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Ta có: \(\left(\frac{2}{3}-x\right)^2=\left(-2x+\frac{1}{3}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{3}-x=-2x+\frac{1}{3}\\\frac{2}{3}-x=2x-\frac{1}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{3}\\3x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-\frac{1}{3}\end{cases}}\)
Ta có 2 trường hợp:
TH1: \(x-2>0\Rightarrow x>0+2\Rightarrow x>2\)
\(x+\frac{2}{3}>0\Rightarrow x>0-\frac{2}{3}\Rightarrow x>\frac{-2}{3}\)
TH2: \(x-2< 0\Rightarrow x< 0+2\Rightarrow x< 2\)
\(x-\frac{2}{3}< 0\Rightarrow x< \frac{2}{3}+0\Rightarrow x< \frac{-2}{3}\)
TH1: \(\Rightarrow x>2\)
Th2: \(\Rightarrow x< \frac{-2}{3}\)
Vậy\(x>2\) hoặc\(x< \frac{-2}{3}\)
\(\frac{x-1}{2}=\frac{2.\left(x-1\right)}{2.2}=\frac{2x-2}{4}\)
\(\frac{y-2}{3}=\frac{3.\left(y-2\right)}{3.3}=\frac{3y-6}{9}\)
Theo t/c dãy tỉ số bằng nhau:
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{50-5}{9}=\frac{45}{9}=5\)
=> \(\frac{x-1}{2}=5\Rightarrow x-1=10\Rightarrow x=11\)
=> \(\frac{y-2}{3}=5\Rightarrow y-2=15\Rightarrow y=17\)
=> \(\frac{z-3}{4}=5\Rightarrow z-3=20\Rightarrow z=23\)
(2x-2)/4 = (3y-6)/9 =(z-3)/4
(2x+3y -z -5)/10 = (50-5)/10 = 4,5
x -1 = 4,5.2 = 9
x = 10
y-2 = 4,5.3 = 13,5
y = 15,5
z-3 = 4,5.4 = 18
z = 21
Vì \(\left|x^2+2x\right|\ge0;\left|y^2-9\right|\ge0\)
Dấu ''='' xảy ra <=> \(x^2+2x=0\Leftrightarrow x\left(x+2\right)=0\Leftrightarrow x=0;x=-2\)
\(y^2-9=0\Leftrightarrow\left(y-3\right)\left(y+3\right)=0\Leftrightarrow y=\pm3\)
Từ \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{3-z}{-4}\)
Ap dụng tính chất của tỉ lệ thức ta có \(\frac{2x-2}{4}=\frac{2x-2+3y-6+3-z}{4+9-4}\)=\(\frac{2x+3y-z-5}{9}\)
Lại có 2x+3y-z=50\(\Rightarrow\frac{2x-2}{4}=\frac{50-5}{9}=5\Rightarrow2x-2=20\Rightarrow x=11\)
Tương tự \(\frac{y-2}{3}=5\Rightarrow y=17\)
\(\frac{z-3}{4}=5\Rightarrow z=23\)
Vậy x=11,y=17,z=23
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau, Ta có:
\(\frac{x-1+y-2-\left(z-3\right)}{2+3-4}\)=\(\frac{2x-2+3y-6-z+3}{4+9-4}\)
=\(\frac{2x-3y-z-2-6+3}{9}\)=\(\frac{2x-3y-z-\left(2+6-3\right)}{9}\)
=\(\frac{2x+3y-z-5}{9}=\frac{50-5}{9}=\frac{45}{9}=5\)
\(\frac{2x-2}{4}=5\)x = 11
\(\frac{3y-6}{9}=5\) y=17
\(\frac{z-3}{4}=5\)
z = 23
Ta co x+2/2=y+3/3=z+4/4 hay x+1=y+1=z+1 => x=y=z
Suy ra: 2x+y+z=11 hay 2x+x+x=11 => 4x=11 => x=11/4
Vay: x^2+y^2+z^2 = (11/4)^2+(11/4)^2+(11/4)^2 =121/16 . 3 = 363/16
\(\left(\frac{2}{3}-x\right)^2=\left(-2x+\frac{1}{3}\right)^2\)
\(\orbr{\begin{cases}\frac{2}{3}-x=-2x+\frac{1}{3}\\\frac{2}{3}-x=2x-\frac{1}{3}\end{cases}}\)
\(\orbr{\begin{cases}-x+2x=\frac{1}{3}-\frac{2}{3}\\-x-2x=-\frac{1}{3}-\frac{2}{3}\end{cases}}\)
\(\orbr{\begin{cases}x=-\frac{1}{3}\\-3x=-1\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{1}{3}\\x=-\frac{1}{3}\end{cases}}\)