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Ta có M = x4 - 2x3 + 3x2 - 2x + 2
= x4 - x3 - x3 + x2 + 2x2 - 2x +2
= x2( x2 - x ) - x( x2 - x ) + 2( x2 - x ) + 2
= ( x2 - x + 2 )( x2 - x ) + 2
= ( 4 + 2 )*2 + 2 = 14
P= 3x2 - [2x2-3x(x-4)] với x=\(\frac{-3}{2}\)
\(\Rightarrow P=\frac{27}{4}-\left[\frac{9}{2}-\frac{99}{4}\right]=\frac{27}{4}+\frac{81}{4}=\frac{108}{4}=27\)
Q=(x2 + y2) (x2y+y3)-y(x4+y4)với x=\(\frac{-1}{2}\) và y=3
\(\Rightarrow Q=\frac{37}{4}.\frac{111}{4}-\frac{3891}{16}=\frac{4107}{16}-\frac{3891}{16}=\frac{216}{16}=\frac{27}{2}\)
1) (2x^2 + 1)(x^2 - 2x - 1)
= 2x^4 - 4x^3 - 2x^2 + x^2 - 2x - 1
= 2x^4 - 4x^3 - x^2 - 2x - 1
2) (x^2 - x^4)/(x^2 - 1 + 1)
= (x^2.(1 - x^2))/(x^2 - 1 + 1)
= (x^2.(1 + x)(1 - x))/x^2
= (1 + x)(1 - x)
3) (3x + y)^3 + x^3 - 3x^2 + 3x + 1
Thay x = 1,1; y = -0,7 vào biểu thức, ta có:
= [3.1,1 + (-0,7)]^3 + 1,1^3 - 3.1,1^2 + 3.1,1 + 1
= 19,577
\(x^2=x+1\Rightarrow x^2-x-1=0\)
\(A=x^4-2x^3-3x^2+4x+4\)
\(=x^4-x^3-x^2-x^3+x^2+x-3x^2+3x+3+1\)
\(=x^2\left(x^2-x-1\right)-x\left(x^2-x-1\right)-3\left(x^2-x-1\right)+1=0-0-0+1=1\)
\(A=4x^2-2\left(y+2,5x^2\right)+x^2-4y\)
\(=4x^2-2y-5x^2+x^2-4y=-6y\)
\(B=\left(x+y\right).\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)-\left(x^5+y^5-8\right)\)
\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5-x^5-y^5+8\)
\(=8\)
Vậy BT B ko phụ thuộc vào biến
câu sau tương tự
\(5x\left(x+1\right)-3\left(x-5\right)+4\left(3x-6\right)=2x^2-7\)
\(\Rightarrow5x^2+5x-3x+15+12x-24=2x^2-7\)
\(\Rightarrow5x^2+14x-9=2x^2-7\Rightarrow5x^2+14x-9-2x^2+7=0\)
\(\Rightarrow3x^2+14x-2=0\)
\(\Rightarrow3\left(x^2+\frac{14}{3}x-\frac{2}{3}\right)=0\Rightarrow x^2+2.x.\frac{7}{3}+\frac{49}{9}-\frac{55}{9}=0\)
\(\Rightarrow\left(x+\frac{7}{3}\right)^2=\frac{55}{9}\Rightarrow x+\frac{7}{3}\in\left\{\sqrt{\frac{55}{9}};-\sqrt{\frac{55}{9}}\right\}\Rightarrow x\in\left\{\sqrt{\frac{55}{9}}-\frac{7}{3};-\sqrt{\frac{55}{9}}-\frac{7}{3}\right\}\)
`Answer:`
`a)`
`A=5(x+1)^2-3(x-3)^2-4(x^2-4)`
`=>A=5(x^2+2x+1)-3(x^2-6x+9)-4x^2+16`
`=>A=5x^2+10x+5-3x^2+18x-27-4x^2+16`
`=>A=(5x^2-3x^2-4x^2)+(10x+18x)+(5-27+16)`
`=>A=-2x^2+28x-6`
`b)`
`B=5(x+1)^2-3(x-3)^2-4(x+2)(x-2)`
`=2x(3x+5)-3(3x+5)-2x(x^2-4x+4)-[(2x)^2-3^2]`
`=6x^2+10x-9x-15-2x^3+8x^2-8x-4x^2+9`
`=(6x^2-4x^2+8x^2)-2x^3+(10x-9x-8x)+(-15+9)`
Thay `x=-7` vào ta được:
`B=10(-7)^2-2(-7)^3-7(-7)-6`
`=>B=10.49-2(-343)+49-6`
`=>B=490+686+49-6`
`=>B=1219`
\(M=x^4-2x^3+3x^2-x+2\)
\(M=x^4-x^3+x^2+2x^2-2x+2\)
\(M=x^2\left(x^2-x\right)-x\left(x^2-x\right)+2\left(x^2-x\right)+2\)
\(M=\left(x^2-x\right)\left(x^2-x+2\right)+2\)
\(M=4.\left(4+2\right)+2\)( Vì \(x^2-x=4\))
\(M=24+2=26\)
Vậy M = 26 khi \(x^2-x=4\)