x/2=y/5 =>x=2/5y

x.y=90 =>y.2/5y=90=>y²=225=>y=15

=>x=90:15

=>x=6

Vậy x=6,y=15

Ta có : \(\frac{x}{2}=\frac{y}{5}\Leftrightarrow2y=5x\Rightarrow y=\frac{2y}{5}\)

Thay \(y=\frac{2y}{5}\)vào biểu thức \(xy=90\); ta được : 

\(\frac{2y}{5}\cdot y=90\Leftrightarrow2y^2=90.5\Leftrightarrow2y^2=450\Leftrightarrow y^2=225\Leftrightarrow y=15\)

Vì \(y=15\Rightarrow x=\frac{2.15}{5}=6\)

Vậy \(x;y=\left[6;15\right]\)

a) \(\left(\frac{1}{3}\right)^2-\left(\frac{3}{4}\right)^3\left(\frac{4}{3}\right)^3=\frac{1}{3^2}-\left(\frac{3}{4}.\frac{4}{3}\right)^3\)

=\(\frac{1}{9}-1^3=\frac{1}{9}-1=\frac{8}{9}\)

b) \(\frac{5^2.5^3}{\left(-5\right)^4}=\frac{5^{2+3}}{5^4}=\frac{5^5}{5^4}=5\)

a) \(3x^2-10x+7\)

\(=3\left(x^2-\frac{10}{3}x+\frac{7}{3}\right)\)

\(=3\left(x^2-\frac{10}{3}x+\frac{25}{9}-\frac{4}{9}\right)\)

\(=3\left[\left(x-\frac{5}{3}\right)^2-\frac{4}{9}\right]\)

\(=3\left[\left(x-\frac{5}{3}\right)^2\right]-\frac{4}{3}\ge\frac{-4}{3}>0\)

b) \(4x^2+9x+5\)

\(=4x^2+9x+\frac{81}{16}-\frac{1}{16}\)

\(=\left(2x+\frac{9}{4}\right)^2-\frac{1}{16}\ge\frac{-1}{16}>0\)

a. \(\frac{x-5}{2000}+\frac{x-4}{1999}+\frac{x-3}{1998}=\frac{x-2}{1997}+\frac{x-1}{1996}+\frac{x}{1995}\)

\(\Leftrightarrow\left(\frac{x-5}{2000}+1\right)+\left(\frac{x-4}{1999}+1\right)+\left(\frac{x-3}{1998}+1\right)=\left(\frac{x-2}{1997}+1\right)+\left(\frac{x-1}{1996}+1\right)+\left(\frac{x}{1995}+1\right)\)

\(\Leftrightarrow\left(x+1995\right)\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}-\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\right)=0\)

\(\Leftrightarrow x+1995=0\)

\(\Leftrightarrow x=-1995\)

CÂU B BẠN LÀM TƯƠNG TỰ NHÉ

\(\frac{x}{3}=\frac{y}{5}\Rightarrow y=\frac{5x}{3}\)

thay vào biểu thức \(3x^2-y^2=8\)

\(\Rightarrow3x^2-\left(\frac{5y}{3}\right)^2=8\)

\(\Leftrightarrow3x^2-\frac{25y^2}{9}=8\)

\(\Leftrightarrow27x^2-25x^2=72\)

\(\Leftrightarrow2x^2=72\)

\(\Leftrightarrow x^2=36\Leftrightarrow\orbr{\begin{cases}x=-6\Rightarrow y=5.\left(-6\right):3=-10\\x=6\Rightarrow y=5.6:3=10\end{cases}}\)

\(\frac{x}{3}=\frac{y}{5}=\frac{3x^2}{27}=\frac{y^2}{25}=\frac{3x^2-y^2}{27-25}=\frac{8}{2}=4\)

\(\Rightarrow3x^2=4\cdot27=108\Rightarrow x^2=\frac{108}{3}=36\)

\(\Rightarrow x=\sqrt{36}=4,-4\)  

Chứng Minh Tương Tự \(y=\sqrt{25}=5,-5\)

Mà X,Y phải cùng dấu nên nếu X=4,Y=5       X=-4, Y=-5

\(7x=3y\Rightarrow\frac{x}{3}=\frac{y}{7}\Rightarrow\frac{4x}{12}=\frac{2y}{14}=\frac{4x-2y}{12-14}=\frac{-6}{2}=-3\)

\(\Rightarrow x=-3\cdot3=-9\)

\(\Rightarrow y=-3\cdot7=-21\)

7x=3y=>\(\frac{7x}{21}=\frac{3y}{21}\)=>\(\frac{x}{3}=\frac{y}{7}\)=>\(\frac{4x}{12}=\frac{2y}{14}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{4x}{12}=\frac{2y}{14}=\frac{4x+2y}{12+14}=\frac{6}{30}=\frac{1}{5}\)

=>\(\frac{x}{3}=\frac{1}{5}\)=>\(x=\frac{1}{5}.3=\frac{3}{5}\)

    \(\frac{y}{7}=\frac{1}{5}\)=>\(y=\frac{1}{5}.7=\frac{7}{5}\)

Vậy \(x=\frac{3}{5};y=\frac{7}{5}\)

Với \(7x=3y\Rightarrow x=\frac{3y}{7}\)

Thay \(x=\frac{3y}{7}\)vào biểu thức \(3x+y=80\); ta được : 

\(\frac{3y.3}{7}+y=80\Leftrightarrow9y+7y=80.7\Leftrightarrow16y=560\Leftrightarrow y=35\)

Vì \(y=35\Rightarrow x=\frac{3.35}{7}=15\)

Vậy ........ 

\(7x=3y\Rightarrow\frac{x}{3}=\frac{y}{7}\Rightarrow\frac{3x}{9}=\frac{y}{7}=\frac{3x+y}{9+7}=\frac{80}{16}=5\)

\(\Rightarrow x=5\cdot3=15\)

\(\Rightarrow y=5\cdot7=35\)