\(x^4+6x^3+13x^2+12x+4\)

\(=x^4+2x^3+4x^3+8x^2+5x^2+10x+2x+4\)

\(=\left(x^4+2x^3\right)+\left(4x^3+8x^2\right)+\left(5x^2+10x\right)+\left(2x+4\right)\)

\(=x^3\left(x+2\right)+4x^2\left(x+2\right)+5x\left(x+2\right)+2\left(x+2\right)\)

\(=\left(x+2\right)\left(x^3+4x^2+5x+2\right)\)

\(=\left(x+2\right)\left(x^3+x^2+3x^2+3x+2x+2\right)\)

\(=\left(x+2\right)\left[x^2\left(x+1\right)+3x\left(x+1\right)+2\left(x+1\right)\right]\)

\(=\left(x+2\right)\left(x+1\right)\left(x^2+3x+2\right)\)

\(=\left(x+2\right)\left(x+1\right)\left(x^2+x+2x+2\right)\)

\(=\left(x+2\right)\left(x+1\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)

\(=\left(x+2\right)\left(x+1\right)\left(x+1\right)\left(x+2\right)\)

\(=\left(x+2\right)^2\left(x+1\right)^2\)

a) \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)

\(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)

b) \(x\left(3x-2\right)-5\left(2-3x\right)=0\)

\(\Rightarrow x\left(3x-2\right)+5\left(3x-2\right)=0\)

\(\Rightarrow\left(3x-2\right)\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=2\\x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(A=10x-x^2+2\)

\(A=-\left(x^2-10x-2\right)\)

\(A=-\left(x^2-2.x.5+25-25-2\right)\)

\(A=-\left(x^2-2.x.5+25-27\right)\)

\(A=-\left(x-5\right)^2+27\)

Vì \(-\left(x-5\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x-5\right)^2+27\le27\)

\(\Rightarrow Amax=27\Leftrightarrow x=5\)

a) \(a^2+25b^2+17+10b-8a=0\)

\(\Rightarrow a^2-8a+16+25b^2+10b+1=0\)

\(\Rightarrow\left(a-4\right)^2+\left(5b+1\right)^2=0\)

Vì \(\left(a-4\right)^2\ge0\) với mọi a

\(\left(5b+1\right)^2\ge0\) với mọi b

\(\Rightarrow\left(a-4\right)^2+\left(5b+1\right)^2\ge0\) với mọi a,b

Mà \(\left(a-4\right)^2+\left(5b+1\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a-4\right)^2=0\\\left(5b+1\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a-4=0\\5b+1=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=4\\5b=-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=4\\b=-\dfrac{1}{5}\end{matrix}\right.\)

cả 2 bạn có số quả táo là

chiều cao là :120/16=8

           ĐS:8

a) \(7x^3y-14x^2y+7xy\)

\(=7xy\left(x^2-2x+1\right)\)

\(=7xy\left(x-1\right)^2\)

b) \(x^2-9+\left(x-3\right)^2\)

\(=\left(x-3\right)\left(x+3\right)+\left(x-3\right)^2\)

\(=\left(x-3\right)\left(x+3+x-3\right)\)

\(=2x\left(x-3\right)\)

c) \(x^3-4x^2-12x+27\)

\(=x^3+3x^2-7x^2-21x+9x+27\)

\(=x^2\left(x+3\right)-7x\left(x+3\right)+9\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

d) \(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xz-yz+2xy-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xz-yz-xy\right)\)

anh can may tinh ko?em cho muon

a) A=1/2+1/14+1/35+1/65+1/99+1/152

 Nhân cả tử và mẫu với 2 ta được :

Ax2=(1/2+1/14+1/35+1/65+1/99+1/152)x2

Ax2=2/4+2/28+2/70+...+2/198+2/304

Ax2=\(\frac{2}{1x4}+\frac{2}{4x7}+\frac{2}{7x11}+...+\frac{2}{16x19}=\frac{2}{3}x\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)

=\(\frac{2}{3}x\left(1-\frac{1}{19}\right)=\frac{2}{3}x\frac{18}{19}=\frac{12}{19}\)

=> A =\(\frac{12}{19}:2=\frac{6}{19}\)

     Vậy A=6/19

Mình mất công làm bài này , các bạn ủng hộ nhé