Giúp mình giải với
a) P(x) = - x4 - 2
b) Q(x) = - x2 - 2x - 2
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a) \(x^4+x^3-8x-8\)
\(=x^3\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x^3+8\right)\left(x+1\right)\)
\(=\left(x+2\right)\left(x^2-2x+4\right)\left(x+1\right)\)
a) \(=x^3\left(x+1\right)-8\left(x+1\right)=\left(x+1\right)\left(x^3-8\right)=\left(x+1\right)\left(x-2\right)\left(x^2+2x+4\right)\)
b) \(=y\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(y-3\right)\)
c) \(=3\left(x-y\right)-a\left(x-y\right)=\left(x-y\right)\left(3-a\right)\)
a)\(x^4+3x^3+x^2+3x=x\left(x^3+3x^2+x+3\right)\)
\(=x\left[x^2\left(x+3\right)+\left(x+3\right)\right]=x\left(x+3\right)\left(x^2+1\right)\)
b) \(x^2+6xy+9y^2-4z^2=\left(x+3y\right)^2-4z^2=\left(x+3y-2z\right)\left(x+3y+2z\right)\)
c) \(=2x\left(x-1\right)-7\left(x-1\right)=\left(x-1\right)\left(2x-7\right)\)
\(a,=x^3\left(x+3\right)+x\left(x+3\right)=x\left(x^2+1\right)\left(x+3\right)\\ b,=\left(x+3y\right)^2-4z^2=\left(x+3y+2z\right)\left(x+3y-2z\right)\\ c,=2x^2-2x-7x+7=\left(x-1\right)\left(2x-7\right)\)
a) \(=2x\left(x-25\right)\)
b) \(=x\left(x-4\right)-\left(x-4\right)=\left(x-4\right)\left(x-1\right)\)
c) \(=x^2-\left(y^2-12y+36\right)=x^2-\left(y-6\right)^2=\left(x-y+6\right)\left(x+y-6\right)\)
d) \(=y\left(x^2+4xz+4yz\right)\)
b: \(\left(x^2+4\right)^2-16x^2\)
\(=\left(x^2-4x+4\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)^2\cdot\left(x+2\right)^2\)
c: \(x^5-x^4+x^3-x^2\)
\(=x^4\left(x-1\right)+x^2\left(x-1\right)\)
\(=x^2\left(x-1\right)\left(x^2+1\right)\)
Lời giải:
a. Bạn xem lại đề
b. \((x^2+4)^2-16x^2=(x^2+4)^2-(4x)^2=(x^2+4-4x)(x^2+4+4x)\)
\(=(x-2)^2(x+2)^2\)
c.
\(x^5-x^4+x^3-x^2=x^4(x-1)+x^2(x-1)=(x^4+x^2)(x-1)\)
\(=x^2(x^2+1)(x-1)\)
\(a,\left(x+y\right)^2-2xy=x^2+2xy+y^2-2xy=x^2+y^2\left(đpcm\right)\\ b,\left(a+b\right)^2-\left(a-b\right)\left(a+b\right)=\left(a+b\right)\left(a+b-a+b\right)=2b\left(a+b\right)\left(đpcm\right)\)
Lời giải:
a.
$A=(x+6)^2-(x+2)^2+2[(x-5)^2-(x-3)^2]$
$=(x+6-x-2)(x+6+x+2)+2[(x-5-x+3)(x-5+x-3)]$
$=4(2x+8)+2(-2)(2x-8)$
$=4(2x+8)-4(2x-8)=4[(2x+8)-(2x-8)]=4.16=64$ không phụ thuộc vào $x$
b.
$B=(x^3-2^3)-(x^3+2^3)=-16$ không phụ thuộc vào $x$
c.
$C=x^4+2x^2-[(x^2+3)^2-(2x)^2]$
$=x^4+2x^2-(x^4+6x^2-4x^2)$
$=x^4+2x^2-(x^4+2x^2)=0$ không phụ thuộc vào $x$
a) Ta có: \(A=\left(x+6\right)^2+2\left(x-5\right)^2-\left(x+2\right)^2-2\left(x-3\right)^2\)
\(=x^2+12x+36+2\left(x^2-10x+25\right)-\left(x^2+4x+4\right)-2\left(x^2-6x+9\right)\)
\(=x^2+12x+36+2x^2-20x+50-x^2-4x-4-2x^2+12x-18\)
\(=34\)
b) Ta có: \(B=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+2\right)\left(x^2-2x+4\right)\)
\(=x^3-8-x^3-8\)
=-16
c) Ta có: \(C=x^4+2x^2-\left(x^2-2x+3\right)\left(x^2+2x+3\right)\)
\(=x^4+2x^2-\left[\left(x^2+3\right)^2-4x^2\right]\)
\(=x^4+2x^2-\left(x^4+6x^2+9\right)+4x^2\)
\(=-9\)
Đáp án: D
A: \(f(x)=x^2+2x-x^2=2x\) → Bậc 1.
B: \(f(x)=x+3\) → Bậc 1.
C: Bậc 4.
x bằng bao nhiêu vậy bạn