~~~~~e)~~~~~
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(x^2+x+1=v\)
Ta có: \(v.\left(v+1\right)-12\)
\(=v^2+v-12\)
\(=v^2-3v+4v-12\)
\(=v\left(v-3\right)+4\left(v-3\right)\)
\(=\left(v-3\right)\left(v+4\right)\)
\(=\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
~~~~~g)~~~~~
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)(nhân cái đầu vs cái cuối, hai cái giữa nhân vs nhau)
Đặt \(x^2+5x+5=t\)
Ta có: \(\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-1-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
~~~~~h)~~~~~
\(\left(x^2+x+1\right)\left(x^2+3x+1\right)+x^2\)
Đặt \(x^2+2x+1=n\)
Ta có: \(\left(n-x\right)\left(n+x\right)+x^2\)
\(=n^2-x^2+x^2\)
\(=n^2\)
\(=\left(x^2+2x+1\right)^2\)
\(=\left(\left(x+1\right)^2\right)^2\)
\(=\left(x+1\right)^4\)
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(Mong là mình làm đúng, chúc you học tốt nha, tíck cho mìk với nhé!)
