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d)
$x^4+2x^3+2x^2+2x+1$
$=(x^4+2x^3+x^2)+(x^2+2x+1)$
$=(x^2+x)^2+(x+1)^2=x^2(x+1)^2+(x+1)^2$
$=(x+1)^2(x^2+1)$
e)
$x^2y+xy^2+x^2z+y^2z+2xyz$
$=xy(x+y)+z(x^2+y^2)+2xyz$
$=xy(x+y)+z(x^2+y^2+2xy)$
$=xy(x+y)+z(x+y)^2=(x+y)(xy+zx+zy)$
f)
$x^5+x^4+x^3+x^2+x+1$
$=(x^5+x^4)+(x^3+x^2)+(x+1)=x^4(x+1)+x^2(x+1)+(x+1)$
$=(x+1)(x^4+x^2+1)$
$=(x+1)[(x^4+2x^2+1)-x^2]$
$=(x+1)[(x^2+1)^2-x^2]=(x+1)(x^2+1-x)(x^2+1+x)$
a)
$x^4-2x^3+2x-1=(x^4-2x^3+x^2)-(x^2-2x+1)$
$=(x^2-x)^2-(x-1)^2$
$=x^2(x-1)^2-(x-1)^2=(x-1)^2(x^2-1)=(x-1)^2(x-1)(x+1)$
$=(x-1)^3(x+1)$
b)
$a^6-a^4+2a^3+2a^2$
$=a^4(a^2-1)+2a^2(a+1)$
$=a^4(a-1)(a+1)+2a^2(a+1)$
$=(a+1)[a^4(a-1)+2a^2]$
$=a^2(a+1)[a^2(a-1)+2]$
$=a^2(a+1)(a^3-a^2+2)=a^2(a+1)[a^2(a+1)-2(a^2-1)]$
$=a^2(a+1)[a^2(a+1)-2(a-1)(a+1)]$
$=a^2(a+1)(a+1)(a^2-2a+2)=a^2(a+1)^2(a^2-2a+2)$
c)
$x^4+x^3+2x^2+x+1$
$=(x^4+2x^2+1)+(x^3+x)$
$=(x^2+1)^2+x(x^2+1)=(x^2+1)(x^2+1+x)$
Mình ko ghi lại đề , bạn ghi ra xong rồi suy ra như mình nha .
1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)
\(=>A=-12x+16\)
2) \(=>B=8x^3+27-8x^3+2=29\)
3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)
4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)
5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)
\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)
\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)
6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)
k cho mik nha ,
Bài 4:
a) Ta có: \(x^3+6x^2+12x+8\)
\(=x^3+2x^2+4x^2+8x+4x+8\)
\(=x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+4x+4\right)\)
\(=\left(x+2\right)^3\)
b) Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-x^2-2x^2+2x+x-1\)
\(=x^2\left(x-1\right)-2x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+1\right)\)
\(=\left(x-1\right)^3\)
c) Ta có: \(1-9x+27x^2-27x^3\)
\(=1-3x-6x+18x^2+9x^2-27x^3\)
\(=\left(1-3x\right)-6x\left(1-3x\right)+9x^2\left(1-3x\right)\)
\(=\left(1-3x\right)\left(1-6x+9x^2\right)\)
\(=\left(1-3x\right)^3\)
d) Ta có: \(x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)
\(=x^3+3\cdot x^2\cdot\frac{1}{2}+3\cdot x\cdot\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3\)
\(=\left(x+\frac{1}{2}\right)^3\)
e) Ta có: \(27x^3-54x^2y+36xy^2-8y^3\)
\(=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot2y+3\cdot3x\cdot\left(2y\right)^2-\left(2y\right)^3\)
\(=\left(3x-2y\right)^3\)
Bài 9 : Tìm x, biết :
a, (x - 2)(x - 3) + (x - 2) - 1 = 0
\(\Leftrightarrow\left(x-2\right)\left(x-3+1\right)-1=0\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2+1\right)\left(x-2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy x ={1; 3}
b, (x + 2)2 - 2x(2x + 3) = (x + 1)2
\(\Leftrightarrow\left(x+2\right)^2-\left(x+1\right)^2-2x\left(2x+3\right)=0\)
\(\Leftrightarrow\left(x+2+x+1\right)\left(x+2-x-1\right)-2x\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3-2x\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(1-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\1-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=\left\{-\frac{3}{2};\frac{1}{2}\right\}\)
c, 6x3 + x2 = 2x
\(\Leftrightarrow6x^3+x^2-2x=0\)
\(\Leftrightarrow x\left(6x^2+x-2\right)=0\)
\(\Leftrightarrow x\left(6x^2+4x-3x-2\right)=0\)
\(\Leftrightarrow x\left[2x\left(3x+2\right)-\left(3x+2\right)\right]=0\)
\(\Leftrightarrow x\left(3x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+2=0\\2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{2}{3}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=\left\{0;-\frac{2}{3};\frac{1}{2}\right\}\)