+1\right)\) = 0
⇒\(\left(x-4\right)\left(3x-2\right)=0\)
⇔\(\left[{}\begin{matrix}x-4=0\\3x-2=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)
vậy \(x\in\left\{4;\dfrac{2}{3}\right\}\)
18 tháng 10 2021
Bài 3:
b: \(x^2+2x+1=\left(x+1\right)^2\)
c: \(x^2-16=\left(x-4\right)\left(x+4\right)\)
d: \(\left(2x-1\right)^2-\left(x+3\right)^2\)
\(=\left(2x-1-x-3\right)\left(2x-1+x+3\right)\)
\(=\left(x-4\right)\left(3x+2\right)\)
12 tháng 10 2021
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
22 tháng 10 2023
2:
a: \(x^2-12x+20\)
\(=x^2-2x-10x+20\)
=x(x-2)-10(x-2)
=(x-2)(x-10)
b: \(2x^2-x-15\)
=2x^2-6x+5x-15
=2x(x-3)+5(x-3)
=(x-3)(2x+5)
c: \(x^3-x^2+x-1\)
=x^2(x-1)+(x-1)
=(x-1)(x^2+1)
d: \(2x^3-5x-6\)
\(=2x^3-4x^2+4x^2-8x+3x-6\)
\(=2x^2\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(2x^2+4x+3\right)\)
e: \(4y^4+1\)
\(=4y^4+4y^2+1-4y^2\)
\(=\left(2y^2+1\right)^2-\left(2y\right)^2\)
\(=\left(2y^2+1-2y\right)\left(2y^2+1+2y\right)\)
f; \(x^7+x^5+x^3\)
\(=x^3\left(x^4+x^2+1\right)\)
\(=x^3\left(x^4+2x^2+1-x^2\right)\)
\(=x^3\left[\left(x^2+1\right)^2-x^2\right]\)
\(=x^3\left(x^2-x+1\right)\left(x^2+x+1\right)\)
g: \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-3\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)\left(x^2+x-2\right)-3\left(x^2+x-2\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x-3\right)\)
\(=\left(x^2+x-3\right)\left(x+2\right)\left(x-1\right)\)
h: \(\left(x^2+2x\right)^2-2\left(x+1\right)^2-1\)
\(=\left(x^2+2x+1-1\right)^2-2\left(x+1\right)^2-1\)
\(=\left[\left(x+1\right)^2-1\right]^2-2\left(x+1\right)^2-1\)
\(=\left(x+1\right)^4-2\left(x+1\right)^2+1-2\left(x+1\right)^2-1\)
\(=\left(x+1\right)^4-4\left(x+1\right)^2\)
\(=\left(x+1\right)^2\left[\left(x+1\right)^2-4\right]\)
\(=\left(x+1\right)^2\left(x+1+2\right)\left(x+1-2\right)\)
\(=\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)\)
i: \(x^2+4xy+4y^2-4\left(x+2y\right)+3\)
\(=\left(x+2y\right)^2-4\left(x+2y\right)+3\)
\(=\left(x+2y\right)^2-\left(x+2y\right)-3\left(x+2y\right)+3\)
\(=\left(x+2y\right)\left(x+2y-1\right)-3\left(x+2y-1\right)\)
\(=\left(x+2y-1\right)\left(x+2y-3\right)\)
j: \(x\cdot\left(x+1\right)\left(x+2\right)\left(x+3\right)-3\)
\(=\left(x^2-3x\right)\left(x^2-3x+2\right)-3\)
\(=\left(x^2-3x\right)^2+2\left(x^2-3x\right)-3\)
\(=\left(x^2-3x+3\right)\left(x^2-3x-1\right)\)
2
10 tháng 1 2023
Bài `1:`
`a)3x^3+6x^2=3x^2(x+2)`
`b)x^2-y^2-2x+2y=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)`
Bài `2:`
`a)(2x-1)^2-25=0`
`<=>(2x-1-5)(2x-1+5)=0`
`<=>(2x-6)(2x+4)=0`
`<=>[(x=3),(x=-2):}`
`b)Q.(x^2+3x+1)=x^3+2x^2-2x-1`
`<=>Q=[x^3+2x^2-2x-1]/[x^2+3x+1]`
`<=>Q=[x^3-x^2+3x^2-3x+x-1]/[x^2+3x+1]`
`<=>Q=[(x-1)(x^2+3x+1)]/[x^2+3x+1]=x-1`
11 tháng 8 2021
Bài 1:
a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left(3-2x+5\right)\)
\(=\left(2x+1\right)\left(8-2x\right)\)
\(=2\left(4-x\right)\left(2x+1\right)\)
b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\left(3x-2\right)\left(3x-6\right)\)
\(=3\left(3x-2\right)\left(x-2\right)\)
11 tháng 8 2021
Bài 2:
a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)
\(=\left(a-b\right)\left(2a-4b\right)\)
\(=2\left(a-b\right)\left(a-2b\right)\)
f: Ta có: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
Bài 2
a) 5x² + 30y
= 5(x² + 6y)
b) x³ - 2x² - 4xy² + x
= x(x² - 2x - 4y² + 1)
= x[(x² - 2x + 1) - 4y²]
= x[(x - 1)² - (2y)²]
= x(x - 1 - 2y)(x - 1 + 2y)
Bài 3:
a: \(2x\left(x-3\right)-x+3=0\)
=>\(2x\left(x-3\right)-\left(x-3\right)=0\)
=>(x-3)(2x-1)=0
=>\(\left[{}\begin{matrix}x-3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
b: \(\left(3x-1\right)\left(2x+1\right)-\left(x+1\right)^2=5x^2\)
=>\(6x^2+3x-2x-1-x^2-2x-1=5x^2\)
=>\(5x^2-x-2=5x^2\)
=>-x-2=0
=>-x=2
=>x=-2