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B1:
a) \(5\left(x^2+y^2\right)-20x^2y^2\)
\(=5\left(x^2-4x^2y^2+y^2\right)\)
b) \(=2\left(x^8-16\right)=2\left(x^4-4\right)\left(x^4+4\right)=2\left(x^2-2\right)\left(x^2+2\right)\left(x^4+4\right)\)
B2:
a) Đặt \(x^2-3x+1=y\)
=> \(y^2-12y+27\)
\(=\left(y^2-12y+36\right)-9\)
\(=\left(y-6\right)^2-3^2\)
\(=\left(y-9\right)\left(y-3\right)\)
\(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)
\(=\left(x+1\right)\left(x-4\right)\left(x^2-3x-10\right)\)
b) Đặt \(x^2+7x+11=t\)
Ta có: \(\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a) \(a^6-b^6=\left(a^2\right)^3-\left(b^2\right)^3=\left(a^2-b^2\right)\left(a^4+a^2b^2+b^{\text{4}}\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(a^{\text{4}}+a^2b^2+b^{\text{4}}\right)\)
c) \(\left(x^2+x\right)^2+2\left(x^2+x\right)+1=\left(x^2+x+1\right)^2\)
e) \(\left(x^2-10x+25\right)-4y^2=\left(x-5\right)^2-\left(2y\right)^2\)
\(=\left(x-5-2y\right)\left(x-5+2y\right)\)
g) \(x^6+27=\left(x^2\right)^3+3^3=\left(x^2+3\right)\left(x^4-3x^2+9\right)\)
Còn lại tớ làm sau nhé, bây h muộn rùi
\(Dat:x^2+x=a\Rightarrow....=a^2-2a-15=\left(a-1\right)^2-4^2=\left(a+3\right)\left(a-7\right)\)
\(=\left(x^2+x+3\right)\left(x^2+x-5\right)\)
\(Dat:x+y=a\Rightarrow....=a^2-a-12=\left(a+3\right)\left(a-4\right)=\left(x+y+3\right)\left(x+y-4\right)\)
a) A= \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
Đặt \(x^2+x=a\) .
Khi đó : \(A=a^2-2a-15=a^2-5a+3a-15\)\(=a\left(a-5\right)+3\left(a-5\right)=\left(a+3\right)\left(a-5\right)\)
Mà \(a=x^2+x\) nên \(A=\left(x^2+x+3\right)\left(x^2+x-5\right)\)
b) B = \(x^2+2xy+y^2-x-y-12\) \(=\left(x+y\right)^2-\left(x+y\right)-12\)
Đặt x+y = z.
Khi đó : \(B=z^2-z-12=z^2-4z+3z-12=z\left(z-4\right)+3\left(z-4\right)\)\(=\left(z+3\right)\left(z-4\right)\)
Mà z = x+y nên B = (x+y+3)(x+y-4)
a) \(x^2\left(x-3\right)+27-9x=0\)
\(x^2\left(x-3\right)+9\left(3-x\right)=0\)
\(x^2\left(x-3\right)-9\left(x-3\right)=0\)
\(\left(x^2-9\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-9=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=9\\x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=3\end{cases}}\Rightarrow x=3\)
vay \(x=3\)
a) \(x^2-4x+3\)
= \(x^2-3x-x+3\)
\(=\left(x^2-3x\right)-\left(x-3\right)\)
\(=x\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-1\right)\left(x-3\right)\)
a)\(x^2-4x+3=x^2-3x-x+3=x\left(x-3\right)-\left(x-3\right)=\left(x-3\right)\left(x-1\right)\)
b)\(x^2+x-6=x^2+3x-2x-6=x\left(x+3\right)-2\left(x+3\right)=\left(x-2\right)\left(x+3\right)\)
c)\(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-3\right)\left(x-2\right)\)
d)\(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
a: \(=\left(x^2-2x\right)^2-\left(x^2-2x\right)-6\)
\(=\left(x^2-2x-3\right)\left(x^2-2x+2\right)\)
\(=\left(x^2-2x+2\right)\left(x-3\right)\left(x+1\right)\)
c: \(=\left(x^2+x+4+3x\right)\left(x^2+x+4+5x\right)\)
\(=\left(x^2+6x+4\right)\left(x^2+4x+4\right)\)
\(=\left(x^2+6x+4\right)\left(x+2\right)^2\)