Khó quá, xé sách

\(a,\sqrt{2x+3}=3-\sqrt{5}\)

\(2x+3=14-6\sqrt{5}\)

\(2x=11-6\sqrt{5}\)

\(x=\frac{11-6\sqrt{5}}{2}\left(TM\right)\)

\(b,5+\sqrt{7x}=11+4\sqrt{7}\)

\(\sqrt{7x}=6+4\sqrt{7}\)

\(7x=148+48\sqrt{7}\)

\(x=\frac{148+48\sqrt{7}}{7}\)

\(c,5\sqrt{x}-10-x+2\sqrt{x}=4-x\)

\(7\sqrt{x}-10-x=4-x\)

\(7\sqrt{x}=14\)

\(\sqrt{x}=2\)

\(x=4\)

bạn ơi thiếu điều kiện

\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)

\(\Leftrightarrow\frac{\left(1+b^2\right)\left(1+ab\right)+\left(1+a^2\right)\left(1+ab\right)-2\left(1+a^2\right)\left(1+b^2\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Leftrightarrow ab^3+b^2+ab+1+a^3b+a^2+ab+1-2a^2b^2-2a^2-2b^2-2\ge0\)

\(\Leftrightarrow a^3b+ab^3-2a^2b^2-a^2-b^2+2ab\ge0\)

\(\Leftrightarrow ab\left(a-b\right)^2-\left(a-b\right)^2\ge0\)

\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)đúng do \(ab\ge1,\left(a-b\right)^2\ge0\).

Do biến đổi tương đương, bất đẳng thức cuối đúng nên bất đẳng thức cần chứng minh cũng đúng. 

Ta có đpcm. 

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

\(P=\frac{\sqrt{x}+\sqrt{y}+x\sqrt{y}+y\sqrt{x}+\sqrt{x}-\sqrt{y}+y\sqrt{x}-x\sqrt{y}}{1-xy}:\frac{1-xy+x+y+2xy}{1-xy}\)

\(P=\frac{2\sqrt{x}+2y\sqrt{x}}{1-xy}.\frac{1-xy}{1+xy+x+y}\)

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

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

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

\(\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}=2\)

dấu "=" xảy ra khi x =1

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

\(< =>MAX:P=1\)

Cảm ơn