X^3-6x^2y+9xy^2
X^3-6x^2y+9xy^2-16x
\(x^3-6x^2y+9xy^2-16x=x\left(x^2-6xy+9y^2-16\right)\)
\(=x\left[\left(x-3y\right)^2-16\right]=x\left(x-3y-4\right)\left(x-3y+4\right)\)
a) x^3+6x^2y+9xy^2-36x b) x^2-xy-x+y c) x^3-4x^2y+4xy^2-36x
tìm đa thức M
a M + (5x^2 - 2xy) = 6x^2 + 9xy - y^2
b (25 x^2y-13xy^2 + y^3)-M= 11 xy^2 - 2y^3
\(a,M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\\ \Rightarrow M=6x^2+9xy-y^2-5x^2+2xy\\ \Rightarrow M=x^2+11xy-y^2\\ b,\left(25x^2y-13xy^2+y^3\right)-M=11xy^2-2y^3\\ \Rightarrow M=25x^2y-13xy^2+y^3-11xy^2+2y^3\\ \Rightarrow M=25x^2y-24xy^2+3y^3\)
A) 7xy^2.(9xy^4+2xy+1) B)1/2xyz(6x^2y^3+4x^3y-3xy) 1)5x^2y+10xy^3 2)15xy^4-9xy 3)5x^2+10xy+5y^2
Tim gia tri x nho nhat
(9xy^2-6x^2y):(-3xy)+(6x^2y+2x^4):(2x^2)
phân tích đa thức thành nhân tử
1/ \(6x^2y-9xy^2+3xy\)
2/ \(\left(4-x\right)^2-16\)
3/ \(x^3+9x^2-4x-36\)
1: \(6x^2y-9xy^2+3xy\)
\(=3xy\left(2x-3y+1\right)\)
2: \(\left(4-x\right)^2-16\)
\(=\left(4-x-4\right)\left(4-x+4\right)\)
\(=-x\cdot\left(8-x\right)\)
3: \(x^3+9x^2-4x-36\)
\(=x^2\left(x+9\right)-4\left(x+9\right)\)
\(=\left(x+9\right)\left(x-2\right)\left(x+2\right)\)
1) \(6x^2y-9xy^2+3xy=3xy\left(2x-3y+1\right)\)
2) \(\left(4-x\right)^2-16=\left(4-x\right)^2-4^2=\left(4-x-4\right)\left(4-x+4\right)=-x\left(8-x\right)\)
3) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)
\(\begin{cases}x^3-6x^2y+9xy^2-4y^3=0\\\sqrt{x-y}+\sqrt{x+y}=2\end{cases}\)
\(\begin{cases}x^3-6x^2y+9xy^2-4y^3=0\left(1\right)\\\sqrt{x-y}+\sqrt{x+y}=2\left(2\right)\end{cases}\)
\(\left(1\right)\Leftrightarrow x^3-2x^2y+xy^2-4y^3+8xy^2-4x^2y=0\)
\(\Leftrightarrow x\left(x^2-2xy+y^2\right)-4y\left(x^2-2xy+y^2\right)=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)\left(x-4y\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x-4y\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x-y\right)^2=0\\x-4y=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=y\\x=4y\end{array}\right.\)
Xét \(x=y\) thay vào (2) ta có:\(\left(2\right)\Leftrightarrow\sqrt{x-x}+\sqrt{x+x}=2\)
\(\Leftrightarrow\sqrt{2x}=2\Leftrightarrow2x=4\Leftrightarrow x=2\).Mà \(\begin{cases}x=2\\x=y\end{cases}\)\(\Rightarrow x=y=2\)
Xét \(x=4y\) thay vào (2) ta có:\(\left(2\right)\Leftrightarrow\sqrt{4y-y}+\sqrt{4y+y}=2\)
\(\Leftrightarrow\sqrt{3y}+\sqrt{5y}=2\)\(\Leftrightarrow\sqrt{y}\left(\sqrt{5}+\sqrt{3}\right)=2\)
\(\Leftrightarrow y\left(\sqrt{15}+4\right)=2\)\(\Leftrightarrow y=\frac{2}{\sqrt{15}+4}\).Mà \(\begin{cases}x=4y\\y=\frac{2}{\sqrt{15}+4}\end{cases}\)\(\Leftrightarrow x=\frac{8}{\sqrt{15}+4}\)
1)x^3 -2x-1
2) x^3 +3x^2 -4
3) x^3y^3 +x^2y^2 +4
4) x^3 + 5x^2 -9xy^2, +5y^3
5) x^4 + x^3 + 6x^2 + 5x + 5
`@` `\text {Ans}`
`\downarrow`
`1)`
`x^3 - 2x - 1`
`= x^3 - x^2 - x + x^2 - x - 1`
`= (x^3 - x^2 - x) + (x^2 - x - 1)`
`= x(x^2 - x - 1) + (x^2 - x - 1)`
`= (x+1)(x^2 - x - 1)`
`2)`
`x^3 +3x^2 -4`
`= x^3 + 4x^2 + 4x - x^2 - 4x - 4`
`= (x^3 + 4x^2 + 4x) - (x^2 + 4x + 4)`
`= x(x^2 + 4x + 4) - (x^2 + 4x + 4)`
`= (x-1)(x^2 + 4x + 4)`
`3)`
`x^3y^3 + x^2y^2 + 4`
`= x^3 y^3 - x^2y^2 + 2x^2y^2 + 2xy - 2xy + 4`
`= (x^3y^3 - x^2y^2 + 2xy) + (2x^2y^2 - 2xy + 4)`
`= xy(x^2y^2 - xy + 2) + 2(x^2y^2 - xy + 2)`
`= (xy+2)(x^2y^2-xy+2)`
`4)`
`x^3 + 5x^2 -9xy^2+5y^3`
Bạn xem lại đề ;-;
`5)`
` x^4 + x^3 + 6x^2 + 5x + 5`
`= x^4 + x^3 + 5x^2 + x^2 + 5x + 5`
`= (x^4 + x^3 + x^2) + (5x^2 + 5x + 5)`
`= x^2(x^2 + x + 1) + 5(x^2 + x + 1)`
`= (x^2 + 5)(x^2 + x + 1)`
Phân tích các đa thức sau thành nhân tử :
1)6x^2y+9xy^2-12xy
2) x^3-2x^2+x
3) x^2-6x+9-y^2