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

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x-y}{2\cdot3-5}=11\)

Do đó: x=33; y=55

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

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

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x-y}{2.3-5}=\dfrac{11}{1}=11\)

\(\dfrac{x}{3}=11\Rightarrow x=33\\ \dfrac{y}{5}=11\Rightarrow y=55\)

what

9/11 x 10/13 + 9/13 x 12/11 - 7/11

= 9/13 x 10/11 + 9/13 x 12/11 - 7/11

= 9/13 x (10/11 + 12/11) - 7/11

= 9/13 x 22/11 - 7/11

= 9/13 x 2 - 7/11

= 18/13 - 7/11

= 198/143 - 91/143

= 107/143

\(\dfrac{9}{11}\times\dfrac{10}{13}+\dfrac{9}{13}\times\dfrac{12}{11}-\dfrac{7}{11}=\dfrac{9}{13}\times\dfrac{10}{11}+\dfrac{9}{13}\times\dfrac{12}{11}-\dfrac{7}{11}=\dfrac{9}{13}\times\left(\dfrac{10}{11}+\dfrac{12}{11}\right)-\dfrac{7}{11}=\dfrac{9}{13}\times2-\dfrac{7}{11}=\dfrac{18}{13}-\dfrac{7}{11}=\dfrac{107}{143}\)

\(\dfrac{1}{x}-\dfrac{y}{11}=-\dfrac{2}{11}\)

\(\dfrac{1}{x}=-\dfrac{2}{11}+\dfrac{y}{11}\)

\(\dfrac{1}{x}=\dfrac{y-2}{11}\)

\(x\left(y-2\right)=11\)

\(\Rightarrow x,\left(y-2\right)\inƯ\left(11\right)=\left\{1,-1,11,-11\right\}\)

có bảng sau :

Vậy ...

\(\dfrac{1}{x}-\dfrac{y}{11}=-\dfrac{2}{11}\Rightarrow11-xy=-2x\)

\(\Leftrightarrow-2x+xy=11\Leftrightarrow x\left(-2+y\right)=11\)

\(\Rightarrow x;y-2\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

\(VT=\sqrt{14}-\sqrt{13}=\dfrac{1}{\sqrt{14}+\sqrt{13}}\)

\(VP=2\sqrt{3}-\sqrt{11}=\sqrt{12}-\sqrt{11}=\dfrac{1}{\sqrt{12}+\sqrt{11}}\)

Ta thấy: \(\sqrt{14}+\sqrt{13}>\sqrt{12}+\sqrt{11}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{14}+\sqrt{13}}< \dfrac{1}{\sqrt{12}+\sqrt{11}}\)

Hay \(VT< VP\)

Vậy \(\sqrt{14}-\sqrt{13}< 2\sqrt{3}-\sqrt{11}\)

Các bạn giúp mình với ạ

\(a,=\left(x+3\right)^3=\left(7+3\right)^3=10^3=1000\\ b,=\left(4-x\right)^3=\left(4-24\right)^3=\left(-20\right)^3=-8000\\ c,=\left(x-1\right)^3=\left(11-1\right)^3=10^3=1000\)