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a) \(x^4+2x^3-4x-4=\left[\left(x^2\right)^2-4\right]+\left(2x^3-4x\right)\)
\(=\left(x^2+2\right)\left(x^2-2\right)+2x\left(x^2-2\right)\)
\(=\left(x^2+2+2x\right)\left(x^2-2\right)\)
a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)=x^2\left(x+1\right)^2-\left(x+2\right)^2\)
\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)=\left(x^2-2\right)\left(x^2+2x+2\right)\)
b) \(x^2+y^2-x^2y^2+xy-x-y=\left(x^2-x^2y^2\right)+\left(y^2-y\right)+\left(xy-x\right)\)
\(=x^2\left(1-y\right)\left(1+y\right)-y\left(1-y\right)-x\left(1-y\right)=\left(1-y\right)\left(x^2+x^2y-y-x\right)\)
\(=\left(1-y\right)\left[\left(x-1\right)x+y\left(x-1\right)\left(x+1\right)\right]=\left(1-y\right)\left(x-1\right)\left(x+xy+y\right)\)
c) Không phân tích được.
1, \(x^2-y^2-2x+2y=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x+y-2\right)\left(x-y\right)\)
2, \(x^2-25+y^2+2xy=\left(x+y\right)^2-5^2=\left(x+y-5\right)\left(x+y+5\right)\)
3, \(x^2y-x^3-9y+9x=x^2\left(y-x\right)-9\left(y-x\right)=\left(x-3\right)\left(x+3\right)\left(y-x\right)\)
4, \(x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
5, \(x^4+8x=x\left(x^3+8\right)=x\left(x+8\right)\left(x^2-8x+64\right)\)
\(1,\)
\(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
\(2,\)
\(x^2-25+y^2+2xy\)
\(=\left(x^2+2xy+y^2\right)-25\)
\(=\left(x+y\right)^2-5^2\)
\(=\left(x+y-5\right)\left(x+y+5\right)\)
\(3,\)
\(x^2y-x^3-9y+9x\)
\(=\left(x^2y-x^3\right)-\left(9y-9x\right)\)
\(=x^2\left(y-x\right)-9\left(y-x\right)\)
\(=\left(x^2-9\right)\left(y-x\right)\)
\(=\left(x-3\right)\left(x+3\right)\left(y-x\right)\)
\(4,\)
\(x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(5,\)
\(x^4-8x\)
\(=x\left(x^3-8\right)\)
\(=x\left(x-2\right)\left(x^2+2x+4\right)\)
a) \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
Vậy tập nghiệm \(S=\left\{-4;0;4\right\}\)
b) \(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x^2+10\right)\left(x-2\right)=0\)
Mà \(x^2+10>0\)nên \(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy tập nghiệm S = { 0;2}
a, x2+2xy+y2+2x+2y-15
<=> (x+y )2+2(x+y)+1-16
Đặt x+y =a
<=> a2+2a+1-42
<=> (a+1)2-42
<=> (a+5)(a-3) =>( x+y+5)(x+y-3)
b, x2-4xy+4y2-2x-4y-35
<=> (x-2y)2-2(x-2y)+1-36
Đặt (x-2y) =b
=> b2-2b+1-62
<=> (b-1)2-62
<=> (b-7)(b+5)=> (x-2y-7)(x-2y+5)
c,
a,A= x^2+2xy+y^2+2x+2y-15
= (x+y)^2+(x+y)-15
Đặt x+y=a, ta có:
A=a^2+2a-15
=a^2+2a+1-16
=(a+1)^2-4^2
=(a+1+4)(a+1-4)
=(a+5)(a-3)
Thay a=x+y, ta có: A=(x+y+5)(x+y-3).
a: 2x^2y-50xy=2xy(x-25)
b: 5x^2-10x=5x(x-2)
c: 5x^3-5x=5x(x^2-1)=5x(x-1)(x+1)
d: \(x^2-xy+x=x\left(x-y+1\right)\)
e: x(x-y)-2(y-x)
=x(x-y)+2(x-y)
=(x-y)(x+2)
f: 4x^2-4xy-8y^2
=4(x^2-xy-2y^2)
=4(x^2-2xy+xy-2y^2)
=4[x(x-2y)+y(x-2y)]
=4(x-2y)(x+y)
f1: x^2ỹ-y^2+y
=(x-y)(x+y)+(x+y)
=(x+y)(x-y+1)
Bạn sai ở dấu bằng thứ 4. Mình làm lại nhé.
\(\left(x+y\right)^4+x^4+y^4\)
\(=\left[\left(x+y\right)^2\right]^2+x^4+y^4\)
\(=\left(x^2+2xy+y^2\right)^2+x^4+y^4\)
\(=x^4+4x^2y^2+y^4+4x^3y+4xy^3+2x^2y^2+x^4+y^4\)
\(=2x^4+4x^3y+6x^2y^2+4xy^3+2y^4\)
\(=2\left(x^4+2x^3y+3x^2y^2+2xy^3+y^4\right)\)
\(=2.\left[\left(x^4+2x^3y+x^2y^2\right)+\left(2x^2y^2+2xy^3\right)+y^4\right]\)
\(=2.\left[\left(x^2+xy\right)^2+2.\left(x^2+xy\right).y^2+\left(y^2\right)^2\right]\)
\(=2.\left(x^2+xy+y^2\right)^2\)
Học tốt nhe.