1: a) \(x^3+10x^2+15x-26\)
\(=\left(x^3-x^2\right)+\left(11x^2-11x\right)+\left(26x-26\right)\)
\(=x^2\left(x-1\right)+11x\left(x-1\right)+26\left(x-1\right)\)
\(=\left(x^2+11x+26\right)\left(x-1\right)\)
b) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\) (1)
Đặt \(x^2+5x+5=y\)
Khi đó (1) trở thành: \(\left(y-1\right)\left(y+1\right)\)
Bài này thiếu đề à bn 
2: Ta có: \(x^2+x=6\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2-2x+3x-6=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{-3;2\right\}\) \(\)
3: a) Ta có: \(A=\left(x+3\right)\left(x-4\right)+7\)
\(=\left(x^2-x-12\right)+7\)
\(=\left[\left(x^2-x+\frac{1}{4}\right)-\frac{49}{4}\right]+7\)
\(=\left(x-\frac{1}{2}\right)^2-\frac{21}{4}\) \(\le-\frac{21}{4}\)
Dấu ''='' xảy ra \(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy GTLN của \(A\) là \(-\frac{21}{4}\) khi \(x=\frac{1}{2}\)
3: b) \(B=3-\left(x-1\right)\left(x-2\right)\)\(=3-\left(x^2-3x+2\right)\)
\(=3-x^2+3x-2\) \(=1-x^2+3x\)
\(=-\left(x^2-3x-1\right)\)
\(=-\left[\left(x^2-3x+\frac{9}{4}\right)-\frac{13}{4}\right]\)
\(=-\left(x-\frac{3}{2}\right)^2+\frac{13}{4}\)
Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\) nên \(-\left(x-\frac{3}{2}\right)^2\le0\)
\(\Rightarrow-\left(x-\frac{3}{2}\right)^2+\frac{13}{4}\le\frac{13}{4}\)
Dấu ''='' xảy ra \(\Leftrightarrow-\left(x-\frac{3}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{3}{2}=0\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy GTLN của \(B\) là \(\frac{13}{4}\) khi \(x=\frac{3}{2}\)
