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b, A=[(a+1)(a+7)][(a+3)(a+5)]+15
=>A=(a2+8a+7)(a2+8a+15)+15
Đặt a2+8a+11= t
=>a2+8a+7= t-4 và a2+8a+15= t+4
=>A=(t-4)(t+4)+15
=>A=t2-16+15
=t2-1=(t-1)(t+1)
Thay t = a2+8a+11
=>A=(a2+8a+11-1)(a2+8a+11+1)
=>A=(a2+8a+10)(a2+8a+12)
a) \(x^2+2xy+7x+7y+y^2+10\)
\(=\left(x+y\right)^2+7\left(x+y\right)+\frac{49}{4}-\frac{9}{4}\)
\(=\left(x+y+\frac{7}{2}\right)^2-\frac{9}{4}\)
\(=\left(x+y+\frac{7}{2}-\frac{3}{2}\right)\left(x+y+\frac{7}{2}+\frac{3}{2}\right)\)
\(=\left(x+y-2\right)\left(x+y+5\right)\)
\(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-5\left(x^2+x\right)-15\)
\(=\left(x^2+x\right)\left(x^2+x+3\right)-5\left(x^2+x+3\right)\)
\(=\left(x^2+x-5\right)\left(x^2+x+3\right)\)
a) \(x^2-25-4xy+4y^2\)
\(=\left(x^2-4xy+4y^2\right)-25\)
\(=\left(x-2y\right)^2-5^2\)
\(=\left(x-2y-5\right)\left(x-2y+5\right)\)
b) \(x^2-8x+15\)
\(=x^2-3x-5x+15\)
\(=x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(x-5\right)\)
a)\(x^2-25-4xy+4y^2\Leftrightarrow\left(x^2-4xy+4y^2\right)-25\)
\(\Leftrightarrow\left(x-2y\right)^2-5^2\)
\(\Leftrightarrow\left(x-2y-5\right)\left(x-2y+5\right)\)
b)\(x^2-8x+15\Leftrightarrow\left(x-3\right)\left(x-5\right)\)
a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)
b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)
c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)
(1+x2)2−4x(1−x2)
= \(-\left(1-x^2\right)^2-4x\left(1-x^2\right)\)
đặt \(\left(1-x^2\right)\)= a
ta có :
- a . a - 4x .a
= a ( - a - 4x )
thay a = \(\left(1+x^2\right)\) ta có
\(\left(1+x^2\right)\left(1-x^2-4x\right)\)
phân tích tiếp nhé !
\(\left(1+x^2\right)^2-4x\left(1-x^2\right)=1+2x^{ }+x^4-4x+4x^3\)\(=\left(x^4+2x^3-x^2\right)+\left(2x^3+4x^2-2x\right)-x^2-2x+1=x^2\left(x^2+2x-1\right)+2x\left(x^2+x-1\right)-\left(x^2+2x-1\right)\)\(\left(x^2+2x-1\right)\left(x^2+2x-1\right)=\left(x^2+2x-1\right)^2\)
Ta có : \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15=\left[\left(x^2+x\right)-2\left(x^2+x\right)+1\right]-16=\left(x^2+x-1\right)^2-4^2\)
\(=\left(x^2+x-5\right)\left(x^2+x+3\right)\)