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Ta có: \(\frac{x}{2}-\frac{2}{y}=\frac{1}{2}\)
=> \(\frac{x}{2}-\frac{1}{2}=\frac{2}{y}\)
=> \(\frac{x-1}{2}=\frac{2}{y}\)
=> (x - 1).y = 2 . 2
=> (x - 1).y = 4 = 1 . 4 = 2. 2 = 4 . 1
Lập bảng :
| x - 1 | 1 | 4 | -1 | -4 | 2 | -2 |
| y | 4 | 1 | -4 | -1 | 2 | -2 |
| x | 2 | 5 | 0 | -3 | 3 | -1 |
Vậy ...
\(\frac{-12}{6}=\frac{x}{5}=\frac{-y}{3}=\frac{z}{-17}=\frac{-t}{-9}\)
=> \(\frac{x}{5}=-2\)
=>x = -10
=> \(\frac{-y}{3}=-2\)
=> -y = -6
=> y = 6
=> \(\frac{z}{-17}=-2\)
=> z = 34
=>\(\frac{-t}{-9}=\frac{t}{9}=2\)
=> t = 18
vậy x = -10 ; y = 6 ; z = 34 ; t = 18
duyệt nha các bn
\(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x+2}{42} +\frac{x+2}{56}+\frac{x+2}{72}=\frac{16}{9}\)
\(\Rightarrow x-2\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\frac{16}{9}\)
\(\Rightarrow x-2\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{8}-\frac{1}{8}+\frac{1}{9}\right)=\frac{16}{9}\)
\(\Rightarrow x-2\left(\frac{1}{3}+\frac{1}{9}\right)=\frac{16}{9}\)
\(\Rightarrow\left(x-2\right)\frac{4}{9}=\frac{16}{9}\)
\(\Rightarrow x-2=4\)
\(\Rightarrow x=6\)
Ta có: \(\frac{-48}{-12}=\frac{12}{x}\Rightarrow x=\frac{\left(-12\right).12}{-48}=3\)
Thế x = 3 \(\Rightarrow\frac{12}{3}=\frac{y^2}{9}\Rightarrow y^2=\frac{12.9}{3}=36\Rightarrow y=\pm6\)
Thế x = 3 \(\Rightarrow\frac{12}{3}=\frac{-256}{t^2}\Rightarrow t^2=\frac{3.\left(-256\right)}{12}=-64\Rightarrow t\in\varnothing\)
Vậy \(x=3;y=\left\{6;-6\right\},t\in\varnothing\)