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

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

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

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

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

\(=\frac{a^2+a\sqrt{a}+2\text{a}+\sqrt{a}+1}{a-2\sqrt{a}+1}\)

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

câu a đã có người làm rồi nên mình không làm

tick cho mình nha

\(VT=\frac{2}{\sqrt{ab}}:\left(\frac{\sqrt{b}-\sqrt{a}}{\sqrt{ab}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

\(=\frac{2}{\sqrt{ab}}.\frac{\left(\sqrt{ab}\right)^2}{\left(\sqrt{b}-\sqrt{a}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

\(=\frac{2\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\) ( vì \(\left(\sqrt{b}-\sqrt{a}\right)^2=\left(\sqrt{a}-\sqrt{b}\right)^2\) )

\(=\frac{2\sqrt{ab}-a-b}{\left(\sqrt{a}-\sqrt{b}\right)^2}=\frac{-\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\) ???? đề sai chăng

3¹¹<4¹¹; 4¹¹=(2²)¹¹=2²²

17¹⁴>16¹⁴; 16¹⁴=(2⁴)¹⁴=2⁵⁶

Vì 2²²<2⁵⁶=> 4¹¹<16¹⁴

=>3¹¹<17¹⁴

ĐKXĐ :  -1 <= x <= 3 

XH : \(\left(-x^2+4x+12\right)-\left(x^2+2x+3\right)=2x+9>0\)

=> VT > 0 

VÌ -1 <=x <=3  => VT = \(\sqrt{x+2}\sqrt{6-x}-\sqrt{x+1}.\sqrt{3-x}\)

Áp dụng BĐT \(\left(ab-cd\right)^2\le\left(a^2-c^2\right)\left(b^2-d^2\right)\) ta có :

\(VT^2=\left(\sqrt{x+2}\sqrt{6-x}-\sqrt{x+1}\sqrt{3-x}\right)^2\ge\left(x+2-x-1\right)\left(6-x-3+x\right)=1.3=3\)

=> VT \(\ge\sqrt{3}\) dấu bằng xảy ra khi \(\left(x+2\right)\left(6-x\right)=\left(x+1\right)\left(3-x\right)\) <=> x = 0 

VP = \(\sqrt{3}-x^2\le\sqrt{3}\)

Dấu bằng xảy ra khi x = 0 

Để VT bằng VP => x = 0 

không cần điều kiện cũng được, giải ra x = 1 hoặc x = 7, lấy ra thay lại xem pt có xác định và thỏa không là được

RA BẬC CAO SAO GIẢI

\(\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}=\sqrt{3}-x^2\)

\(\Leftrightarrow\sqrt{-x^2+4x+12}=\sqrt{3}-x^2+\sqrt{-x^2+2x+3}\)

\(VP=\sqrt{-x^2+4x+12}=\sqrt{-\left(x-2\right)^2+16}\le4\)

\(VT=\sqrt{3}-x^2+\sqrt{-x^2+2x+3}=\sqrt{3}-x^2+\sqrt{-\left(x-1\right)^2+4}\)

\(\le\sqrt{3}+2