Phân tích các đa thức sau thành nhân tử BẰNG CÁCH ĐẶT BIẾN PHỤ :
a) (x^2+x)^2 - 2x^2 - 2x - 15
b) (x^2+2x)^2 + 9x^2 + 18x+20
c) (x^2+3x+1)(x^2+3x +2)- 6
d) (x+2)(x+3)(x+4)(x+5) - 24
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a) \(\left(x^2+x\right)^2-2x^2-2x-15\)
\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
Đặt x2 + x = a
\(=a^2-2a-15\)
\(=a^2-2a+1-16\)
\(=\left(a-1\right)^2-16\)
\(=\left(a-1\right)^2-4^2\)
\(=\left(a-1-4\right)\left(a-1+4\right)\)
\(=\left(a-5\right)\left(a+3\right)\)
\(=\left(x^2+x-5\right)\left(x^2+x+3\right)\)
b) \(\left(x^2+2x\right)^2+9x^2+18x+20\)
\(=\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20\)
Đặt x2 + 2x = a
\(=a^2+9a+20\)
\(=a^2+4a+5a+20\)
\(=a\left(a+4\right)+5\left(a+4\right)\)
\(=\left(a+4\right)\left(a+5\right)\)
\(=\left(x^2+2x+4\right)\left(x^2+2x+5\right)\)
c) \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt x2 + 3x + 1 = a
\(=a\left(a+1\right)-6\)
\(=a^2+a-6\)
\(=a^2-2a+3a-6\)
\(=a\left(a-2\right)+3\left(a-2\right)\)
\(=\left(a-2\right)\left(a+3\right)\)
\(=\left(x^2+3x+1-2\right)\left(x^3+3x+1+3\right)\)
\(=\left(x^2+3x-1\right)\left(x^2+3x+4\right)\)
d) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt x2 + 7x + 11 = a
\(=\left(a-1\right)\left(a+1\right)-24\)
\(=a^2-1-24\)
\(=a^2-25\)
\(=\left(a-5\right)\left(a+5\right)\)
\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(=\left(x^2+7x-6\right)\left(x^2+7x+16\right)\)