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Phân tích các đa thức thành nhân tử:
a. x6-y6
b.x35+x34+x33+.......+x2+x+1
c.x2-6xy+9y2-9
e.(x-9)(x-7)+1
Ta có : x35 + x34 + x33 +.......+ x2 + x + 1
= (x35 + x34 ) + (x33 + x32) +.......+ (x3 + x2) + (x + 1)
= x34(x + 1) + x32(x + 1) + .... + x2(x + 1) + (x + 1)
= (x + 1) ( x34 + x32 + ..... + x2 + 1)
Ta có : (x - 9)(x - 7) + 1
= x2 - 16x + 63 + 1
= x2 - 16x + 64
= x2 - 2.x.8 + 82
= (x - 8)2
a.x\(^6\)-y\(^6\)=(x\(^2\))\(^3\)+(y\(^2\))\(^3\) =(x\(^2\)+y\(^2\))(x\(^4\)-x\(^2\)y\(^2\)+y\(^4\))
b.x\(^{35}\)+x\(^{34}\)+......+x+1 =x\(^{34}\).(x+1)+......+(x+1)
=(x\(^{34}\)+x\(^{32}\)+......+x\(^2\)+1)(x+1)
c.x\(^2\)-6xy+9y\(^2\)-9 =x\(^2\)-2.3xy+(3y)\(^2\)-3\(^2\) =(x-3y)\(^2\)-3\(^2\)
=(x-3y-3)(x-3y+3)
d.(x-9)(x-7)+1 =x\(^2\)-7x-9x+63+1 =x\(^2\)-16x+64
=x\(^2\)-2.8x+8\(^2\) =(x-8)\(^2\)
a) \(2a^{n+2}b^n-18a^nb^{n+2}\)
\(=2a^nb^n\left(a^2-9b^2\right)\)
\(=2a^nb^n\left(a-3b\right)\left(a+3b\right)\)
\(3x^2-2x-1\)
\(=3x^2-3x+x-1\)
\(=3x.\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right).\left(3x+1\right)\)
\(9x^2-4y^2-4xy-x^2\)
\(=\left(3x\right)^2-\left(2y+x\right)^2\)
\(=\left(2x-2y\right)\left(4x+2y\right)\)
\(=4.\left(x-y\right)\left(2x+y\right)\)
Bài 1:
a) \(3x^2-9x=3x\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
Bài 2:
a) \(101^2-1=\left(101-1\right)\left(101+1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2\)
\(=\left(67+33\right)^2=100^2=10000\)
Bài 3:
\(x\left(x-3\right)+2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Vậy \(x=-2\)hoặc \(x=3\)
B1:
a) \(3x^2-9x=3x.\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3+y\right).\left(x+3-y\right)\)
B2:
a) \(101^2-1=\left(101+1\right).\left(101-1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2=\left(67+33\right)^2=100^2=10000\)
B3:
\(x\left(x-3\right)+2\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
phân tích đa thức sau thành nhân tử:
a) x5 + x + 1
b) x2 - 4xy + 4y2 - 2x + 4y - 35
c) x4 - 5x2y2 + 4y2
\(a,3\left(x+4\right)-x^2-4x\)
\(=3\left(x+4\right)-\left(x^2+4x\right)\)
\(=3\left(x+4\right)-x\left(x+4\right)\)
\(=\left(3-x\right)\left(x+4\right)\)
\(a,3\left(x+4\right)-x^2-4x\)
\(=3\left(x+4\right)-\left(x^2+4x\right)\)
\(=3\left(x+4\right)-x\left(x+4\right)\)
\(=\left(3-x\right),\left(x+4\right)\)