Phân tích đa thức thành nhân tử
a, x^2 +2022*x-x*y-2022*y
b, x^2-9y^2-4x+4
c, 4a^2-12ab+9b^2-25
d, x^3+8-(2+x).(5-2x)
Phân tích đa thức sau thành nhân tử
a, \(2^3\)+ 4\(^2\)+6x
b,x\(^2\)-4
c,x\(^2\)-10+25
d,x\(^3\)-4
e,x\(^2\)+xy-3x-3y
g,x\(^2\)-y\(^2\)-4x+4
Giup mk vs ạ bạn nào nhanh mk sẽ vote ạ
Đề bạn có mấy chỗ thiếu mk bổ sung nha
\(a,2^3+4^2+6x=8+16+6x=6x+24=x\left(x+4\right)\\ b,x^2-4=\left(x-2\right)\left(x+2\right)\\ c,x^2-10x+25=\left(x-5\right)^2\\ d,x^3-4x=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\\ e,x^2+xy-3x-3y=x\left(x+y\right)-3\left(x+y\right)=\left(x-3\right)\left(x+y\right)\\ g,x^2-y^2-4x+4=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)
Tick plzz
a: Ta có: \(2x^3+4x^2+6x\)
\(=2x\left(x^2+2x+3\right)\)
b: \(x^2-4=\left(x-2\right)\left(x+2\right)\)
c: \(x^2-10x+25=\left(x-5\right)^2\)
d: \(x^3-4x=x\left(x-2\right)\left(x+2\right)\)
e: \(x^2+xy-3x-3y\)
\(=x\left(x+y\right)-3\left(x+y\right)\)
\(=\left(x+y\right)\left(x-3\right)\)
g: \(x^2-4x+4-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-y-2\right)\left(x+y-2\right)\)
Bài 5. Phân tích các đa thức thành nhân tử
a) (x2-4x)2-8(x2-4x)+15 b) (x2+2x)2+9x2+18x+20
c) ( x+1)(x+2)(x+3)(x+4)-24 d) (x-y+5)2-2(x-y+5)+1
Bài 6. Phân tích các đa thức thành nhân tử
a) x2y+x2-y-1 b) (x2+x)2+4(x2+x)-12
c) (6x+5)2(3x+2)(x+1)-6
phân tích đa thức thành nhân tử
a)25x^2-4a^2+12ab-9b^2
b)x^3+x^2y-xy^2-y^3
\(a.25^2-4a^2+12ab-9b^2\\ =25^2-\left(4a^2+12ab-9b^2\right)\\ =25^2-\left(2a-3b\right)^2\\ =\left(25-2a+3b\right)\left(25+2a-3b\right)\\ b.x^3+x^2y-xy^2-y^3\\ =x^2\left(x+y\right)-y^2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-y^2\right)\\ =\left(x+y\right)\left(x+y\right)\left(x-y\right)\\ =\left(x+y\right)^2\left(x-y\right)\)
a: Ta có: \(25x^2-4a^2+12ab-9b^2\)
\(=25x^2-\left(2a-3b\right)^2\)
\(=\left(5x-2a+3b\right)\left(5x+2a-3b\right)\)
b: Ta có: \(x^3+x^2y-xy^2-y^3\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+xy\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)^2\)
bài 1:phân tích đa thức thành nhân tử
a,x4 +5x2 +9
b,x4 + 3x2 +4
c,2x4 - x2 -1
Bài 2:tìm x biết
a,(x+1) (x+2)(x+3)(x+4)= 120
b,(x-4x+3)(x2+6x +8) +24
Bài 1:
\(a,x^4+5x^2+9\\=(x^4+6x^2+9)-x^2\\=[(x^2)^2+2\cdot x^2\cdot3+3^2]-x^2\\=(x^2+3)^2-x^2\\=(x^2+3-x)(x^2+3+x)\)
\(b,x^4+3x^2+4\\=(x^4+4x^2+4)-x^2\\=[(x^2)^2+2\cdot x^2\cdot2+2^2]-x^2\\=(x^2+2)^2-x^2\\=(x^2+2-x)(x^2+2+x)\)
\(c,2x^4-x^2-1\\=2x^4-2x^2+x^2-1\\=2x^2(x^2-1)+(x^2-1)\\=(x^2-1)(2x^2+1)\\=(x-1)(x+1)(2x^2+1)\)
Bài 2:
\(a,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=120\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\cdot\left[\left(x+2\right)\left(x+3\right)\right]=120\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=120\) (1)
Đặt \(x^2+5x+5=y\), khi đó (1) trở thành:
\(\left(y-1\right)\left(y+1\right)=120\)
\(\Leftrightarrow y^2-1=120\)
\(\Leftrightarrow y^2=121\)
\(\Leftrightarrow\left[{}\begin{matrix}y=11\\y=-11\end{matrix}\right.\)
+, TH1: \(y=11\Leftrightarrow x^2+5x+5=11\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow x^2-x+6x-6=0\)
\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\left(\text{nhận}\right)\)
+, TH2: \(y=-11\Leftrightarrow x^2+5x+5=-11\)
\(\Leftrightarrow x^2+5x+16=0\)
\(\Leftrightarrow\left[x^2+2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]-\dfrac{25}{4}+16=0\)
\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)
Ta thấy: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}\ge\dfrac{39}{4}>0\forall x\)
Mà \(\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)
\(\Rightarrow\) loại
Vậy \(x\in\left\{1;-6\right\}\).
\(b,\) Đề thiếu vế phải rồi bạn.
phân tích đa thức thành nhân tử
a,x^2+6x+8 b,3x^2-2(x-y)^2-3y^2 c,4x^2-9y^2+4x-6y
d,x(x+1)^2+x(x-5)-5(x+1)^2 e,2xy-x^2+3y^2-4y+1 f,4x^16+81
g,x^8+x^4+1
a) x² + 6x + 8
= x² + 2x + 4x + 8
= (x² + 2x) + (4x + 8)
= x(x + 2) + 4(x + 8)
= (x + 2)(x + 4)
b) 3x² - 2(x - y)² - 3y²
= (3x² - 3y²) - 2(x - y)²
= 3(x² - y²) - 2(x - y)²
= 3(x + y)(x - y) - 2(x - y)²
= (x - y)[3(x + y) - 2(x - y)]
= (x - y)(3x + 3y - 2x + 2y)
= (x - y)(x + 5y)
c) 4x² - 9y² + 4x - 6y
= (4x² - 9y²) + (4x - 6y)
= (2x - 3y)(2x + 3y) + 2(2x - 3y)
= (2x - 3y)(2x + 3y + 2)
d) x(x + 1)² + x(x - 5) - 5(x + 1)²
= [x(x + 1)² - 5(x + 1)²] + x(x - 5)
= (x + 1)²(x - 5) + x(x - 5)
= (x - 5)[(x + 1)² + x]
= (x - 5)(x² + 2x + 1 + x)
= (x - 5)(x² + 3x + 1)
e) 2xy - x² + 3y² - 4y + 1
= -x² + 2xy - y² + 4y² - 4y + 1
= -(x² - 2xy + y²) + (4y² - 4y + 1)
= -(x - y)² + (2y - 1)²
= (2y - 1)² - (x - y)²
= (2y - 1 - x + y)(2y - 1 + x - y)
= (3y - x - 1)(x + y - 1)
f) 4x¹⁶ + 81
= (2x⁸)² + 2.2x⁸.9 + 9² - 2.2x⁸.9
= (2x⁸ + 9)² - 36x⁸
= (2x⁸ + 9) - (6x⁴)²
= (2x⁸ + 9 - 6x⁴)(2x⁸ + 9 + 6x⁴)
= (2x⁸ - 6x⁴ + 9)(2x⁸ + 6x⁴ + 9)
Phân tích đa thức thành nhân tử
a, 7x\(^5\)-14\(x^3\)y +21x\(^2\)y
b, 4x\(^2\)-20x+25
a. 7x5 - 14x3y + 21x2y
= 7x2(x3 - 2xy + 3y)
b. 4x2 - 20x + 25
= (2x)2 - 2x.2.5 + 52
= (2x - 5)2
a) \(7x^5-14x^3y+21x^2y\)
\(=7x^2\left(x^3-2xy+3y\right)\)
b) \(4x^2-20x+25\)
\(=\left(2x\right)^2-2.2x.5+5^2\)
\(\left(2x-5\right)^2\)
Phân tích các đa thức sau thành nhân tử
a) \(^{ }3xy-6xy^2\)
b) \(^{ }3x^3+6x^2+3x\)
c) \(^{ }x^3-x^2+2\)
d) \(^{ }x^2+4x+4-y^2\)
e) \(^{ }x^3+4x^2+4x\)
f) \(^{ }x^2+2x+1-9y^2\)
g) \(^{ }6x^2-12x\)
h) \(^{ }x^3+2x^2-x\)
i) \(^{ }x^2-2xy+y^2-9\)
j) \(^{ }x^2+x-6\)
k) \(^{ }2x^3+2x^2y-4xy^2\)
l) \(^{ }x^3-4x^2-12x+27\)
a) \(3xy-6xy^2=3xy\left(1-2y\right)\)
b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)
c) \(x^3-x^2+2\)
d) \(x^2+4x+4-y^2=\left(x^2+4x+4\right)-y^2=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\)
e) \(x^3+4x^2+4x=x\left(x^2+4x+4\right)=x\left(x+2\right)^2\)
f) \(x^2+2x+1-9y^2=\left(x+1\right)^2-\left(3y\right)^2=\left(x-3y+1\right)\left(x+3y+1\right)\)
g) \(6x^2-12x=6x\left(x-2\right)\)
h) \(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
i) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
k) \(2x^3+2x^2y-4xy^2=2x\left(x^2+xy-2y^2\right)\)
l) \(x^3-7x^2+9x+3x^2-21x+27=x\left(x^2-7x+9\right)+3\left(x^2-7x+9\right)=\left(x+3\right)\left(x^2-7x+9\right)\)
Phân tích đa thức thành nhân tử
a) 4x+4y
b) x^2-6xy+9y^2
c) x^3-x-x^2+1
a) \(4\left(x+y\right)\)
b) \(\left(x-3y\right)^2\)
c) \(x^3-x-x^2+1=x\left(x^2-1\right)-\left(x^2-1\right)=\left(x^2-1\right)\left(x-1\right)=\left(x-1\right)\left(x+1\right)\left(x-1\right)\)
a) \(4 (x + y)\)
b) \((x - 3y)^2\)
c) \(x^3 - x - x^2 + 1 = x (x^2 - 1) - (x^2 - 1) = (x^2 - 1) (x - 1) = (x - 1) (x + 1) (x - 1)\)
phân tích đa thức thành nhân tử
a) (x+y)2-8(x+y)+12
b) (x2+2x)2-2x2-4x-3
c) (x2+x)2-2(x2+x)-15
a/ \(\left(x+y\right)^2-8\left(x+y\right)+12\)
\(=\left(x+y\right)\left(x+y-8+12\right)\)
\(=\left(x+y\right)\left(x+y+4\right)\)
==========
b/\(\left(x^2+2x\right)^2-2x^2-4x-3\)
\(=\left(x^2+2x\right)^2-\left(2x^2+4x\right)-3\)
\(=\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3\)
\(=\left(x^2+2x\right)\left(x^2+2x-5\right)\)
===========
c/ \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(=\left(x^2+x\right)\left(x^2+x-2-15\right)\)
\(=\left(x^2+x\right)\left(x^2+x-17\right)\)
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