a) Ta có: \(\frac{x}{5}=\frac{y}{4}\)

\(\Leftrightarrow\frac{x^2}{25}=\frac{y^2}{16}\)

Ta có: \(x^2-y^2=1\)

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

\(\frac{x^2}{25}=\frac{y^2}{16}=\frac{x^2-y^2}{25-16}=\frac{1}{9}\)

Do đó:

\(\left\{{}\begin{matrix}\frac{x^2}{25}=\frac{1}{9}\\\frac{y^2}{16}=\frac{1}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=\frac{25}{9}\\y^2=\frac{16}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{\frac{5}{3};-\frac{5}{3}\right\}\\y\in\left\{\frac{4}{3};-\frac{4}{3}\right\}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{5}{3};-\frac{5}{3}\right\}\) và \(y\in\left\{\frac{4}{3};-\frac{4}{3}\right\}\)

b) Ta có: \(\frac{x}{y}=\frac{2}{3}\)

\(\Leftrightarrow\frac{x}{2}=\frac{y}{3}\)

\(\Leftrightarrow\frac{x^2}{4}=\frac{y^2}{9}\)

Ta có: \(x^2+y^2=208\)

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

\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{x^2+y^2}{4+9}=\frac{208}{13}=16\)

Do đó:

\(\left\{{}\begin{matrix}\frac{x^2}{4}=16\\\frac{y^2}{9}=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=64\\y^2=144\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{8;-8\right\}\\y\in\left\{12;-12\right\}\end{matrix}\right.\)

Vậy: \(x\in\left\{8;-8\right\}\) và \(y\in\left\{12;-12\right\}\)

a/ Đặt: \(\frac{x}{5}=\frac{y}{4}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\end{matrix}\right.\)

Ta có: \(x^2-y^2=1\)

\(\Rightarrow\left(5k\right)^2-\left(4k\right)^2=1\)

\(\Rightarrow25k^2-16k^2=1\)

\(\Rightarrow k^2\left(25-16\right)=1\)

\(\Rightarrow k^29=1\)

\(\Rightarrow k^2=\frac{1}{9}\)

\(\Rightarrow k=\pm\sqrt{\frac{1}{9}}=\pm\frac{1}{3}\)

*Với: \(k=\frac{1}{3}\)

\(\left\{{}\begin{matrix}x=5k\\y=4k\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5.\frac{1}{3}=\frac{5}{3}\\y=4.\frac{1}{3}=\frac{4}{3}\end{matrix}\right.\)

*Với: \(k=-\frac{1}{3}\)

\(\left\{{}\begin{matrix}x=5k\\y=4k\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5.\left(-\frac{1}{3}\right)=-\frac{5}{3}\\y=4\left(-\frac{1}{3}\right)=-\frac{4}{3}\end{matrix}\right.\)

Vậy:..................

b/ \(\frac{x}{y}=\frac{2}{3}\Rightarrow\frac{x}{2}=\frac{y}{3}\)

Đặt: \(\frac{x}{2}=\frac{y}{3}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)

Ta có: \(x^2+y^2=208\)

\(\Rightarrow\left(2k\right)^2+\left(3k\right)^2=208\)

\(\Rightarrow4k^2+9k^2=208\)

\(\Rightarrow k^2\left(4+9\right)=208\)

\(\Rightarrow k^213=208\)

\(\Rightarrow k^2=208:13=16\)

\(\Rightarrow k=\pm4\)

*Với k = 4

\(\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2.4=8\\y=3.4=12\end{matrix}\right.\)

*Với k = -4

\(\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2.\left(-4\right)=-8\\y=3.\left(-4\right)=-12\end{matrix}\right.\)

Vậy...................

\(\Leftrightarrow\frac{xy-27}{9y}=\frac{1}{18}\)

\(\Leftrightarrow2xy-54=y\)

\(\Leftrightarrow y\left(2x-1\right)=54\)

Xét bảng ra

mik đag cần gấp các bn giải nhanh dùm mik nha

\(ĐK:\left\{{}\begin{matrix}x\ge0\\x\ne9\\x\ne25\end{matrix}\right.\)

\(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{2x}{9-x}-1\right):\left(\frac{\sqrt{x}-1}{x-3\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)-2x-\left(x-9\right)}{x-9}:\frac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\frac{-2x-3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{-\sqrt{x}+5}\)

\(=\frac{-\left(2\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\cdot\frac{\sqrt{x}}{5-\sqrt{x}}\)

\(=\frac{-\left(2\sqrt{x}-3\right)\cdot\sqrt{x}}{5-\sqrt{x}}=\frac{-2x+3\sqrt{x}}{5-\sqrt{x}}\)

ĐKXĐ :\(x\) > 0 , x\(\ne9\)

\(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{2x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}-1\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{2}{\sqrt{x}}\right)=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+2x-\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)=\)\(\frac{3\sqrt{x}-x+2x-9+x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x-3}\right)}{\sqrt{x}-1-2\sqrt{x}+6}=\frac{2x-3\sqrt{x}-9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{5-\sqrt{x}}=\frac{\left(\sqrt{x}+3\right)\left(2\sqrt{x}-3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{5-\sqrt{x}}=\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\sqrt{x}-5}\)

\(1.M=\dfrac{2\sqrt{y}}{x-y}+\dfrac{1}{\sqrt{x}-\sqrt{y}}+\dfrac{1}{\sqrt{x}+\sqrt{y}}=\dfrac{2\sqrt{y}+\sqrt{x}+\sqrt{y}+\sqrt{x}-\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{2}{\sqrt{x}-\sqrt{y}}\left(x\ge0;y\ge0;x\ne y\right)\)

\(2a.N=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\left(x\ge0;x\ne9\right)\)

b. Thay x = 49 ( thỏa mãn ĐKXĐ ) vào biểu thức N , ta có :

\(N=\dfrac{3\sqrt{49}}{\sqrt{49}-3}=\dfrac{21}{4}\)

\(3.\dfrac{\sqrt{5}}{1-\sqrt{3}}-\sqrt{3}+\dfrac{1}{1+\sqrt{3}}=\dfrac{5\left(1+\sqrt{3}\right)+1-\sqrt{3}}{1-3}-\sqrt{3}=\dfrac{6+4\sqrt{3}+2\sqrt{3}}{-2}=\dfrac{6\left(\sqrt{3}+1\right)}{-2}=-3\left(\sqrt{3}+1\right)\)

\(P=\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{-x+x\sqrt{x}+6}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(P=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{x+\sqrt{x}-2}+\frac{-x+x\sqrt{x}+6}{x+\sqrt{x}-2}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{x+\sqrt{x}-2}\)

\(P=\frac{x-\sqrt{x}-x+x\sqrt{x}+6-x-3\sqrt{x}-2}{x+\sqrt{x}-2}\)

\(P=\frac{-x+x\sqrt{x}+4-4\sqrt{x}}{x+\sqrt{x}-2}\)

\(P=\frac{x\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{\left(\sqrt{x}-1\right)\left(x-4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\)

\(P=\sqrt{x}-2\)

@Trần Ngọc Thảo

\(Q=\frac{\left(x+27\right)\cdot P}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\ge6\)

\(\Leftrightarrow Q=\frac{\left(x+27\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\ge6\)

\(\Leftrightarrow\frac{x+27}{\sqrt{x}+3}\ge6\)

\(\Leftrightarrow x+27\ge6\left(\sqrt{x}+3\right)\)

\(\Leftrightarrow x+27-6\sqrt{x}-18\ge0\)

\(\Leftrightarrow x-6\sqrt{x}+9\ge0\)

\(\Leftrightarrow\left(\sqrt{x}-3\right)^2\ge0\)( luôn đúng )

Vậy \(x\ge0\)thì bất phương trình luôn đúng

\(\)