Phân tích đa thức a^6-a^4+2a^3+2a^2 thành nhân tử

\(a^6-a^4+2a^3+2a^2\)

\(\Rightarrow a^2.\left(a^4-a^2+2a+2\right)\)

\(\Rightarrow\) \(a^2.\left(a^2.\left(a^2-1\right)+2\left(a+1\right)\right)\)

\(\Rightarrow a^2.\left(a^2.\left(a-1\right).\left(a+1\right)+2\left(a+1\right)\right)\)

\(\Rightarrow a^2.\left(a+1\right)\left(a^2.\left(a-1\right)+2\right)\)

\(\Rightarrow a^2.\left(a+1\right).\left(a^3-a^2+2\right)\)

\(\Rightarrow a^2.\left(a+1\right).\left(a^3+a^2-2a^2+2\right)\)

\(\Rightarrow a^2.\left(a+1\right).\left(a^2.\left(a+1\right)-2\left(a^2-1\right)\right)\)

\(\Rightarrow a^2.\left(a+1\right).\left(a^2.\left(a+1\right)-2\left(a-1\right).\left(a+1\right)\right)\)

\(\Rightarrow a^2.\left(a+1\right).\left(a+1\right).\left(a^2-2\left(a-1\right)\right)\)

\(\Rightarrow a^2.\left(a+1\right).\left(a+1\right).\left(a^2-2a+2\right)\)

\(\Rightarrow a^2.\left(a+1\right)^2.\left(a^2-2a+2\right)\)

a) x⁵ + x³ - x² - 1 = ( x⁵ + x³ ) - ( x² + 1)

= x³ . ( x² + 1 ) - ( x² + 1 )

= ( x² + 1 ) . ( x³ - 1 )

= ( x² + 1 ) . ( x - 1 ) . ( x² + x + 1 )

b) x² - x - 12 = x² - 4x + 3x - 12

= x . ( x - 4 ) + 3 . ( x - 4 )

= ( x - 4 ) . ( x + 3 )

c) 4x⁴ + 4x²y² - 8y⁴ = 4x⁴ - 4x²y² + 8x²y² - 8y⁴

= 4x² . ( x² - y² ) + 8y² . ( x² - y² )

= ( x² - y² ) . ( 4x² + 8y² )

= 4 . ( x - y ) . ( x + y ) . ( x² + 2y² )

Phân tích đa thức thành nhân tử

a,x³+x²y+y²z-xyz+y³

b,bc(b+c)+ca(c-a)-ab(a+b)

c,a²(b-c)+b²(c-a)+c²(a-b)

c) \(a^2\left(b-c\right)+b^2c-ab^2+ac^2-bc^2\)

\(=a^2\left(b-c\right)+b^2c-bc^2-ab^2+ac^2\)

\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)\)

\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b+c\right)\left(b-c\right)\)

\(=\left(b-c\right)\left(a^2+bc-ab-ac\right)\)

\(=\left(b-c\right)\left[a\left(a-c\right)-b\left(a-c\right)\right]\)

\(=\left(b-c\right)\left(a-c\right)\left(a-b\right)\)

phân tích đa thức thanh nhân tử 1/4x^2 - 5xy + 25y^2

Phân tích đa thức thành nhân tử

\(\dfrac{1}{4}x^2-5xy+25y^2\)

\(=\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.5y+\left(5y\right)^2\)

\(=\left(\dfrac{1}{2}x+5y\right)^2\)

phan tich da thuc thanh nhan tu -25x^6-y^8+10x^3y^4

\(-25x^6-y^8+10x^3y^4\\ =-\left(25x^6-10x^3y^4+y^8\right)\\ =-\left[\left(5x^3\right)^2-2\cdot5x^3\cdot y^4+\left(y^4\right)^2\right]\\ \\ =\left(5x^3-y^4\right)^2\)

phan tich da thuc thanh nhan tu xy+xz+3+3y

phân tích đa thức thành nhân tử:

x³+4x²+x-6

Phân tích đa thức thành nhân tử

2x² - 7xy + 3y² + 5xz + 2z²- 5yz

\(2x^2-7xy+3y^2+5xz+2z^2-5yz\)

\(=\left(2x^2+4xz-6xy\right)-\left(xy+2yz-3y^2\right)+\left(2z^2+xz-3yz\right)\)

\(=2x\left(x-3y+2z\right)-y\left(x-3y+2z\right)+z\left(x-3y+2z\right)\)

\(=\left(2x-y+z\right)\left(x-3y+2z\right)\)

\(x^3+4x^2+x-6\Leftrightarrow x^3+5x^2+6x-x^2-5x-6\)

\(\Leftrightarrow x\left(x^2+5x+6\right)-\left(x^2+5x+6\right)\Leftrightarrow\left(x-1\right)\left(x^2+5x+6\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+2x+3x+6\right)\Leftrightarrow\left(x-1\right)\left(x\left(x+2\right)+3\left(x+2\right)\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+3\right)\)

Phân tích đa thức thành nhân tử:

a, x⁵ - x⁴ - x³ + x²

b, x⁴ - 3x³ - x + 3

Phân tích đa thức sau thành nhân tử :

3(x² -x +1)^{2 -} 2x² + 2x - 3

phân tích đa thức thành nhân tử

27x³-27x²+18x-4

2x³-x²+5x+3

(x²-3)²+16

\(3\left(x^2-x+1\right)^2-2x^2+2x-3\)

\(=3\left(x^2-x+1\right)^2-3-2x\left(x-1\right)\)

\(=3\left[\left(x^2-x+1-1\right)\left(x^2-x+1+1\right)\right]-2x\left(x-1\right)\)

\(=3\left(x^2-x\right)\left(x^2-x+2\right)-2x\left(x-1\right)\)

\(=3x\left(x-1\right)\left(x^2-x+2\right)-2x\left(x-1\right)\)

\(=\left(x-1\right)\left(3x^3-3x^2+6x\right)-2x\left(x-1\right)\)

\(=\left(x-1\right)\left(3x^3-3x^2+6x-2x\right)\)

\(=\left(x-1\right)\left(3x^3-3x^2+4x\right)\)

\(a,x^5-x^4-x^3+x^2\\ =x^2\left(x^3-x^2-x+1\right)\\ =x^2\left[x^2\left(x-1\right)-\left(x-1\right)\right]\\ =x^2\left(x-1\right)\left(x^2-1\right)\\ =x^2\left(x-1\right)^2\left(x+1\right)\\ b,x^4-3x^3-x+3\\ =x^3\left(x-3\right)-\left(x-3\right)\\ =\left(x-3\right)\left(x^3-1\right)\\ =\left(x-3\right)\left(x-1\right)\left(x^2+x+1\right)\)

Các bạn trên làm ko triệt để gì cả

a)\(27x^3-27x^2+18x-4\)

\(=27x^3-18x^2+12x-9x^2+6x-4\)

\(=3x\left(9x^2-6x+4\right)-\left(9x^2-6x+4\right)\)

\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)

b)\(2x^3-x^2+5x+3\)

\(=2x^3-2x^2+6x+x^2-x+3\)

\(=2x\left(x^2-x+3\right)+\left(x^2-x+3\right)\)

\(=\left(2x+1\right)\left(x^2-x+3\right)\)

c)\(\left(x^2-3\right)^2+16\)

\(=x^4-6x^2+25\)

\(=\left(x^4+10x^2+25\right)-16x^2\)

\(=\left(x^2+5\right)^2-\left(4x\right)^2\)

\(=\left(x^2+5-4x\right)\left(x^2+5+4x\right)\)

Phân tích thành nhân tử

A) x^2 +5x-6

B ) 5x^2+5xy-x-y

C) 7x-6x^2-2

D) x^2+4x+3

E) 2x+3x-5

F) 16x-5x^3

Tìm x

A) 5x(x-1)=x-1

B) 2(x+5)-x^2-5x=0

Cho a+b+c=0.Chứng minh a^3+b^3+c^3 = 3abc

2.

a. \(5x\left(x-1\right)=x-1\Leftrightarrow5x\left(x-1\right)-x+1=0\Leftrightarrow5\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

Vậy \(x=1\) hoặc \(x=\dfrac{1}{5}\)

b. \(2\left(x+5\right)-x^{^2}-5x=0\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Vậy x=-5 hoặc x=2

c. thay \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\) ta có:

\(a^{^3}+b^{^3}+c^{^3}=3abc^{^3}\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\left[\left(a+b\right)^3+c^{^3}\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^{^2}-c\left(a+b\right)+c^{^2}\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^{^2}+2ab+b^{^2}-ac-bc+c^{^2}-3ab\right)...\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^{^2}+b^{^2}+c^{^2}-ab-ac-bc\right)=0\) luôn đúng vì a+b+c=0

\(\Rightarrow a^{^3}+b^{^3}+c^{^3}=3abc\left(đpcm\right)\)

Phân tích thành nhân tử:

(x²+x) ²+3(x²+x) +1

Help me, mình cần gấp

\(\left(x^2+x\right)^2+3\left(x^2+x\right)+1\)

\(=\left(x^2+x\right)^2+2.\left(x^2+x\right).1,5+\left(1,5\right)^2-1,25\)

\(=\left(x^2+x+1,5\right)^2-\sqrt{1,25}^2\)

\(=\left(x^2+x+1,5-\sqrt{1,25}\right)\left(x^2+x+1,5+\sqrt{1,25}\right)\)

Chúc bạn học tốt!!!