Tìm \(x,\)\(y\)biết rằng \(x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\)
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\(x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}=4\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{1}{x}+\frac{1}{x^2}+y^2-2\cdot y\cdot\frac{1}{y}+\frac{1}{y^2}=0\)
\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{x}=0\\y-\frac{1}{y}=0\end{cases}\Leftrightarrow\hept{\begin{cases}\orbr{\begin{cases}x=-1\\x=1\end{cases}}\\\orbr{\begin{cases}y=-1\\y=1\end{cases}}\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{x}=0\\y-\frac{1}{y}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\orbr{\begin{cases}x=-1\\x=1\end{cases}}\\\orbr{\begin{cases}y=1\\y=-1\end{cases}}\end{cases}}\)\(x-\frac{1}{x}=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)
\(y-\frac{1}{y}=0\Leftrightarrow\orbr{\begin{cases}y=1\\y=-1\end{cases}}\)
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=> \(\left(x^2-2+\frac{1}{x^2}\right)+\left(y^2-2+\frac{1}{y^2}\right)=0\)
=> \(\left(x^2-2x.\frac{1}{x}+\frac{1}{x^2}\right)+\left(y^2-2.y.\frac{1}{y}+\frac{1}{y^2}\right)=0\)
=> \(\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)
<=> \(x-\frac{1}{x}=0;y-\frac{1}{y}=0\)
=> \(x^2=1;y^2=1\)
=> x = 1 hoặc -1
y = 1 hoặc -1
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1, ta co \(\frac{x}{5}=\frac{y}{6}=\frac{x}{20}=\frac{y}{24}\)
\(\frac{y}{8}=\frac{z}{7}=\frac{y}{24}=\frac{z}{21}\)
=>\(\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y-z}{20+24-21}=\frac{69}{23}=3\)
=>\(x=3\cdot20=60\)
\(y=3\cdot24=72\)
\(z=3\cdot21=63\)
3. ta co \(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}=\frac{x+y-z+t}{15-7+3-1}=\frac{10}{10}=1\)
=> \(x=1\cdot15=15\)
\(y=1\cdot7=7\)
\(z=1\cdot3=3\)
\(t=1\cdot1=1\)
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Bài 1:
\(\frac{x}{2}\) = \(\frac{y}{3}\) , \(\frac{y}{4}\) = \(\frac{z}{5}\) và x + y - z = 10
\(\frac{x}{2}\) = \(\frac{y}{3}\) --> \(\frac{x}{2.4}\) = \(\frac{y}{3.4}\) => \(\frac{x}{8}\) = \(\frac{y}{12}\)
\(\frac{y}{4}\) = \(\frac{z}{5}\) --> \(\frac{y}{4.3}\) = \(\frac{z}{5.3}\) => \(\frac{y}{12}\) = \(\frac{z}{15}\)
=> \(\frac{x}{8}=\frac{y}{12}\) = \(\frac{z}{15}\)
- 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
=> \(\frac{x}{8}\) = 2 --> x = 16
\(\frac{y}{12}=2\) --> y = 24
\(\frac{z}{15}=2\) --> z = 30
Vậy x = 16 ; y = 24 ; z = 30
Bài 2:
\(\frac{x}{2}=\frac{y}{5}\) và x . y = 10
Đặt \(\frac{x}{2}=\frac{y}{5}=k\)
Ta có: x = 2 . k ; y = 5 . k
x . y = 10 => 2k . 5k = 10
=> 10 . \(^{k^2}\) = 10
=> \(^{k^2}\) = 1 --> k = -1 hoặc k = 1
k = 1 ta có \(\frac{x}{2}=\frac{y}{5}=1\) --> x = 2 ; y = 5
k = -1 ta có \(\frac{x}{2}=\frac{y}{5}=-1\) --> x = -2 ; y = -5
Bài 1:
\(\frac{x}{2}=\frac{y}{3}\Rightarrow\)\(\frac{x}{8}=\frac{y}{12}\)
\(\frac{y}{4}=\frac{z}{5}\Rightarrow\)\(\frac{y}{12}=\frac{z}{15}\)
=> \(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)
Áp dụng tc của dãy tỉ số bằng nhau ta có:
\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)
=>\(\begin{cases}x=16\\y=24\\z=30\end{cases}\)
Bài 2:
Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow x=2k;y=5k\)
Có: xy=10
\(\Leftrightarrow2k\cdot5k=10\)
\(\Leftrightarrow k^2=1\Leftrightarrow\left[\begin{array}{nghiempt}k=1\\k=-1\end{array}\right.\)
Với k=1 thì x=2 ; y=5
Với k=-1 thì x=-2 ; y=-5
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Bài 1: \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12};\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{20}\)
=>\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2z}{18}=\frac{3y}{36}\)
Áp dụng tính chất của dãy tỉ số bằng nhau: \(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2z}{18}=\frac{3y}{36}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)
=>x=27;z=36;z=60
Bài 2: \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=5k\end{cases}}\Rightarrow xy=2k.5k=10k^2=40\Rightarrow k^2=4\Rightarrow\hept{\begin{cases}k=-2\\k=2\end{cases}}\)
+)k=-2 => x=-4;y=-5
+)k=2 => x=4;y=5
Vậy x=-4;y=-5 hoặc x=4;y=5
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(x-\(\frac{1}{2}\) )(y+\(\frac{1}{3}\) )(z-2)=0 và x+2=y+3=z+4
<=> x-\(\frac{1}{2}\)=0 hoặc y+\(\frac{1}{3}\)=0 hoặc z-2=0
+,với z-2=0
=>z=2
=>x+2=y+3=2+4
=>x+2=y+3=6
=. x=4;y=3
+,x-\(\frac{1}{2}\)=0
=>x=\(\frac{1}{2}\)
=>\(\frac{1}{2}\)+2=y+3=z+4
=>\(\frac{5}{2}\)=y+3=z+4
=>y=\(\frac{-1}{2}\);z=\(\frac{-3}{2}\)
+,với y+\(\frac{1}{3}\)=0
=>y=\(\frac{-1}{3}\)
=>x+2=\(\frac{-1}{3}\)+3=z+4
=>x+2=\(\frac{8}{3}\)=z+4
=>x=\(\frac{2}{3}\);z=\(\frac{4}{3}\)
Vậy khi x-\(\frac{1}{2}\)=0 thì x=\(\frac{1}{2}\);y=\(\frac{-1}{2}\);z=\(\frac{-3}{2}\)
khi y+\(\frac{1}{3}\)=0 thì x=\(\frac{2}{3}\);y=\(\frac{-1}{3}\);z=\(\frac{4}{3}\)
khi z-2=0 thì x=4;y=3;z=2
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\(x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\)
\(\Leftrightarrow\)\(\left(x^2+\frac{1}{x^2}-2\right)+\left(y^2+\frac{1}{y^2}-2\right)=0\)
\(\Leftrightarrow\)\(\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x-\frac{1}{x}=0\\y-\frac{1}{y}=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x^2=1\\y^2=1\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=\pm1\\y=\pm1\end{cases}}\)
\(^{x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4}\)
\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)+\left(y^2+\frac{1}{y^2}\right)-2-2=0\)
\(\Leftrightarrow\left(x^2-2.x.\frac{1}{x}+\frac{1}{x^2}\right)+\left(y^2-2.y.\frac{1}{y}+\frac{1}{y^2}\right)=0\)
\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)
Mặt khác: \(\left(x-\frac{1}{x}\right)^2\ge0\)\(\forall\)x\(\ne\)0
\(\left(y-\frac{1}{y}\right)^2\ge0\)\(\forall\)y \(\ne\)0
Từ hai điều trên \(\Rightarrow\)\(\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2\ge0\)\(\forall\)x,y \(\ne\)0
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