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Ta có:
\(\frac{t}{x}=\frac{4}{3};\frac{y}{z}=\frac{3}{2};\frac{z}{x}=\frac{1}{6}\)
\(\Rightarrow\frac{t}{x}:\frac{y}{z}:\frac{z}{x}=\frac{4}{3}:\frac{3}{2}:\frac{1}{6}\)
\(\Rightarrow\frac{t}{x}.\frac{z}{y}.\frac{x}{z}=\frac{4}{3}.\frac{2}{3}.\frac{1}{6}\)
\(\Rightarrow\frac{t}{y}=\frac{4}{27}\)
\(\Rightarrow t:y=4:27\)
Ta có:\(\frac{z}{1}=\frac{x}{6}\Rightarrow\frac{z}{2}=\frac{x}{12}\left(1\right)\)
\(\frac{y}{3}=\frac{z}{2}\left(2\right)\)
\(\frac{t}{4}=\frac{x}{3}\Rightarrow\frac{t}{16}=\frac{x}{12}\left(3\right)\)
Từ (1),(2) và (3)
\(\Rightarrow\frac{t}{16}=\frac{y}{3}=\frac{x}{12}=\frac{z}{2}\)
\(\Rightarrow\frac{t}{y}=\frac{16}{3}\) hay t:y=16:3
\(\frac{t}{x}=\frac{4}{3}\Rightarrow\frac{t}{4}=\frac{x}{3}\Rightarrow t=\frac{4x}{3}\)
\(\frac{z}{x}=\frac{1}{6}\Rightarrow\frac{z}{1}=\frac{x}{6}\Rightarrow z=\frac{x}{6}\)
\(\frac{y}{z}=\frac{3}{2}\Rightarrow\frac{y}{3}=\frac{z}{2}\Rightarrow y=\frac{3z}{2}\)
\(\Rightarrow\frac{t}{y}=\frac{\frac{4x}{3}}{\frac{3z}{2}}=\frac{4x}{3}\cdot\frac{2}{3z}=\frac{8x}{9z}=\frac{8x}{9\cdot\frac{x}{6}}=\frac{8x}{\frac{3x}{2}}=8x\cdot\frac{2}{3x}=\frac{16}{3}\)
Ta có: \(t:x=4:3=>\frac{t}{4}=\frac{x}{3}=>\frac{t}{8}=\frac{x}{6}\)
Tương tự
=> \(z=\frac{t}{8}\)
\(\frac{y}{3}=\frac{\frac{t}{8}}{2}=>\frac{y}{3}=\frac{t}{16}=>\frac{t}{y}=\frac{16}{3}\)
T=4; y=3
t:y=4:3=tứ chia tam = tám chia tư
=2
đ/s:.................
**** nha
bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
ta có: \(\frac{6x}{2}=\frac{y}{5}=\frac{3x}{1}\Rightarrow\frac{x}{1}=\frac{y}{15}\)
\(\frac{y}{3}=\frac{z}{2}\Rightarrow\frac{y}{15}=\frac{z}{10}\)
\(\Rightarrow\frac{x}{1}=\frac{y}{15}=\frac{z}{10}=\frac{2x}{2}=\frac{3y}{45}\)
ADTCDTSBN
có: \(\frac{2x}{2}=\frac{3y}{45}==\frac{2x+3y}{2+45}=\frac{3z}{47}\)
=> \(\Rightarrow\frac{z}{10}=\frac{3z}{47}\Rightarrow47z=30z\Rightarrow47z-30z=0\Rightarrow17z=0\Rightarrow z=0\)
=> \(\frac{x}{1}=\frac{y}{15}=\frac{z}{10}=\frac{0}{10}=0\)
=> x = 0; y = 0
KL: ...