\(25^1;5^2;25^2\)

Ta có \(\frac{x-y}{z}=\frac{3y}{x-z}=\frac{x}{y}=\frac{x-y+3y+x}{z+x-z+y}=\frac{2x+2y}{x+y}=\frac{2\left(x+y\right)}{x+y}=2\)(dãy tỉ số bằng nhau)

=> x = 2y (đpcm)

Khi đó \(\frac{x-y}{z}=2\Leftrightarrow x-y=2z\Rightarrow2y-y=2z\Rightarrow y=2z\)(đpcm)

dpcm là j vậy

Đề bài yêu cầu gì nhỉ ?

-5.83 ( sai thì xin thông cảm )

\(1,75-\left|x\right|=3,21\)

\(\Leftrightarrow\left|x\right|=1,75-3,21\)

\(\Leftrightarrow\left|x\right|=-1,46\) ( Vô lí )

Không tồn tại x thỏa mãn đề.

1,75-|x|=3,21

1,75-x=3,21   hoặc    1,75-x=-3,21

x=1,75-3,21  hoặc   x=1,75+3,21

x=-1,46   hoặc    x=4,96

vậy  x=-1,46  hoặc  4,96

Đề bài là gì thế bạn nhờ?

a) \(\frac{a^2m-a^2n-b^2n+b^2m}{a^2+b^2}=\frac{a^2\left(m-n\right)+b^2\left(m-n\right)}{a^2+b^2}\)

\(=\frac{\left(m-n\right)\left(a^2+b^2\right)}{a^2+b^2}=m-n\)

b) \(\frac{\left(ab+bc+cd+ad\right)abcd}{\left(c+d\right)\left(a+b\right)+\left(b-c\right)\left(a-b\right)}\)

\(=\frac{\left[b.\left(a+c\right)+d.\left(a+c\right)\right].abcd}{ac+bc+da+db+ab-b^2-ca+bc}\)

\(=\frac{\left(a+c\right)\left(d+b\right)abcd}{2bc+da+db+ab-b^2}\)

giúp mik

a, |x - 2| < 3

=> \(\hept{\begin{cases}x-2< 3\\x-2>-3\end{cases}}\)

=> \(\hept{\begin{cases}x< 5\\x>-1\end{cases}}\)

=> x\(\in\){4 ; 3 ; 2 ; 1 ; 0}

b, |x + 1| > 2

=> \(\orbr{\begin{cases}x+1\ge2\\x+1\le-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x\ge1\\x\le-3\end{cases}}\)

a. Vì \(\left|x+\frac{1}{2}\right|\ge0\forall x;\left|y-\frac{3}{4}\right|\ge0\forall y;\left|z-1\right|\ge0\forall z\)

\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|y-\frac{3}{4}\right|+\left|z-1\right|\ge0\forall x;y;z\)

Dấu "=" xảy ra <=> | x + 1/2 | = 0 ; | y - 3/4 | = 0 ; | z - 1 | = 0

<=> x = - 1/2 ; y = 3/4 ; z = 1

b. Vì \(\left|x-\frac{3}{4}\right|\ge0\forall x;\left|\frac{2}{5}-y\right|\ge0\forall y\left|x-y+z\right|\ge0\forall x;y;z\)

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

Dấu "=" xảy ra <=> | x - 3/4 | = 0 ; | 2/5 - y | = 0 ; | x - y + z | = 0

<=> x = 3/4 ; y = 2/5 ; z = - 7/20

a) Ta có \(\hept{\begin{cases}\left|x+\frac{1}{2}\right|\ge0\forall x\\\left|y-\frac{3}{4}\right|\ge0\forall y\\\left|z-1\right|\ge0\forall z\end{cases}}\Rightarrow\left|x+\frac{1}{2}\right|+\left|y-\frac{3}{4}\right|+\left|z-1\right|\ge0\forall x;y;z\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+\frac{1}{2}=0\\y-\frac{3}{4}=0\\z-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{3}{4}\\z=1\end{cases}}\)

Vậy x = -1/2 = y = 3/4 ; z = 1 

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

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

Vậy x = 3/4 ; y = 2/5 ; z = -7/20