\(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\Rightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0\)

\(\Rightarrow\left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)^2=1\end{matrix}\right.\)

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

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

\(\Rightarrow\left(x+3\right)\left(y+7\right)=\left(x+5\right)\left(y+5\right)\)

\(\Rightarrow xy+7x+3y+21=xy+5x+5y+25\)

\(\Rightarrow2x-2y=4\Rightarrow x-y=2\)

e giúp đc ko '-'sama

full hd :))

Đặt:

\(\dfrac{a}{b}=\dfrac{c}{d}=t\Leftrightarrow\left\{{}\begin{matrix}a=bt\\c=dt\end{matrix}\right.\)

Khi đó:

\(\dfrac{ac}{bd}=\dfrac{bt.dt}{bd}=\dfrac{t^2bd}{bd}=t^2\)

\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2t^2+d^2t^2}{b^2+d^2}=\dfrac{t^2\left(b^2+d^2\right)}{b^2+d^2}=t^2\)

Vậy.....

\(\left(x+3\right)\left(x-4\right)-x\left(x-3\right)=12\)

\(\Rightarrow x\left(x-4\right)+3\left(x-4\right)-x\left(x-3\right)=12\)

\(\Rightarrow x^2-4x+3x-12-x^2+3x=12\)

\(\Rightarrow\left(x^2-x^2\right)+\left(3x+3x-4x\right)-12=12\)

\(\Rightarrow2x-12=12\)

\(\Rightarrow2x=24\Rightarrow x=12\)

Chị cx học Tê Tiêu ạ,A mấy ạ

neu chuyen cho 2 hop = nhau thi moi hop co so kg che la :

1,6:2=6,8 kg

luc dau hop thu nhat co so kg che la : 

6,8+1,2=8 kg 

luc dau hop thu 2 co so kg che la : 

13,6-8=5,6 kg

**** nhe

t làm câu 1.you viết đề sai nhé

\(\left(x+1\right)^{x+2}=\left(x+1\right)^{x+4}\)

\(\Rightarrow\left(x+1\right)^{x+4}-\left(x+1\right)^{x+2}=0\)

\(\Rightarrow\left(x+1\right)^{x+2}\left[\left(x+1\right)^2-1\right]=0\)

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

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

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

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