\(A=\left(\sqrt{22}+7\sqrt{2}\right)\sqrt{30-7\sqrt{11}}\)

\(2A=\left(\sqrt{44}+7\sqrt{4}\right)\sqrt{60-2.7\sqrt{11}}\)

\(2A=\left(2\sqrt{11}+14\right)\sqrt{7^2-2.7\sqrt{11}+\left(\sqrt{11}\right)^2}\)

\(2A=\left(2\sqrt{11}+14\right)\sqrt{\left(7-\sqrt{11}\right)^2}\)

\(2A=\left(2\sqrt{11}+14\right)\left|7-\sqrt{11}\right|\)

\(2A=\left(2\sqrt{11}+14\right)\left(7-\sqrt{11}\right)\)

\(A=\left(7+\sqrt{11}\right)\left(7-\sqrt{11}\right)\)

\(A=49-11=38\)

49, \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{9+6\sqrt{2}+2}-\sqrt{9-6\sqrt{2}+2}\)

\(=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}=\left|3+\sqrt{2}\right|-\left|3-\sqrt{2}\right|=2\sqrt{2}\)

50, \(\sqrt{3+2\sqrt{2}}+\sqrt{\left(\sqrt{2}-2\right)^2}=\sqrt{2+2\sqrt{2}+1}+\sqrt{\left(2-\sqrt{2}\right)^2}\)

\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}=\left|\sqrt{2}+1\right|+\left|2-\sqrt{2}\right|=3\)

51, \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{5-2\sqrt{5}\sqrt{3}+3}-\sqrt{5-2\sqrt{5}\sqrt{3}+3}\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left|\sqrt{5}-\sqrt{3}\right|-\left|\sqrt{5}-\sqrt{3}\right|=-2\sqrt{3}\)

52, \(\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}-\sqrt{4-4\sqrt{2}+2}\)

\(=\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}=\left|\sqrt{2}+1\right|-\left|2-\sqrt{2}\right|=-1\)

47659:9

M giải luôn nha

\(\frac{1}{2}=\frac{x^2}{\left(y+1^2\right)}+\)\(\frac{y^2}{\left(x+1\right)^2}\) \(\ge\frac{2xy}{\left(x+1\right)\left(y+1\right)}\)

\(\Leftrightarrow\left(x+1\right)\left(y+1\right)\ge4xy\)

\(\Leftrightarrow3xy\le x+y+1\)

Dấu " = " xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}\frac{x^2}{\left(y+1\right)^2}=\frac{y^2}{\left(x+1\right)^2}\\3xy=x+y+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=y\\3x^2-2x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=y=1\left(tm\right)\\x=y=-\frac{1}{3}\left(tm\right)\end{cases}}\)

Vậy ( x ; y ) ......

a, tự tìm tự vẽ 

b, Ta có : \(\hept{\begin{cases}y=x^2\\y=-x+2\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2+x-2=0\\y=-x+2\end{cases}}\)

\(\left(1\right)\Rightarrow\Delta=1+8=9>0\)

\(x_1=\frac{-1-3}{2}=-2;x_2=\frac{-1+3}{2}=1\)

Với x = -2 => \(y=2+2=4\)

Với x = 1 => \(-1+2=1\)

Vậy giao điểm của 2 đồ thị trên là A ( -2 ; 4 ) ; B ( 1 ; 1 )

đk: \(-3\le x< 3\)

Ta có: 

\(A=\frac{x^2+5x+x\sqrt{9-x^2}+6}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}=\frac{\left(x^2+5x+6\right)+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)

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

\(=\frac{\left[\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right]\sqrt{x+3}}{\left(x\sqrt{3-x}+\left(x+2\right)\sqrt{3+x}\right)\sqrt{3-x}}\)

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

t post lại lên thôi Dung nhá :))

\(\hept{\begin{cases}2x-11y=-7\\10x+11y=31\end{cases}}\Leftrightarrow\hept{\begin{cases}12x=24\\y=\frac{2x+7}{11}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)

Ad làm hơi rắc rối , hệ số y đối nhau cộng lại thì dễ hơn kìa :v

\(P\left(\sqrt{x}+1\right)=4\sqrt{x}+7\)

\(P\sqrt{x}+P=4\sqrt{x}+7\)

\(P\sqrt{x}+P-4\sqrt{x}-7=0\)

\(P\sqrt{x}-4\sqrt{x}+P-7=0\)

\(\sqrt{x}\left(P-4\right)=7-P\)

\(\sqrt{x}=\frac{7-P}{P-4}\)\(\left(P\ne4\right)\)

Vì \(\sqrt{x}\ge0=>\frac{7-P}{P-4}\ge0\)

TH1 :\(\hept{\begin{cases}7-P\ge0\\P-4>0\end{cases}}\)\(\hept{\begin{cases}7\ge P\\P>4\end{cases}}\)

\(7\ge P>4\)

Vì P nguyên dương nên \(P\in\left(7,6,5\right)\)

\(\sqrt{x}\in\left(0,\frac{1}{2},2\right)\)

\(x\in\left(0,\frac{1}{4},4\right)\)

TH2:\(\hept{\begin{cases}7-P\le0\\P-4< 0\end{cases}}\)

\(\hept{\begin{cases}7\le P\\P< 4\end{cases}}\)

Vậy \(x\in\left(0,\frac{1}{4},2\right)\)

x= 4,0000000000000000

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

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

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

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

\(\frac{3}{\sqrt{x}-2}\)

\(b,\frac{3}{\sqrt{x}-2}=Q=1\)

\(3=\sqrt{x}-2\)

\(5=\sqrt{x}\)

\(\sqrt{25}=\sqrt{x}\)

\(< =>x=5\)