Bài 1:

a)  \(\left(n+2\right)^2-\left(n-2\right)^2=n^2+4n+4-\left(n^2-4n+4\right)=8n\)    \(⋮\)\(8\)   (đpcm)

b)  \(\left(n+7\right)^2-\left(n-5\right)^2=n^2+14n+49-\left(n^2-10n+25\right)=24n-24\)\(⋮\)\(24\) (đpcm)

Bài 2:

mk biến đổi về pt tích sau đó bạn giải nốt nhé

a)   \(\left(x-4\right)^2-36=0\)

<=>  \(\left(x-4-6\right)\left(x-4+6\right)=0\)

<=>  \(\left(x-10\right)\left(x+2\right)=0\)

................

b)  \(4x^2-12x=-9\)

<=> \(4x^2-12x+9=0\)

<=>  \(\left(2x-3\right)^2=0\)

..............

c) \(\left(x+8\right)^2=121\)

<=>  \(\left(x+8\right)^2-121=0\)

<=>  \(\left(x+8+11\right)\left(x+8-11\right)=0\)

<=> \(\left(x+19\right)\left(x-3\right)=0\)

...................

Bài 3:

a)  \(31,8^2-2\times31,8\times21,8+21,8^2\)

    \(=\left(31,8-21,8\right)^2=10^2=100\)

b)  mạo phép chỉnh đề

\(58,2^2+2\times58,2\times41,8+41,8^2\)

\(=\left(58,2+41,8\right)^2=100^2=10000\)

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

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

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

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

Bài 3) 

Ta có :

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

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

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

\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

P/s tham khảo nha

hok tốt

\(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4-9x^2}\)

\(=\frac{3x+2}{9x^2-4}-\frac{3x-2}{9x^2-4}+\frac{3x-6}{9x^2-4}\)

\(=\frac{3x+2-3x+2+3x-6}{9x^2-4}\)

\(=\frac{3x-2}{9x^2-4}\)

\(=\frac{1}{3x+2}\)

\(\frac{18}{\left(x-3\right)\left(x^2-9\right)}-\frac{3}{x^2-6x+9}-\frac{x^2}{x^2-9}\)

\(=\frac{18}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\) \(-\frac{3\left(x+3\right)}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\)\(-\frac{x^2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x-3\right)}\)

\(=\frac{18-3x-9-x^3+3x^2}{\left(x-3\right)^2\left(x+3\right)}\)

\(=\frac{-x^3+3x^2-3x+9}{\left(x-3^2\right)\left(x+3\right)}\)

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

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

học tốt

\(P=a^5-a\)

\(=a\left(a^4-1\right)=a\left(a^2-1\right)\left(a^2+1\right)\)

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

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

\(=5\left(a-1\right)a\left(a+1\right)+\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)\)

Nhân thấy   \(5\left(a-1\right)a\left(a+1\right)⋮5\);    \(\left(a-1\right)a\left(a+1\right)⋮3!=6\)

=>   \(5\left(a-1\right)a\left(a+1\right)⋮30\)

                    \(\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)⋮5!\)

=>   \(\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)⋮30\)

Vậy P chia hết cho 30

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

Tự cm tiếp

a)  \(A=a^3-b^3-c^3-3abc\)

\(=\left(a-b\right)^3+3ab\left(a-b\right)-c^3-3abc\)

\(=\left(a-b-c\right)\left[\left(a-b\right)^2+c\left(a-b\right)+c^2\right]+3ab\left(a-b-c\right)\)

\(=\left(a-b-c\right)\left(a^2-2ab+b^2+ac-bc+c^2+3ab\right)\)

\(=\left(a-b-c\right)\left(a^2+b^2+c^2+ab+ac-bc\right)\)

b)  \(B=a^2b^2\left(a-b\right)-c^2b^2\left(c-b\right)+a^2c^2\left(c-a\right)\)

\(=a^2b^2\left(a-b\right)+c^2b^2\left(b-c\right)+a^2c^2\left(c-a\right)\)

\(=a^2b^2\left(a-b\right)+c^2b^2\left(b-c\right)-a^2c^2\left[\left(a-b\right)+\left(b-c\right)\right]\)

\(=a^2b^2\left(a-b\right)+c^2b^2\left(b-c\right)-a^2c^2\left(a-b\right)-a^2c^2\left(b-c\right)\)

\(=a^2\left(a-b\right)\left(b^2-c^2\right)+c^2\left(b-c\right)\left(b^2-a^2\right)\)

\(=a^2\left(a-b\right)\left(b-c\right)\left(b+c\right)+c^2\left(b-c\right)\left(b-a\right)\left(b+a\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a^2b+a^2c-bc^2-ac^2\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(ab+bc+ca\right)\)

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

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=1-3xy\)

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

\(=3\left(1-2xy\right)-2\left(1-3xy\right)\)

\(=3-6xy-2+6xy\)

\(=1\)

\(A=x^3+3x^2+3x\)

   \(=x^3+3x^2+3x+1-1\)

   \(=\left(x+1\right)^3-1\)

   \(=\left(99+1\right)^3-1\)

   \(=999999\)

\(A=x^3+3.x^2+3.x\)

    \(=99^3+3.99^2+3.99\)

\(=99\left(99^2+3.99+3\right)\)

\(=99.111111\)

\(=999999\)

Chúc bạn học tốt!

\(x^2+y^2+z^2=xy+yz+zx\)

<=>   \(2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

<=>  \(\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)

<=>  \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

<=>  \(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}}\) <=>  \(\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}}\)<=>  \(x=y=z\)  (đpcm)

      \(x^2+y^2+z^2=xy+xz+yz\)

\(\Rightarrow2x^2+2y^2+2z^2=2xy+2xz+2yz\)

\(\Rightarrow2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)

\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)=0\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)(1)

Ta có: \(\hept{\begin{cases}\left(x-y\right)^2\ge0\forall x;y\\\left(y-z\right)^2\ge0\forall y;z\\\left(x-z\right)^2\ge0\forall x;z\end{cases}\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\forall x;y;z}\) (2)

Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}\Rightarrow\hept{\begin{cases}x=y\\y=z\\x=z\end{cases}\Rightarrow}x=y=z}\)

Chúc bạn học tốt.