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\(a,3x\left(x-\frac{2}{3}\right)=0
\)
\(\)TH1:
3x=0
x=0:3
x=0
TH2
\(x-\frac{2}{3}=0
\)
\(x=0+\frac{2}{3}=\frac{2}{3}\)
Vậy x={0;\(\frac{2}{3}\)}
34 +14 :x=25
\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}\)
\(x=\frac{1}{4}:-\frac{7}{20}\)
\(x=-\frac{20}{28}\)
\(x=-\frac{5}{7}\)
\(3\frac{1}{5}-x=1\frac{3}{5}+\frac{7}{10}\)
\(\frac{16}{5}-x=\frac{8}{5}+\frac{7}{10}\)
\(\frac{16}{5}-x=\frac{23}{10}\)
\(x=\frac{23}{10}-\frac{16}{5}\)
\(x=-\frac{9}{10}\)
Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\) => \(\left(\dfrac{x}{3}\right)^2=\left(\dfrac{y}{4}\right)^2=\left(\dfrac{z}{5}\right)^2\)
=> \(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{25}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{25}=\dfrac{2x^2+y^2-z^2}{2.9+16-25}=\dfrac{9}{18+16-25}=\dfrac{9}{9}=1\)
=> \(\left\{{}\begin{matrix}\dfrac{x^2}{9}=1\Rightarrow\dfrac{x}{3}=1\Rightarrow x=3\\\dfrac{y^2}{16}=1\Rightarrow\dfrac{y}{4}=1\Rightarrow y=4\\\dfrac{z^2}{25}=1\Rightarrow\dfrac{z}{5}=1\Rightarrow z=5\end{matrix}\right.\)
Vậy x = 3, y = 4, z = 5
Đặt x/3=y/4=z/5=k
=>x=3k; y=4k; z=5k
Ta có: \(2x^2+y^2-z^2=9\)
\(\Leftrightarrow18k^2+16k^2-25k^2=9\)
\(\Leftrightarrow9k^2=9\)
\(\Leftrightarrow k^2=1\)
TH1: k=1
=>x=3; y=4; z=5
TH2: k=-1
=>x=-3; y=-4; z=-5
\(\frac{1}{2}-\left\{\frac{2}{3}.x-\frac{1}{3}\right\}=\frac{2}{3}\)
\(\frac{2}{3}.x-\frac{1}{3}=\frac{1}{3}-\frac{2}{3}\)
\(\frac{2}{3}.x-\frac{1}{3}=\frac{-1}{3}\)
\(\frac{2}{3}.x=\frac{-1}{3}+\frac{1}{3}\)
\(\frac{2}{3}.x=0\)
\(x=0:\frac{2}{3}\)
\(x=0\)
Vậy \(x=0\)
\(\left[9-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
có 9 số 1 có 9 số hạng
\(=\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=\left[\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=1\)
Đặt \(x=\frac{y}{2}=\frac{z}{3}=k\left(k\in Q\right)\)\(\Rightarrow x=k;y=2k;z=3k\)
Thế (1) vào biểu thức trên
\(\Rightarrow2\left(x^2+y^2\right)-z^2=9\)
\(\Leftrightarrow2\left[\left(k\right)^2+\left(2k\right)^2\right]-\left(3k\right)^2=9\)
\(\Rightarrow2\left(k^2+4k^2\right)-9k^2=9\)
\(\Rightarrow2k^2+8k^2-9k^2=9\)
\(\Rightarrow k^2=9\)
\(\Rightarrow k=\hept{\begin{cases}3\\-3\end{cases}}\)
Với k = 3
\(\Rightarrow x=3;y=3.2=6;z=3.3=9\)
Với k = -3
\(\Rightarrow x=-3;y=-3.2=-6;z=-3.3=-9\)
Đặt x = y/2 = z/3 = k= \(\hept{\begin{cases}x=1.k\\y=2.k\\z=3.k\end{cases}}\)
2 ( x2+y2) - z2 = 9 => .....( mình mới làm được đến đấy thôi! )