Dễ thôi, kẻ tam giác ABC

Vẽ đường thẳng xy qua A song song với BC

CM 2 góc B,C so le trong với 2 góc BAx và CAy và 2 góc vừa rồi với BAC có tổng là 180 độ

=> đpcm

Ta có: \(\frac{1}{x}-\frac{1}{y}-\frac{1}{z}=1\)

\(\Leftrightarrow\left(\frac{1}{x}-\frac{1}{y}-\frac{1}{z}\right)^2=1\)

\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{yz}-\frac{1}{xy}-\frac{1}{zx}\right)=1\)

\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\cdot\frac{x-y-z}{xyz}=1\)

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

a) Ta có: \(\frac{8n+5}{4n+1}=\frac{\left(8n+2\right)+3}{4n+1}=2+\frac{3}{4n+1}\)

Để BT nguyên

=> \(\frac{3}{4n+1}\inℤ\)<=> \(4n+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Mà \(4n+1\equiv1\left(mod4\right)\)

=> \(4n+1\in\left\{1;-3\right\}\Rightarrow n\in\left\{0;-1\right\}\)

b) Ta có: \(7^6+7^5-7^4\)

\(=7^4\left(7^2+7-1\right)\)

\(=7^4\cdot55⋮55\)

=> đpcm

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

\(\orbr{\begin{cases}\frac{2}{3}-x=-2x+\frac{1}{3}\\\frac{2}{3}-x=2x-\frac{1}{3}\end{cases}}\)   

\(\orbr{\begin{cases}-x+2x=\frac{1}{3}-\frac{2}{3}\\-x-2x=-\frac{1}{3}-\frac{2}{3}\end{cases}}\)    

\(\orbr{\begin{cases}x=-\frac{1}{3}\\-3x=-1\end{cases}}\)   

\(\orbr{\begin{cases}x=\frac{1}{3}\\x=-\frac{1}{3}\end{cases}}\)

a ) 0,5 + 7/2 = 1/2 + 7/2 = 8/2 = 4

c ) 0,75 . 3/10 = 3/4 . 3/10 = 9/40

d) 3/2 : 25/6 = 3/2 . 6/25 = 9/25

\(\left(x-\frac{1}{2}\right)\left(x+\frac{2}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=0\\x+\frac{2}{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{2}{3}\end{cases}}\)

Vậy x=1/2 hoặc x=-2/3

( x -\(\frac{1}{2}\)) ( x +\(\frac{2}{3}\)) = 0

<=> \(\orbr{\begin{cases}x-\frac{1}{2}=0\\x+\frac{2}{3}=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{2}{3}\end{cases}}\)

Bài 1:

a) \(\frac{x-1}{0-2}=\frac{1,2}{1,5}\)

\(\Leftrightarrow\frac{1-x}{2}=\frac{4}{5}\)

\(\Leftrightarrow5-5x=8\)

\(\Leftrightarrow x=-\frac{3}{5}\)

b) Ta có: \(x=\frac{y}{2}=\frac{z}{3}=\frac{4x-3y+2z}{4-6+6}=\frac{16}{4}=4\)

\(\Rightarrow\hept{\begin{cases}x=4\\y=8\\z=12\end{cases}}\)

Bài 1:

c) \(2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\Leftrightarrow\frac{x}{21}=\frac{y}{14}\)

\(5y=7z\Leftrightarrow\frac{y}{7}=\frac{z}{5}\Leftrightarrow\frac{y}{14}=\frac{z}{10}\)

\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2\)

\(\Rightarrow\hept{\begin{cases}x=42\\y=28\\z=20\end{cases}}\)

d) \(x:y:z=3:5:2\Leftrightarrow\frac{x}{3}=\frac{y}{5}=\frac{z}{2}=\frac{5x-7y+5z}{15-35+10}=\frac{124}{-10}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{186}{5}\\y=-62\\z=-\frac{124}{5}\end{cases}}\)