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\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\) hay \(\frac{x}{18}=\frac{y}{16}=\frac{z}{15}\) => \(\frac{3x}{54}=\frac{4y}{64}=\frac{5z}{75}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{3x}{54}=\frac{4y}{64}=\frac{5z}{75}=\frac{3x-4y+5z}{54-64+75}=\frac{65}{65}=1\)
suy ra: \(\frac{3x}{54}=1\) => \(x=18\)
\(\frac{4y}{64}=1\) => \(y=16\)
\(\frac{5z}{75}=1\) => \(z=15\)
\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\)
\(\Leftrightarrow\frac{x}{\frac{2}{3}}=\frac{y}{\frac{3}{4}}=\frac{z}{\frac{4}{5}}\Rightarrow\frac{3x}{\frac{2}{3}.3}=\frac{4y}{\frac{3}{4}.4}=\frac{5z}{\frac{4}{5}.5}\)
\(\Leftrightarrow\frac{3x}{2}=\frac{4y}{3}=\frac{5z}{4}\)
ÁP DỤNG TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU:
\(\Leftrightarrow\frac{3x}{2}-\frac{4y}{3}+\frac{5z}{5}\Rightarrow\frac{3x-4y+5z}{2-3+5}=\frac{65}{4}\)
\(\Rightarrow\frac{3x}{2}=\frac{65}{4}\Rightarrow3x=\frac{65}{4}.2\Rightarrow3x=\frac{65}{2}\Rightarrow x=\frac{65}{6}\)
\(\Rightarrow\frac{4y}{3}=\frac{65}{4}\Rightarrow4y=\frac{65}{4}.3\Rightarrow4y=\frac{195}{4}\Rightarrow y=\frac{195}{16}\)
\(\Rightarrow\frac{5z}{5}=\frac{65}{4}\Rightarrow5z=\frac{65}{4}.5\Rightarrow5z=\frac{325}{4}\Rightarrow z=\frac{65}{4}\)
# chúc bạn học tốt #
![](https://rs.olm.vn/images/avt/0.png?1311)
a
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\left(1\right)\)
\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\left(2\right)\)
từ (1) và (2) => \(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)
đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=k\Rightarrow x=15k,y=20k,z=24k\)
thay x=15k, y=20k, z=24k vào M ta có:
\(M=\frac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}=\frac{30k+60k+96k}{45k+80k+120k}=\frac{186k}{245k}=\frac{186}{245}\)
vậy M=\(\frac{186}{245}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo đề ta có: \(x:y:z=3:4:5\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)
Đặt: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\left(k\inℕ^∗\right)\)
Suy ra: \(x=3k;y=4k;z=5k\) Thay vào biểu thức P ta có:
\(P=\frac{3k+8k+15k}{6k+12k+20k}+\frac{6k+12k+20k}{9k+16k+25k}+\frac{9k+16k+25k}{12k+20k+30k}\)
\(P=\frac{26k}{38k}+\frac{38k}{50k}+\frac{50k}{62k}=\frac{13}{19}+\frac{19}{25}+\frac{25}{31}=\frac{33141}{14725}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\)
\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{24}=k\)
\(\Rightarrow x=15k;y=20k;z=24k\)
\(\frac{2x+3y+4z}{3x+4y+5z}=\frac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}=\frac{30k+60k+96k}{45k+80k+120k}=\frac{186k}{245k}=\frac{186}{245}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\hept{\begin{cases}\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\\\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\end{cases}}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)\(\Leftrightarrow\frac{2x}{30}=\frac{3y}{60}=\frac{4z}{96}=\frac{2x+3y+4z}{30+60+96}=\frac{2x+3y+4z}{186}\)(theo tính chất dãy tỉ số bằng nhau).(1)
= \(\frac{3x}{45}=\frac{4y}{80}=\frac{5z}{120}=\frac{3x+4y+5z}{45+80+120}=\frac{3x+4y+5z}{245}\)(theo tính chất dãy tỉ số bằng nhau). (2)
Từ (1) và (2) \(\Rightarrow\frac{2x+3y+4z}{186}=\frac{3x+4y+5z}{245}\Rightarrow\frac{2x+3y+4z}{3x+4y+5z}=\frac{186}{245}\)
ta có: \(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{3x}{54}=\frac{4y}{64}=\frac{4z}{60}.\)
ADTCDTSBN
...
bn tự lm típ nha
\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\)
hay \(\frac{x}{90}=\frac{y}{80}=\frac{z}{75}\)<=> \(\frac{3x}{270}=\frac{4y}{320}=\frac{4z}{300}\)
Áp dụng TCDTS bằng nhau ta có:
\(\frac{3x}{270}=\frac{4y}{320}=\frac{4z}{300}=\frac{3x-4y+4z}{270-320+300}=\frac{65}{250}=0,26\)
đến đây bạn tự tính ra nhé