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Gọi x/3=y/7=k nên x=3k; y=7k
mà x*y=84
nên 3k*7k=84
21*k2=84
k2=84/21
k2=4 nên k=2 hoặc k=-2
Nếu k=2 thì x=3*2=6; y=7*2=14
Nếu k=-2 thì x=-2*3=-6; y=-2*7=-14
Vậy cặp số x;y là: (6;14);(-6;-14)
a) Ta có:
\(\frac{x}{3}=\frac{y}{7}\) và \(x.y=84.\)
Đặt \(\frac{x}{3}=\frac{y}{7}=k.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=7k\end{matrix}\right.\)
+ Có: \(x.y=84\)
\(\Rightarrow3k.7k=84\)
\(\Rightarrow21.k^2=84\)
\(\Rightarrow k^2=84:21\)
\(\Rightarrow k^2=4\)
\(\Rightarrow k^2=\left(\pm2\right)^2\)
\(\Rightarrow k=\pm2.\)
+ TH1: \(k=2.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.2=6\\y=7.2=14\end{matrix}\right.\)
+ TH2: \(k=-2.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.\left(-2\right)=-6\\y=7.\left(-2\right)=-14\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(6;14\right),\left(-6;-14\right).\)
Chúc bạn học tốt!
a )
Ta có :
\(\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
\(\Rightarrow\frac{4\left(1+5y\right)}{20x}=\frac{5\left(1+7y\right)}{20x}\)
\(\Rightarrow\frac{4+20y}{20x}=\frac{5+35y}{20x}\)
\(\Rightarrow4+20y=5+35y\)
\(\Rightarrow35y-20y=4-5\)
\(\Rightarrow15y=4-5\)
\(\Rightarrow15y=-1\)
\(\Rightarrow y=-\frac{1}{15}\)
Lại có :
\(\frac{1+3y}{12}=\frac{1+5y}{5x}\)
\(\Rightarrow\frac{1+3.-\frac{1}{15}}{12}=\frac{1+5.-\frac{1}{15}}{5x}\)
\(\Rightarrow\frac{1-\frac{1}{5}}{12}=\frac{1-\frac{1}{3}}{5x}\)
\(\Rightarrow\frac{4}{5}:12=\frac{4}{3}:5x\)
\(\Rightarrow\frac{1}{15}=\frac{4}{3}:5x\)
\(\Rightarrow5x=\frac{4}{3}:\frac{1}{15}\)
\(\Rightarrow5x=20\)
\(\Rightarrow x=4\)
Vậy \(x=4;y=-\frac{1}{15}\)
a) Xét \(\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
\(\Rightarrow\frac{4x\left(1+5y\right)}{20x}=\frac{5\left(1+7y\right)}{20x}\)
\(\Rightarrow4x\left(1+5y\right)=5\left(1+7y\right)\)
\(\Rightarrow4+20y=5+35y\)
\(\Rightarrow35y-20y=4-5\)
\(\Rightarrow15y=-1\)
\(\Rightarrow y=\frac{-1}{15}\)
Xét \(\frac{1+3y}{12}=\frac{1+5y}{5x}\)
\(\Rightarrow\frac{1+3.\frac{-1}{15}}{12}=\frac{1+5.\frac{-1}{15}}{5x}\)
\(\Rightarrow\frac{1+\frac{-1}{5}}{12}=\frac{1+\frac{-1}{3}}{5x}\)
\(\Rightarrow\frac{\frac{4}{5}}{12}=\frac{\frac{2}{3}}{5x}\)
\(\Rightarrow\frac{4}{5}:12=\frac{2}{3}:5x\)
\(\Rightarrow\frac{1}{15}=\frac{2}{3}:5x\)
\(\Rightarrow5x=\frac{2}{3}:\frac{1}{15}\)
\(\Rightarrow5x=\frac{30}{3}\)
\(\Rightarrow x=\frac{30}{3}:5\)
\(\Rightarrow x=\frac{30}{3}.\frac{1}{5}\)
\(\Rightarrow x=2\)
Vậy x = 2 ; y = \(\frac{-1}{15}\)
\(a)\) \(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+15}=0\)
\(\Leftrightarrow\)\(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+5}.\left(x-3\right)^{10}=0\)
\(\Leftrightarrow\)\(\left(x-3\right)^{x+5}.\left[1-\left(x-3\right)^{10}\right]=0\)
Trường hợp 1 :
\(\left(x-3\right)^{x+5}=0\)
\(\Leftrightarrow\)\(\left(x-3\right)^{x+5}=0^{x+5}\)
\(\Leftrightarrow\)\(x-3=0\)
\(\Leftrightarrow\)\(x=3\)
Trường hợp 2 :
\(1-\left(x-3\right)^{10}=0\)
\(\Leftrightarrow\)\(\left(x-3\right)^{10}=1\)
\(\Leftrightarrow\)\(\left(x-3\right)^{10}=1^{10}\)
\(\Leftrightarrow\)\(x-3=1\)
\(\Leftrightarrow\)\(x=4\)
Vậy \(x=3\) hoặc \(x=4\)
Chúc bạn học tốt ~
a) Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\)
\(\Leftrightarrow\hept{\begin{cases}x=3k\\y=4k\\z=5k\end{cases}}\)
Khi đó : \(\left(3k\right)^2+2.\left(4k\right)^2+4.\left(5k\right)^2=141\)
\(\Leftrightarrow141k^2=141\)
\(\Leftrightarrow k^2=1\)
\(\Leftrightarrow k=\pm1\)
TH1 \(\hept{\begin{cases}x=3\\y=4\\z=5\end{cases}}\)
TH2 \(\hept{\begin{cases}x=-3\\y=-4\\z=-5\end{cases}}\)
Vậy.....
a)
Theo đề bài ta có: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\) và \(x^2+2y^2+4z^2=141\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{x^2}{3^2}=\frac{2y^2}{2.4^2}=\frac{4z^2}{4.5^2}=\frac{x^2+2y^2+4z^2}{9+32+100}=\frac{141}{141}=1\)
\(\frac{x}{3}=1\Rightarrow x=3.1=3\)
\(\frac{y}{4}=1\Rightarrow y=4.1=4\)
\(\frac{z}{5}=1\Rightarrow z=5.1=5\)
Vậy x = 3
y=4
z=5
c)\(\left|2x+3\right|=x+2\)
Đk:\(x+2\ge0\Rightarrow x\ge-2\)
TH1:2x+3=x+2
\(\Rightarrow2x-x=2-3\)
\(\Rightarrow x=-1\)(Thỏa mãn đk )
TH2:2x+3=-x-2
\(\Rightarrow2x+x=-2+3\)
\(\Rightarrow3x=1\)
\(\Rightarrow x=\frac{1}{3}\)(Thỏa mãn đk)
Vậy x=-1 hoặc x=1/3
a) \(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+15}=0\)
\(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+5}\cdot\left(x-3\right)^{10}=0\)
\(\left(x-3\right)^{x+5}\cdot\left[1-\left(x-3\right)^{10}\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^{x+5}=0\\1-\left(x-3\right)^{10}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^{10}=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\\left(x-3\right)^{10}=\left(\pm1\right)^{10}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=\left\{4;2\right\}\end{cases}}\)
Vậy........
\(\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Leftrightarrow7\left(x-3\right)=5\left(x+5\right)\)
\(\Leftrightarrow7x-21=5x+25\)
\(\Leftrightarrow7x-5x=25+21\)
\(\Leftrightarrow2x=46\)
\(\Leftrightarrow x=23\)
Vậy ..........