bài 1: Phân tích đa thức sau thành nhân tử
a) (x2-2x)2+9x2+18x+20
b) a2-b2+4ab-9
c) 16x2(x-y)2-10y(y-x)3
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
11 tháng 11 2021
a) x2-xy+5y-25
= x(2-y)+ 5(y-2)
= x(2-y)-5(2-y)
= (x-5)(2-y)
H9
HT.Phong (9A5)
CTVHS
22 tháng 8 2023
a) \(4x^2-1\)
\(=\left(2x\right)^2-1^2\)
\(=\left(2x-1\right)\left(2x+1\right)\)
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) \(9x^2-\dfrac{1}{4}\)
\(=\left(3x\right)^2-\left(\dfrac{1}{2}\right)^2\)
\(=\left(3x-\dfrac{1}{2}\right)\left(3x+\dfrac{1}{2}\right)\)
d) \(\left(x-y\right)^2-4\)
\(=\left(x-y\right)^2-2^2\)
\(=\left(x-y-2\right)\left(x-y+2\right)\)
e) \(9-\left(x-y\right)^2\)
\(=3^2-\left(x-y\right)^2\)
\(=\left(3+x-y\right)\left(3-x+y\right)\)
f) \(\left(x^2+4\right)^2-16x^2\)
\(=\left(x^2+4\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+4\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)^2\left(x+2\right)^2\)
CM
17 tháng 11 2019
a) (x - 1)(x + l)(x - 2)(x - 4). b) (x - 2)( x 2 + 4).
c) 2y(3 x 2 + y 2 ). d) 2(x + y + z) ( a - b ) 2 .
24 tháng 8 2021
a. \(x^2\left(x-3\right)^2-\left(x-3\right)^2-x^2+1\)
\(=\left(x-3\right)^2\left(x^2-1\right)-\left(x^2-1\right)\)
\(=\left[\left(x-3\right)^2-1\right]\left(x^2-1\right)\)
\(=\left(x-3+1\right)\left(x-3-1\right)\left(x+1\right)\left(x-1\right)\)
\(=\left(x-2\right)\left(x-4\right)\left(x+1\right)\left(x-1\right)\)
b. \(x^3-2x^2+4x-8\)
\(=\left(x^3+4x\right)-\left(2x^2+8\right)\)
\(=x\left(x^2+4\right)-2\left(x^2+4\right)\)
\(=\left(x-2\right)\left(x^2+4\right)\)
c. \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x^3-3x^2y+3xy^2-y^3\right)\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)
\(=6x^2y+2y^3\)
\(=2y\left(3x^2+y^2\right)\)
d. \(2a^2\left(x+y+z\right)-4ab\left(x+y+z\right)+2b^2\left(x+y+z\right)\)
\(=\left(2a^2-4ab+2b^2\right)\left(x+y+z\right)\)
\(=2\left(a^2-2ab+b^2\right)\left(x+y+z\right)\)
\(=2\left(a-b\right)^2\left(x+y+z\right)\)
5 tháng 9 2021
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
5 tháng 9 2021
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
23 tháng 10 2021
Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
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: \(x\cdot\left(x+1\right)\left(x+2\right)\left(x+3\right)-3\)
\(=\left(x^2-3x\right)\left(x^2-3x+2\right)-3\)
\(=\left(x^2-3x\right)^2+2\left(x^2-3x\right)-3\)
\(=\left(x^2-3x+3\right)\left(x^2-3x-1\right)\)
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: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
KV
9 tháng 12 2023
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)
KV
9 tháng 12 2023
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