\(\frac{x}{5}=\frac{y}{4}\)biết x.y=20
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Đặt
x/5=y/4=k
khi đó:
x=5k
y=4k
Ta lại có:
x.y=4k.5k=20k^2=20
=> K=+-1
Khi k=1
Khi k=-1
Giải ra nhé
\(\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{2x}{2.3}=\frac{5y}{5.2}=\frac{2x}{6}=\frac{5y}{10}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{2x}{6}=\frac{5y}{10}=\frac{2x+5y}{6+10}\)\(=\frac{32}{16}=2\)
\(\frac{2x}{6}=2\Rightarrow2x=12\Rightarrow x=6\)
\(\frac{5y}{10}=2\Rightarrow5y=20\Rightarrow y=4\)
Vậy ..
ta có: x/3 =y/2 => 2x/6 = 5y/10
áp dụng tính chất dãy tỉ số bằng nhau ta có:
2x/6 = 5y/10 = 2x + 5y/ 6 + 10 = 32/16 = 2
=> x = 3 . 2 = 6 ; y = 2 . 2 = 4
vậy ( x , y ) = ( 6 ; 4 )
Ta có:
\(\frac{x}{4}=\frac{y}{7}\) và \(x.y=84.\)
Đặt \(\frac{x}{4}=\frac{y}{7}=k\Rightarrow\left\{{}\begin{matrix}x=4k\\y=7k\end{matrix}\right.\)
Có: \(x.y=84\)
=> \(4k.7k=84\)
=> \(28.k^2=84\)
=> \(k^2=84:28\)
=> \(k^2=3\)
=> \(k^2=\left(\pm\sqrt{3}\right)^2\)
=> \(k=\pm\sqrt{3}.\)
TH1: \(k=\sqrt{3}.\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.\sqrt{3}=4\sqrt{3}\\y=7.\sqrt{3}=7\sqrt{3}\end{matrix}\right.\)
TH2: \(k=-\sqrt{3}.\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.\left(-\sqrt{3}\right)=4\left(-\sqrt{3}\right)\\y=7.\left(-\sqrt{3}\right)=7\left(-\sqrt{3}\right)\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(4\sqrt{3};7\sqrt{3}\right);\left[4\left(-\sqrt{3}\right);7\left(-\sqrt{3}\right)\right].\)
Chúc bạn học tốt!
A = 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2
1/2^2 < 1/1*2
1/3^2 < 1/2*3
1/4^2 < 1/3*4
...
1/100^2 < 1/99*100
=> A < 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/99*100
=> A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
=> A < 1 - 1/100
=> A < 1
minh deo can ban k dau :((
\(a,\frac{1}{2}x+\frac{3}{5}(x-2)=3\)
\(\Rightarrow\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)
\(\Rightarrow\left[\frac{1}{2}+\frac{3}{5}\right]x=3+\frac{6}{5}\)
\(\Rightarrow\left[\frac{5}{10}+\frac{6}{10}\right]x=\frac{21}{5}\)
\(\Rightarrow\frac{11}{10}x=\frac{21}{5}\)
\(\Rightarrow x=\frac{21}{5}:\frac{11}{10}=\frac{21}{5}\cdot\frac{10}{11}=\frac{21}{1}\cdot\frac{2}{11}=\frac{42}{11}\)
Vậy x = 42/11
\(\frac{1}{x}-\frac{1}{y}=\frac{1}{x}.\frac{1}{y}\)
\(=>\frac{y-x}{xy}=\frac{1}{xy}\)
\(=>xy^2-x^2y=xy\)
\(=>xy^2-x^2y-xy=0\)
\(=>x.\left(y^2-xy-y\right)=0\)
\(=>\orbr{\begin{cases}x=0\\y^2-xy-y=0\end{cases}}\)
Ta thấy \(y^2-xy-y=0\)
\(=>y.\left(y-x-y\right)=0\)
\(=>\orbr{\begin{cases}y=0\left(2\right)\\y-y=0\end{cases}}\)
Từ 1 và 2 => x = y = 0
\(\frac{1}{x}-\frac{1}{y}=\frac{1}{x}.\frac{1}{y}\)
\(\Rightarrow\frac{y-x}{xy}=\frac{1}{xy}\)
\(\Rightarrow y-x=1\)
Vậy x,y có dạng \(\hept{\begin{cases}x=y-1\\y=x+1\end{cases}}\)với \(y\ne1;x\ne-1;x\ne0;y\ne0\)
1.
a) \(\frac{11}{2}-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3\)
\(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3-\frac{11}{2}\)
\(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=-\frac{5}{2}\)
\(\left|2x+-\frac{3}{2}\right|=-\frac{2}{3}:\left(-\frac{5}{2}\right)\)
\(\left|2x+-\frac{3}{2}\right|=\frac{4}{15}\)
\(\Rightarrow\left|2x+-\frac{3}{2}\right|\in\text{{}\frac{4}{15};-\frac{4}{15}\)}
Nếu, \(2x+\left(-\frac{3}{2}\right)=\frac{4}{15}\)
\(2x=\frac{53}{30}\)
\(x=\frac{53}{60}\)
Nếu, \(2x+\left(-\frac{3}{2}\right)=-\frac{4}{15}\)
\(2x=\frac{37}{30}\)
\(x=\frac{37}{60}\)
Vậy \(x\in\text{{}\frac{53}{60};\frac{37}{60}\)}
b) \(\left|\frac{2}{7}x-\frac{1}{5}\right|-\left|-x+\frac{4}{9}\right|=0\)
\(\left|\frac{2}{7}x-\frac{1}{5}\right|=\left|-x+\frac{4}{9}\right|\)
\(\Rightarrow\left|\frac{2}{7}x-\frac{1}{5}\right|\in\text{{}-x+\frac{4}{9};-\left(x+\frac{4}{9}\right)\)}
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-x+\frac{4}{9}\)
\(x=\frac{203}{405}\)
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-\left(-x+\frac{4}{9}\right)\)
\(\frac{2}{7}x-\frac{1}{5}=x-\frac{4}{9}\)
\(\frac{2}{7}x-x=\frac{1}{5}-\frac{4}{9}\)
\(-\frac{5}{7}x=-\frac{11}{45}\)
\(x=\frac{77}{225}\)
Vậy \(x\in\text{{}\frac{203}{405};\frac{77}{225}\)}
Áp dung tính chất của DTSBN,ta có :
\(\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}=\frac{x+y}{x+y-z}\)(1)
=>\(\frac{x+y}{z}=\frac{x+y}{x+y-z}\)=>z=x+y-z =>2z = x + y
Thay vào (1) =>\(\frac{2z}{z}=\frac{x}{y}\)=> \(2=\frac{x}{y}\)=>y=2x (ĐPCM)
\(\frac{x-2}{2}-\frac{1+x}{3}=\frac{4-3x}{4}-1\)
\(\Leftrightarrow\frac{3\left(x-2\right)-2\left(1+x\right)}{6}=\frac{4-3x-4}{4}\)
\(\Leftrightarrow\frac{3x-6-2-2x}{6}=-\frac{3x}{4}\)
\(\Leftrightarrow\frac{x-8}{6}=-\frac{3x}{4}\)
\(\Leftrightarrow4x-32=-18x\)
\(\Rightarrow x=\frac{16}{11}\)
\(x\cdot y=20\)\(\Rightarrow x=\frac{20}{y}\)
thay vào biểu thức ta đc
\(\frac{\frac{20}{y}}{5}=\frac{y}{4}\)
\(\frac{20}{y}\cdot4=5y\)
\(\frac{80}{y}=5y\)
\(80=5y^2\)
\(16=y^2\)
\(\Rightarrow y=4\)
\(\Rightarrow x=\frac{20}{4}=5\)
hok tốt
\(\frac{x}{5}=\frac{y}{4}\) biết x.y=20
Ta có:\(\frac{x}{5}=\frac{y}{4}=\frac{x.y}{5.4}=\frac{20}{20}=1\)
*\(\frac{x}{5}=1\Rightarrow x=5;\frac{y}{4}=1\Rightarrow y=4\)