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Đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\)
\(\Rightarrow x=2k+1;y=3k+2;z=4k+3\)
Thay vào,ta được:
\(2\left(2k+1\right)+3\left(3k+2\right)-\left(4k+3\right)=50\)
\(\Leftrightarrow4k+2+9k+6-4k-3=50\)
\(\Leftrightarrow9k+5=50\)
\(\Leftrightarrow9k=45\)
\(\Leftrightarrow k=5\)
\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}=\frac{5x-5}{10}=\frac{3y+9}{12}=\frac{4z-20}{24}\)
\(=\frac{5x-5-3y-9-4z+20}{10-12-24}=\frac{\left(5x-3y-4z\right)+\left(20-5-9\right)}{26}=\frac{46+6}{26}=2\)
\(\Rightarrow x=2\cdot2+1=5\)
\(y=4\cdot2-3=5\)
\(z=2\cdot6+5=17\)
Câu c tương tự như câu 1
1/ Ta có: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Leftrightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)\(=\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}\)\(=\frac{\left(2x+3y-z\right)-5}{9}=\frac{45}{9}=5\)
\(\Rightarrow\)x=11;y=17;z=23
2/ Theo bài ra, ta có: \(\frac{2x}{3}=\frac{2y}{4}=\frac{4z}{5}\)\(\Leftrightarrow\frac{x}{\frac{3}{2}}=\frac{y}{2}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+2+\frac{5}{4}}\)\(=\frac{49}{\frac{19}{4}}=\frac{196}{19}\)
\(\Rightarrow\)x=\(\frac{294}{19};y=\frac{392}{19};z=\frac{245}{19}\)
Ta có: \(\frac{x}{3}\)=\(\frac{y}{4}\)=> \(\frac{x}{15}\)=\(\frac{y}{20}\)
\(\frac{y}{5}\)=\(\frac{z}{6}\)=> \(\frac{y}{20}\)=\(\frac{z}{24}\) Vậy \(\frac{x}{15}\)=\(\frac{y}{20}\)=\(\frac{z}{24}\)
đặt \(\frac{x}{15}\)=\(\frac{y}{20}\)=\(\frac{z}{24}\)=k => x=15k; y=20k; z=24k
Thay x=15k; y=20k ; z=24k vào Biểu thức M ta có:
M=\(\frac{2x+3y+4z}{3x+4y+5z}\)=\(\frac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}\)=\(\frac{k\left(30+60+96\right)}{k\left(45+80+120\right)}\)=\(\frac{186}{245}\)
Theo bài ra ta có : \(\frac{x}{3}=\frac{y}{4}\Leftrightarrow x=\frac{3y}{4}\) ; \(\frac{y}{5}=\frac{z}{6}\Leftrightarrow z=\frac{6y}{5}\), Vậy ta có : \(M=\frac{2x+3y+z}{3x+4y+5z}=\frac{2.\frac{3y}{4}+3y+4.\frac{6y}{5}}{3.\frac{3y}{4}+4y+5.\frac{6y}{5}}=\frac{\frac{93y}{10}}{\frac{49y}{4}}=\frac{93}{10}.\frac{4}{49}=\frac{186}{245}\)
Vì \(\frac{x}{3}\) = \(\frac{y}{4}\) => \(\frac{x}{15}\) = \(\frac{y}{20}\)
\(\frac{y}{5}\) = \(\frac{z}{6}\) => \(\frac{y}{20}\) = \(\frac{z}{24}\)
nên \(\frac{x}{15}\) = \(\frac{y}{20}\) = \(\frac{z}{24}\)
Đặt \(\frac{x}{15}\) = \(\frac{y}{20}\) = \(\frac{z}{24}\) = k
=> x = 15k; y = 20k và z = 24k
Thay vào M ta đc:
M = \(\frac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}\)
= \(\frac{30k+60k+96k}{45k+80k+120k}\)
= \(\frac{\left(30+60+96\right)k}{\left(45+80+120\right)k}\)
= \(\frac{186k}{245k}\) = \(\frac{186}{245}\)
Vậy M = \(\frac{186}{245}\).
Giải:
Ta có: \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\)
\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)
Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=k\)
\(\Rightarrow\hept{\begin{cases}x=15k\\y=20k\\z=24k\end{cases}}\)
\(\Rightarrow M=\frac{2x+3y+4z}{3x+4y+5z}=\frac{30k+60k+96k}{45k+80k+120k}=\frac{\left(30+60+96\right)k}{\left(45+80+120\right)k}\)
bạn tự tính nốt nhé
Mình giải tiếp cho:
\(M=\frac{\left(30+60+96\right)k}{\left(45+80+120\right)k}=\frac{186k}{245k}=\frac{186}{245}\)
Vậy \(M=\frac{186}{245}\)
Đặt \(\frac{2x}{3}=\frac{3y}{5}=\frac{5z}{6}=k\)
\(\Rightarrow\hept{\begin{cases}x=\frac{3k}{2}\\y=\frac{5k}{3}\\z=\frac{6k}{5}\end{cases}}\)
\(\Rightarrow3x-4y+3z=\frac{3.3k}{2}-\frac{4.5k}{3}+\frac{3.6k}{5}=-59\)
\(\Rightarrow\frac{9k}{2}-\frac{20k}{3}+\frac{18k}{5}=-59\)
\(\Rightarrow k.\left(\frac{9}{2}-\frac{20}{3}+\frac{18}{5}\right)=-59\)
\(\Rightarrow k.\frac{43}{30}=-59\)
=> k = -1770/43
Số lớn khiếp , còn lại tự nhân lên rồi tìm x,y,z nha