bạn giải đi bạn

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:

\(P=\frac{3k+8k+15k}{6k+12k+20k}+\frac{6k+12k+20k}{9k+16k+25k}+\frac{9k+16k+25k}{12k+20k+30k}\)

\(P=\frac{26k}{38k}+\frac{38k}{50k}+\frac{50k}{62k}=\frac{13}{19}+\frac{19}{25}+\frac{25}{31}=\frac{33141}{14725}\)

\(3x=4y;2y=5z\)

\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{2}\)

\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{15};\dfrac{y}{15}=\dfrac{z}{6}\)

\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{6}\)

\(\Rightarrow\dfrac{2x}{40}=\dfrac{3y}{45}=\dfrac{5z}{30}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{2x}{40}=\dfrac{3y}{45}=\dfrac{5z}{30}\)

\(=\dfrac{2x+3y-5z}{40+45-30}=\dfrac{55}{55}=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=1.20=20\\y=1.15=15\\z=1.6=6\end{matrix}\right.\)

Tương tự

Ta có :

\(2x+3y-5z=55\)

\(3x=4y;2y=5z\)

\(\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{2}=\dfrac{z}{5}\)

\(\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12};\dfrac{y}{12}=\dfrac{z}{16}\)

\(\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}=\dfrac{2x+3y-5z}{2.19+3.12-2.16}=\dfrac{55}{22}=\dfrac{5}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{45}{2}\\\dfrac{y}{12}=\dfrac{5}{2}\Leftrightarrow x=30\\\dfrac{z}{16}=\dfrac{5}{2}\Leftrightarrow z=40\end{matrix}\right.\)

Vậy ..............

\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\)  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 #

Ta có:\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\Rightarrow\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}\Rightarrow\frac{3x}{\frac{9}{2}}=\frac{4y}{\frac{16}{3}}=\frac{5z}{\frac{25}{4}}\)

                Áp dụng tính chất của dãy tỉ số bằng nhau:

          \(\frac{3x}{\frac{9}{2}}=\frac{4y}{\frac{16}{3}}=\frac{5z}{\frac{25}{4}}=\frac{3x-4y+5z}{\frac{9}{2}-\frac{16}{3}+\frac{25}{4}}=\frac{65}{\frac{65}{12}}=65.\frac{12}{65}=12\)

                                    \(\frac{x}{\frac{3}{2}}=12\Rightarrow x=12.\frac{3}{2}=18\)

                                    \(\frac{y}{\frac{4}{3}}=12\Rightarrow y=12.\frac{4}{3}=16\)

                                    \(\frac{z}{\frac{5}{4}}=12\Rightarrow z=12.\frac{5}{4}=15\)

                                                     Vậy \(x=18,y=16,z=15\)

\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\left(1\right)\)

\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\left(2\right)\)

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=\(\frac{186}{245}\)

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

\(c,\frac{2x}{5}=\frac{3y}{10}=\frac{z}{12}\)và x + y + z = 107

Ta có : \(\frac{2x}{5}=\frac{3y}{10}=\frac{z}{12}\Leftrightarrow\frac{x}{\frac{5}{2}}=\frac{y}{\frac{10}{3}}=\frac{z}{12}=\frac{x+y+z}{\frac{5}{2}+\frac{10}{3}+12}=\frac{107}{\frac{107}{6}}=107\cdot\frac{6}{107}=6\)

Vậy : \(\hept{\begin{cases}\frac{2x}{5}=6\\\frac{3y}{10}=6\\\frac{z}{12}=6\end{cases}}\Leftrightarrow\hept{\begin{cases}x=15\\x=20\\z=72\end{cases}}\)

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) \(\Rightarrow\frac{2x+3y+4z}{186}=\frac{3x+4y+5z}{245}\Rightarrow\frac{2x+3y+4z}{3x+4y+5z}=\frac{186}{245}\)