ta có :

\(|x-\frac{2}{3}|+|x+y+\frac{3}{4}|+|y-z-\frac{5}{6}|=0\Leftrightarrow\hept{\begin{cases}x-\frac{2}{3}=0\\x+y+\frac{3}{4}=0\\y-z-\frac{5}{6}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=-\frac{17}{12}\\z=-\frac{9}{4}\end{cases}}\)

giúp mình với ạ

a) 3x = 4 + 2

3x = 6

=> x = 2;

Ta có : \(\hept{\begin{cases}\left|x-\frac{3}{4}\right|\ge0\forall x\\\left|\frac{2}{5}-y\right|\ge0\forall y\\\left|x-y+z\right|\ge0\forall x;y;z\end{cases}}\Leftrightarrow\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|\ge0\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-\frac{3}{4}=0\\\frac{2}{5}-y=0\\x-y+z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{2}{5}\\z=-\frac{7}{20}\end{cases}}\)

Vậy x = 3/4 ; y = 2/5 ; z = -7/20 

\(\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|=0\)

Ta có: \(\left|x-\frac{3}{4}\right|;\left|\frac{2}{5}-y\right|;\left|x-y+z\right|\ge0\Rightarrow\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|\ge0\)

Mà \(\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|=0\)

\(\Rightarrow\hept{\begin{cases}x-\frac{3}{4}=0\\\frac{2}{5}-y=0\\x-y+z=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{2}{5}\\\frac{3}{4}-\frac{2}{5}+z=0\Rightarrow z=\frac{-7}{20}\end{cases}}\)

Theo đề ra, ta có: \(\frac{x}{y}=\frac{2}{4}=\frac{1}{2}\Rightarrow\frac{x}{1}=\frac{y}{2}\)

Áp dụng tính chất dãy tỷ số bằng nhau

\(\frac{x}{1}=\frac{y}{2}=\frac{x-y}{1-2}=\frac{-8}{-1}=8\)

\(\Rightarrow x=8.1=8\)

\(\Rightarrow y=8.2=16\)

Ta có : 3^x + 3^{x + 3} = 756

<=> 3^x(1 + 3³) = 756 

<=> 3^x.28 = 756 

<=> 3^x = 27

<=> 3^x = 3³ 

<=> x = 3

Vậy x = 3 

a) tco:

aNQ+QNP=180

Mà 2 góc nàu ở vtri TCP=>aa'//bb'