Tìm các số nguyên x,y,z biết:
\(\frac{-4}{8}=\frac{x}{-10}\frac{-7}{y}=\frac{z}{-24}\)
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-6/12=-1/2
suy ra : x=(-1/2)*8=-4
y=(-7)/(-1/2)=14
z=(-1/2)*(-18)=9
\(\frac{2^{12}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)
\(=\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6+3^5\right)}\)
\(=\frac{3^5-3^4}{3^6+3^5}=\frac{3^4.\left(3-1\right)}{3^5\left(3+1\right)}\)
\(=\frac{3^4.2}{3^5.4}=\frac{3^4.2}{3^4.3.4}=\frac{2}{12}=\frac{1}{6}\)
P/s: Hoq chắc ạ (: Ms lp 6 lm đại
\(\frac{x}{2}=\frac{y}{3}\)
\(\Leftrightarrow\frac{x}{8}=\frac{y}{12}\)(1)
\(\frac{y}{4}=\frac{z}{5}\)
\(\Leftrightarrow\frac{y}{12}=\frac{z}{15}\)(2)
Từ (1) (2)
\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)
\(\Rightarrow\hept{\begin{cases}x=2.8\\y=2.12\\z=2.15\end{cases}\Rightarrow}\hept{\begin{cases}x=16\\y=24\\z=30\end{cases}}\)
ta có: \(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{12}\)
\(\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\)
\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)
ADTCDTSBN
có: \(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)
\(\Rightarrow\frac{x}{8}=2\Rightarrow x=16\)
y/12 = 2 => y = 24
z/15 = 2 => z = 30
KL: x = 16; y=24;z=30
Ta có :
\(\frac{x}{2}=\frac{y}{3}\)\(\Rightarrow\)\(\frac{x}{8}=\frac{y}{12}\)
\(\frac{y}{4}=\frac{z}{5}\)\(\Rightarrow\)\(\frac{y}{12}=\frac{z}{15}\)
Suy ra : \(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)
Do đó :
\(\frac{x}{8}=2\)\(\Rightarrow\)\(x=2.8=16\)
\(\frac{y}{12}=2\)\(\Rightarrow\)\(y=2.12=24\)
\(\frac{z}{15}=2\)\(\Rightarrow\)\(z=2.15=30\)
Vậy \(x=16\)\(;\)\(y=24\) và \(z=30\)
Chúc bạn học tốt ~
\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{12}\)
\(\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\)
\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)
\(\Rightarrow\hept{\begin{cases}x=16\\y=24\\z=30\end{cases}}\)
a) Ta có \(\frac{x-1}{2}\)\(=\)\(\frac{y-2}{3}\)\(=\)\(\frac{z-3}{4}\)\(=\)\(\frac{2x-2}{4}\)\(=\)\(\frac{3y-6}{9}\)\(=\)\(\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}\)\(=\)\(\frac{\left(2x+3y-z\right)-5}{9}\)\(=\)\(\frac{50-5}{9}\)\(=\)5 Do đó x \(=\)5\(\times\)2\(+\)1\(=\)11 y\(=\)5\(\times\)3\(+\)2\(=\)17 z\(=\)5\(\times\)4\(+\)3\(=\)23
Ta có : \(\frac{-4}{8}=\frac{-1}{2}\)
+) \(\frac{-1}{2}=\frac{x}{-10}\Rightarrow x.2=-1.\left(-10\right)\)
\(\Rightarrow x.2=10\Rightarrow x=10\div2=5\in Z\)
+) \(\frac{-1}{2}=\frac{-7}{y}\Rightarrow-1.y=-7.2\)
\(\Rightarrow-1.y=-14\Rightarrow-y=-14\Rightarrow y=14\in Z\)
+) \(\frac{-1}{2}=\frac{z}{-24}\Rightarrow z.2=-1.\left(-24\right)\)
\(\Rightarrow z.2=24\Rightarrow z=24\div2=12\in Z\)
Vậy x = 5 ; y = 14 ; z = 12