đk: \(x,y\ge0\)

Đặt \(\hept{\begin{cases}\sqrt{x}+\sqrt{y}=a\\\sqrt{xy}=b\end{cases}}\) với \(a,b\ge0\)

\(\Rightarrow x+y=\left(\sqrt{x}+\sqrt{y}\right)^2-2\sqrt{xy}=a^2-2b\)

Khi đó \(HPT\Leftrightarrow\hept{\begin{cases}a+4b=16\\a^2-2b=10\end{cases}}\)

Đến đây thì dễ dàng rồi: \(HPT\Leftrightarrow\hept{\begin{cases}b=\frac{16-a}{4}\\a^2-2b=10\end{cases}}\)

\(\Leftrightarrow a^2-\frac{16-a}{2}=10\)

\(\Leftrightarrow2a^2+a-36=0\)

\(\Leftrightarrow\left(2a^2-8a\right)+\left(9a-36\right)=0\)

\(\Leftrightarrow\left(a-4\right)\left(2a+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=4\\a=-\frac{9}{2}\left(ktm\right)\end{cases}}\Rightarrow\hept{\begin{cases}a=4\\b=\frac{16-4}{4}=3\end{cases}}\)

Gọi \(\sqrt{x},\sqrt{y}\) là 2 nghiệm của PT \(t^2-4t+3=0\)

\(\Leftrightarrow\left(t-1\right)\left(t-3\right)=0\Leftrightarrow\orbr{\begin{cases}t=1\\t=3\end{cases}}\Leftrightarrow\left(\sqrt{x};\sqrt{y}\right)\in\left\{\left(1;3\right);\left(3;1\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(1;9\right);\left(9;1\right)\right\}\)

a) ĐK : \(x\ge1\)

pt <=> \(\sqrt{3^2\left(x-1\right)}-\frac{1}{2}\sqrt{2^2\left(x-1\right)}=2\)

<=> \(\left|3\right|\sqrt{x-1}-\frac{1}{2}\cdot\left|2\right|\sqrt{x-1}=2\)

<=> \(3\sqrt{x-1}-1\sqrt{x-1}=2\)

<=> \(2\sqrt{x-1}=2\)

<=> \(\sqrt{x-1}=1\)

<=> \(x-1=1\)=> \(x=2\)( tm )

b) \(3x-\sqrt{49-14x+x^2}=15\)

<=> \(\sqrt{x^2-14x+49}=3x-15\)

<=> \(\sqrt{\left(x-7\right)^2}=3x-15\)

<=> \(\left|x-7\right|=3x-15\)(1)

Với x < 7

(1) <=> 7 - x = 3x - 15

     <=> -x - 3x = -15 - 7

     <=> -4x = -22

     <=> x = 11/2 ( tm )

Với x ≥ 7

(1) <=> x - 7 = 3x - 15

      <=> x - 3x = -15 + 7

      <=> -2x = -8

      <=> x = 4 ( ktm )

Vậy x = 11/2

a) \(ĐKXĐ:x\ge1\)

\(\sqrt{9x-9}-\frac{1}{2}\sqrt{4x-4}=2\)

\(\Leftrightarrow\sqrt{9.\left(x-1\right)}-\frac{1}{2}.\sqrt{4\left(x-1\right)}=2\)

\(\Leftrightarrow3\sqrt{x-1}-\frac{1}{2}.2\sqrt{x-1}=2\)

\(\Leftrightarrow3\sqrt{x-1}-\sqrt{x-1}=2\)

\(\Leftrightarrow2\sqrt{x-1}=2\)

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)\(\Leftrightarrow x=2\)( thỏa mãn ĐKXĐ )

Vậy phương trình có nghiệm là \(x=2\)

b) \(3x-\sqrt{49-14x+x^2}=15\)

\(\Leftrightarrow3x-\sqrt{\left(7-x\right)^2}=15\)

\(\Leftrightarrow3x-\left|7-x\right|=15\)

+) TH1: Nếu \(7-x< 0\)\(\Leftrightarrow x>7\)

thì \(3x-\left(x-7\right)=15\)

\(\Leftrightarrow3x-x+7=15\)\(\Leftrightarrow2x=8\)

\(\Leftrightarrow x=4\)( không thỏa mãn )

+) TH2: Nếu \(7-x\ge0\)\(\Leftrightarrow x\le7\)

thì \(3x-\left(7-x\right)=15\)

\(\Leftrightarrow3x-7+x=15\)

\(\Leftrightarrow4x=22\)\(\Leftrightarrow x=\frac{22}{4}\)( thỏa mãn ĐKXĐ )

Vậy nghiệm của phương trình là \(x=\frac{22}{4}\)

Áp dụng các BĐT quen thuộc ta được:

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

\(x^2+xy+y^2=\frac{1}{2}\left(x^2+2xy+y^2\right)+\frac{1}{2}\left(x^2+y^2\right)\ge\frac{1}{2}\left(x+y\right)^2+\frac{1}{4}\left(x+y\right)^2\)

Khi đó: \(\sqrt{\frac{x^2+y^2}{2}}+\sqrt{\frac{x^2+xy+y^2}{3}}\ge\sqrt{\frac{\left(x+y\right)^2}{4}}+\sqrt{\frac{\left(x+y\right)^2}{4}}\)

\(=\frac{\left|x+y\right|}{2}+\frac{\left|x+y\right|}{2}=\left|x+y\right|\ge x+y\)

Dấu "=" xảy ra khi: \(x=y\ge0\)

=> đpcm

Ta có :

\(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)

\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)\)

\(=4-a=VP\)

=> đpcm

Bổ sung ĐK \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)dùm mình nhé ;-;

a) \(\sqrt{36}.\sqrt{121}+\sqrt[3]{-64}-\sqrt[3]{125}\)

\(=6.11+\left(-4\right)-5=66-9=57\)

b) \(\sqrt{75}+\sqrt{\left(\sqrt{3}-2\right)^2}-30\sqrt{\frac{3}{25}}\)

\(=\sqrt{25.3}+\left|\sqrt{3}-2\right|-30.\frac{\sqrt{3}}{\sqrt{25}}\)

\(=5\sqrt{3}+2-\sqrt{3}-30.\frac{\sqrt{3}}{5}\)

\(=5\sqrt{3}+2-\sqrt{3}-6\sqrt{3}=2-2\sqrt{3}\)

c) \(\sqrt{11-4\sqrt{7}}-\frac{12}{1+\sqrt{7}}=\sqrt{7-4\sqrt{7}+4}-\frac{12}{1+\sqrt{7}}\)

\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\frac{12}{1+\sqrt{7}}=\left|\sqrt{7}-2\right|-\frac{12}{1+\sqrt{7}}\)

\(=\left(\sqrt{7}-2\right)-\frac{12}{\sqrt{7}+1}=\frac{\left(\sqrt{7}-2\right)\left(\sqrt{7}+1\right)}{\sqrt{7}+1}-\frac{12}{\sqrt{7}+1}\)

\(=\frac{5-\sqrt{7}}{\sqrt{7}+1}-\frac{12}{\sqrt{7}+1}=\frac{-7-\sqrt{7}}{\sqrt{7}+1}\)

\(=\frac{-\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}+1}=-\sqrt{7}\)