vì x/5=y/4=> 4x=5y mà x.y=20=>

x=5,y=4

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 ) 

Quá dễ

X/3 =y/2 suy ra 2x/3.2 =5y/2.5

=32/16 =2

X/3=2 suy ra x = 6

Y/2=2 suy ra y =4 

  mik nhé

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!

help me

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 \(y\ne1;x\ne-1;x\ne0;y\ne0\)

Ta có\(\frac{1}{x}-\frac{1}{y}=\frac{1}{x}.\frac{1}{y}\)

\(\Leftrightarrow\frac{y-x}{xy}=\frac{1}{xy}\)

\(\Leftrightarrow y-x=1\Rightarrow y=x+1\)

Vậy..................

Ko có giá trị cụ thể nha

\(\frac{4}{x}=\frac{-y}{6}=0,5\)

\(=>\frac{4}{x}=\frac{-y}{6}=\frac{1}{2}\)

\(\frac{4}{x}=\frac{-y}{6}=\frac{3}{6}\)

=>y=-3

\(\frac{-4}{-x}=\frac{-3}{6}\)

\(\frac{-12}{-3x}=\frac{-12}{24}\)

=>-3x=24

    x=24:(-3)

   Vậy   x=-8

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)

\(\frac{x-2}{2}-\frac{1+x}{3}=\frac{4-3x}{4}-1\)

\(\Leftrightarrow\frac{3\left(x-2\right)-2\left(1+x\right)}{6}=\frac{4-3x-4}{4}\)

\(\Leftrightarrow\frac{3x-6-2-2x}{6}=-\frac{3x}{4}\)

\(\Leftrightarrow\frac{x-8}{6}=-\frac{3x}{4}\)

\(\Leftrightarrow4x-32=-18x\)

\(\Rightarrow x=\frac{16}{11}\)