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a: \(=\dfrac{27}{10}\cdot\dfrac{5}{9}\cdot x^4y^2\cdot xy=\dfrac{3}{2}x^3y^3\)
bậc là 6
b: \(=\dfrac{1}{3}x^3y\cdot x^2y^2=\dfrac{1}{3}x^5y^3\)
Bậc là 8
c: \(=-2x^2y\cdot\dfrac{1}{4}x\cdot y^6z^3=-\dfrac{1}{2}x^3y^7z^3\)
Bậc là 13
Câu 1 :
a, \(4x^4y^2.9x^2y^4z^2=36x^6y^6z^2\)
b, bậc 14 ; hệ số 36
biến x^6y^6z^2
a) (x-y)(x4+x3y+x2y2+xy3+y4) = x(x4+x3y+x2y2+xy3+y4)-y(x4+x3y+x2y2+xy3+y4) =(x5+x4y+x3y2+x2y2+xy4)-(x4y+x3y2+x2y2+xy4+y5) = x5+x4y+x3y2+x2y2+xy4-x4y-x3y2-x2y2-xy4-y5 =x5-y5⇒Điều cần chứng minh
Các câu b d tương tự
`@` `\text {Ans}`
`\downarrow`
`a)`
\(P(x) = 5x^3 + 3 - 3x^2 + x^4 - 2x - 2 + 2x^2 + x\)
`= x^4 + 5x^3 + (-3x^2 + 2x^2) + (-2x+x) + (3-2)`
`= x^4 + 5x^3 - x^2 - x + 1`
\(Q(x) = 2x^4 + x^2 + 2x + 2 - 3x^2 - 5x + 2x^3 - x^4\)
`= (2x^4 - x^4) + 2x^3 + (x^2 - 3x^2) + (2x-5x) + 2`
`= x^4 + 2x^3 - 2x^2 - 3x +2`
`b)`
`P(x)+Q(x) = (x^4 + 5x^3 - x^2 - x + 1) + (x^4 + 2x^3 - 2x^2 - 3x +2)`
`= x^4 + 5x^3 - x^2 - x + 1 + x^4 + 2x^3 - 2x^2 - 3x +2`
`= (x^4+x^4)+(5x^3 + 2x^3) + (-x^2 - 2x^2) + (-x-3x) + (1+2)`
`= 2x^4 + 7x^3 - 3x^2 - 4x + 3`
`P(x)-Q(x)=(x^4 + 5x^3 - x^2 - x + 1) - (x^4 + 2x^3 - 2x^2 - 3x +2)`
`= x^4 + 5x^3 - x^2 - x + 1 - x^4 - 2x^3 + 2x^2 + 3x -2`
`= (x^4 - x^4) + (5x^3 - 2x^3) + (-x^2+2x^2)+(-x+3x)+(1-2)`
`= 3x^3 + x^2 + 2x - 1`
`Q(x)-P(x) = (x^4 + 2x^3 - 2x^2 - 3x +2)-(x^4 + 5x^3 - x^2 - x + 1)`
`= x^4 + 2x^3 - 2x^2 - 3x +2-x^4 - 5x^3 + x^2 + x - 1`
`= (x^4-x^4)+(2x^3 - 5x^3)+(-2x^2+x^2)+(-3x+x)+(2-1)`
`= -3x^3 - x^2 - 2x + 1`
`@` `\text {Kaizuu lv u.}`
b: 4x^2-20x+25=(x-3)^2
=>(2x-5)^2=(x-3)^2
=>(2x-5)^2-(x-3)^2=0
=>(2x-5-x+3)(2x-5+x-3)=0
=>(3x-8)(x-2)=0
=>x=8/3 hoặc x=2
c: x+x^2-x^3-x^4=0
=>x(x+1)-x^3(x+1)=0
=>(x+1)(x-x^3)=0
=>(x^3-x)(x+1)=0
=>x(x-1)(x+1)^2=0
=>\(x\in\left\{0;1;-1\right\}\)
d: 2x^3+3x^2+2x+3=0
=>x^2(2x+3)+(2x+3)=0
=>(2x+3)(x^2+1)=0
=>2x+3=0
=>x=-3/2
a: =>x^2(5x-7)-3(5x-7)=0
=>(5x-7)(x^2-3)=0
=>\(x\in\left\{\dfrac{7}{5};\sqrt{3};-\sqrt{3}\right\}\)
a) Ta có: \(\left(5x-2y\right)\left(x^2-xy+1\right)\)
\(=5x^3-5x^2y+5x-2x^2y+2xy^2-2y\)
\(=5x^3-7x^2y+2xy^2+5x-2y\)
b) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
\(=\left(x^2-1\right)\left(x+2\right)\)
\(=x^3+2x^2-x-2\)
c) Ta có: \(\dfrac{1}{2}x^2y^2\cdot\left(2x+y\right)\left(2x-y\right)\)
\(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)\)
\(=2x^4y^2-\dfrac{1}{2}x^2y^4\)
\(A\left(x\right)+B\left(x\right)-C\left(x\right)\)
\(=\left(-7+2x^2+x^4+3x^5-x^3\right)+\left(-x+x^4+2x^3-7\right)-\left(2x-x^4-3x^3\right)\)
\(=3x^5+3x^4+4x^3+2x^2-3x-14\)
Ta có: \(x+y-2=0\Rightarrow x+y=2\)
Và P=\(x^4+2x^3y-2x^3+x^2y^2-2x^2y-x\left(x+y\right)+2x+3\)
\(=\left(x^4+2x^3y+x^2y^2\right)-\left(2x^3+2x^2y\right)-\left[x\left(x+y\right)-2x\right]+3\)
\(=x^2\left(x+y\right)^2-2x^2\left(x+y\right)-x\left(x+y-2\right)+3\)
\(=x^2\cdot2^2-2x^2\cdot2-x\cdot0+3=3\) (thế x+y=2,x+y-2=0)
Vậy P=3