!!!!!!!!!!

Ta có:\(\left(x-y+z\right)\left(x+y+z\right)=\left[\left(x+z\right)-y\right]\left[\left(x+z\right)+y\right]\)

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

Xong rồi đấy,chúc bạn học tốt

Ta có; n⁵-n=n(n⁴-1)

=n(n²-1)(n²-4+5)

=n(n-1)(n+1)(n²-4)+5n(n-1)(n+1)

=n(n-1)(n+1)(n-2)(n+2)+5n(n-1)(n+1)

Vì n(n-1)(n+1) là tích 3 số tự nhiên liên tiếp nên n(n-1)(n+1) chia hết cho 2 và 3 (1) => 5n(n-1)(n+1) chia hết cho 30 (2)

CÓ: n(n-1)(n+1)(n-2)(n+2) là tích 5 số tự nhiên liên tiếp nên n(n-1)(n+1)(n-2)(n+2) chia hết cho 5 

Mà n(n-1)(n+1) chia hết cho 2 và 3 => n(n-1)(n+1)(n-2)(n+2) chia hết cho 30 (3)

Từ (1),(2),(3) => n(n-1)(n+1)(n-2)(n+2)+5n(n-1)(n+1) chia hết cho 30 hay n⁵-n chia hết cho 30 (đpcm)

56x64=3584

ủng hộ nhé~~

56x64

=(60-4)x(60+4)

=3600-16

=3584

help me

sr mình chưa học lớp 8

\(a,\)

\(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}.\frac{4x^2-4x+16}{x^2-4}\right):\frac{16}{x+2}.\frac{x^2+3x+2}{x^2+x+1}\)\(ĐKXĐ:x\ne\pm2\)

\(A=[\frac{4x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right).4\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x+2\right)}]:\frac{16}{x+2}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)

\(A=[\frac{4x}{x+2}-\frac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}].\frac{x+2}{16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)

\(A=\frac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}.\frac{x+2}{16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)

\(A=\frac{16\left(x+2\right)}{\left(x+2\right)^2.16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)

\(A=\frac{-\left(x+1\right)}{x^2+x+1}\)

\(B=\frac{x^2+x-2}{x^3-1}\)\(ĐKXĐ:x\ne1\)

\(B=\frac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(B=\frac{x+2}{x^2+x+1}\)

\(b,\)

Ta có:

\(A+B=\frac{-\left(x+1\right)}{x^2+x+1}+\frac{x+2}{x^2+x+1}\)

\(=\frac{-x-1+x+2}{x^2+x+1}\)

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

\(\Rightarrow A+B=\frac{1}{x^2+x+1}=\frac{1}{x^2+2.x.\left(\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{1}{\left(x+\frac{1}{2}\right)^2}+\frac{3}{4}\)

Vì:\(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

\(\Rightarrow\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{1}{\frac{3}{4}}\)

\(\Rightarrow A+B\le\frac{4}{3}\)

\(\Rightarrow GTLN\)của \(A+B=\frac{4}{3}\Leftrightarrow x+\frac{1}{2}=0\)

                                                        \(\Leftrightarrow x=\frac{-1}{2}\left(TMĐK\right)\)

Vậy........

d, (3x-1)³=(-2x+1)³ <=> 3x-1=-2x+1 <=> 5x=2 <=> x=2/5

e, Vì \(\hept{\begin{cases}x^2\ge0\\x^4\ge0\end{cases}\Rightarrow x^2+x^4\ge0}\)

Mà x²+x⁴=0

=>x=0

f, Vì \(\hept{\begin{cases}x^2\ge0\\\left|x\right|\ge0\end{cases}\Rightarrow x^2+\left|x\right|\ge0\Rightarrow x^2+\left|x\right|+1\ge1>0}\)

Vậy k có x thỏa mãn đề bài

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

\(\Rightarrow\)\(3x-1=-2x+1\)

\(3x+2x=1+1\)

\(5x=2\)

\(x=\frac{2}{5}\)

Vậy \(x=\frac{2}{5}\)

\(x^2+x^4=0\)

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

\(\Rightarrow\orbr{\begin{cases}x^2=0\\1+x^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^2=-1\left(v\text{ô}l\text{ý}\right)\end{cases}}}\)

Vậy \(x=0\)

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

Ta có:

\(\hept{\begin{cases}x^2\ge0\forall x\\\left|x\right|\ge0\forall x\end{cases}\Rightarrow x^2+\left|x\right|+1\ge1\forall x}\)

Mà \(x^2+\left|x\right|+1=0\)

\(\Rightarrow\)x không có giá trị

Vậy x không có giá trị

a, \(\frac{1}{\left(2x-3\right)^2}=9\Leftrightarrow\left(\frac{1}{2x-3}\right)^2=3^2\Leftrightarrow\frac{1}{2x-3}=3\Leftrightarrow1=6x-9\Leftrightarrow x=\frac{5}{3}\)

b, \(\frac{1}{\left(x-1\right)^3}=-\frac{1}{27}\Leftrightarrow\left(\frac{1}{x-1}\right)^3=\left(\frac{-1}{3}\right)^3\Leftrightarrow\frac{1}{x-1}=\frac{1}{-3}\Leftrightarrow x-1=-3\Leftrightarrow x=-2\)

c, \(\left(x-1\right)^2=\left(2x-5\right)^2\Leftrightarrow\orbr{\begin{cases}x-1=2x-5\\x-1=-2x+5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}}\)