Bài 1: Tìm x, y, z (hoặc a, b, c)
1, \(3\left(x-1\right)=2\left(y-2\right);4\left(y-2\right)=3\left(z-3\right)\) và \(2x+3y-z=50\)
2, \(\frac{3x-2y}{37}=\frac{5y-3z}{15}=\frac{2z-5x}{2}\) và \(10x-3y-2z=-4\)
3, \(\frac{a}{b}=\frac{8}{5};\frac{b}{c}=\frac{2}{7}\) và \(a+b+c=61\)
4, \(ab=\frac{1}{2};bc=\frac{2}{3};ac=\frac{3}{4}\)
5, \(15x=-10y=6z\) và \(xyz=-3000\)
6, \(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\) và \(xyz=22400\)
Please help me😭😭
Câu 1:
\(3\left(x-1\right)=2\left(y-2\right)\Leftrightarrow3x-3=2y-4\Leftrightarrow3x=2y-1\)
\(4\left(y-2\right)=3\left(z-3\right)\Leftrightarrow4y-8=3z-9\Leftrightarrow4y=3z-1\)
Lại có:
\(3x=2y-1\Leftrightarrow6x=4y-2=3z-1-2=3z-3\)
\(\Rightarrow6x=4y-2=3z-3\)
\(\Rightarrow6x=3z-3\Leftrightarrow2x=z-1\)
\(\Rightarrow2x+3y-z=z-1+3y-z=3y-1=50\Leftrightarrow3y=51\Leftrightarrow y=17\)\(\Rightarrow\left\{{}\begin{matrix}x=11\\z=23\end{matrix}\right.\)
Câu 3:
\(\frac{a}{b}=\frac{8}{5}\Leftrightarrow\frac{a}{8}=\frac{b}{5}\Leftrightarrow\frac{1}{2}.\frac{a}{8}=\frac{1}{2}.\frac{b}{5}\Leftrightarrow\frac{a}{16}=\frac{b}{10}\) (1)
\(\frac{b}{c}=\frac{2}{7}\Leftrightarrow\frac{b}{2}=\frac{c}{7}\Leftrightarrow\frac{1}{5}.\frac{b}{2}=\frac{1}{5}.\frac{c}{7}\Leftrightarrow\frac{b}{10}=\frac{c}{35}\) (2)
Từ (1) và (2)
\(\Rightarrow\frac{a}{16}=\frac{b}{10}=\frac{c}{35}=k\)\(\Rightarrow\left\{{}\begin{matrix}a=16k\\b=10k\\c=35k\end{matrix}\right.\)
\(\Rightarrow a+b+c=16k+10k+35k=61k=61\Rightarrow k=1\)
\(\Rightarrow\left\{{}\begin{matrix}a=16k=16\\b=10k=10\\c=35k=35\end{matrix}\right.\)
Câu 4:
\(ab=\frac{1}{2};bc=\frac{2}{3}\Leftrightarrow abbc=\frac{1}{2}.\frac{2}{3}\Leftrightarrow ab^2c=\frac{1}{3}\)
Lại có:
\(ac=\frac{3}{4}\Leftrightarrow b^2=\frac{4}{9}\Leftrightarrow\left[{}\begin{matrix}b=\frac{2}{3}\\b=-\frac{2}{3}\end{matrix}\right.\)
TH1: \(b=\frac{2}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}a=\frac{3}{4}\\c=1\end{matrix}\right.\)
TH2:\(b=-\frac{2}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}a=-\frac{3}{4}\\c=-1\end{matrix}\right.\)
Câu 6:
\(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\Leftrightarrow\frac{x-30}{40}=\frac{y-15}{20}=\frac{z-21}{28}\Leftrightarrow\frac{x}{40}-\frac{3}{4}=\frac{y}{20}-\frac{3}{4}=\frac{z}{28}-\frac{3}{4}\)\(\Leftrightarrow\frac{x}{40}=\frac{y}{20}=\frac{z}{28}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=40k\\y=20k\\z=28k\end{matrix}\right.\)
\(\Rightarrow xyz=40k20k28k=\text{22400}k^3=22400\Leftrightarrow k^3=1\Leftrightarrow k=1\)\(\Rightarrow\left\{{}\begin{matrix}x=40k=40\\y=20k=20\\z=28k=28\end{matrix}\right.\)
