a)\(\frac{\sqrt{a-2\sqrt{ab}+b}}{\sqrt{\sqrt{a}-\sqrt{b}}}=\frac{\sqrt{\left(\sqrt{a}-\sqrt{b}\right)^2}}{\sqrt{\sqrt{a}-\sqrt{b}}}=\sqrt{a}-\sqrt{b}\) (vì a > b > 0)

b) \(\frac{\sqrt{x-3}}{\sqrt{\sqrt{x}+\sqrt{3}}}:\frac{\sqrt{\sqrt{x}-\sqrt{3}}}{\sqrt{3}}=\frac{\sqrt{3}.\sqrt{x-3}}{\sqrt{\left(\sqrt{x}+\sqrt{3}\right)\left(\sqrt{x}-\sqrt{3}\right)}}=\frac{\sqrt{3\left(x-3\right)}}{\sqrt{x-3}}=\sqrt{3}\)

c) \(2y^2\sqrt{\frac{x^4}{4y^2}}=2y^2\cdot\frac{x^2}{-2y}=-x^2y\) (vì y < 0)

d) \(\frac{y}{x}\cdot\sqrt{\frac{x^2}{y^4}}=\frac{y}{x}\cdot\frac{x}{y^2}=\frac{1}{y}\)(vì x > 0)

e) \(5xy\cdot\sqrt{\frac{25x^2}{y^6}}=5xy\cdot\frac{-5x}{y^3}=\frac{-25x^2}{y^2}\) (Vì x < 0, y > 0)

Với x ≥ 0

⇒ √xx ≤ 2√xx

⇔ √xx + 1 ≤ 2√xx + 1

Vậy A lớn nhất khi dấu bằng xảy ra √xx = 2√xx

⇔ x = 0

MaxAMaxA=1 khi x = 0

Ta có: \(A-1=\frac{\sqrt{x}+1}{2\sqrt{x}+1}-1=\frac{\sqrt{x}+1-2\sqrt{x}-1}{2\sqrt{x}+1}=-\frac{\sqrt{x}}{2\sqrt{x}+1}\le0\)

\(\Rightarrow A\le1\)

Vậy A_max=1 \(\Leftrightarrow x=0\)

x² + y²  \(\ge\)2

<=> x² + y² + 1  \(\ge\)x + y + xy

<=> 2x² + 2y² + 2  \(\ge\)2x + 2y + 2xy

<=> 2x² + 2y² + 2 - 2x - 2y - 2xy  \(\ge\)0

<=> (x² - 2xy + y²) + (x² - 2x + 1) + (y² - 2y + 1)  \(\ge\)0

<=> (x - y)² + (x - 1)² + (y - 1)²  \(\ge\)0 (đúng) (Dấu "=" xảy ra <=> x = y = 1)

=> BĐT được chứng minh

\(=\sqrt{7}-\sqrt{4-2.2\sqrt{7}+7}\)

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

\(=\sqrt{7}-\sqrt{7}+2=2\)

\(\sqrt{7}-\sqrt{11-4\sqrt{7}}\)

\(=\sqrt{7}-\sqrt{\sqrt{7}^2-2\cdot\sqrt{7}\cdot2+2^2}\)

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

\(=\sqrt{7}-\left|\sqrt{7}-2\right|\)

\(=\sqrt{7}-\sqrt{7}+2=2\)

ΔABC vuông tại B, áp dụng hệ thức lượng trong tam giác vuông ta có :

\(\frac{1}{BH^2}=\frac{1}{AB^2}+\frac{1}{BC^2}\Rightarrow BH=\sqrt{\frac{\left(AB\cdot BC\right)^2}{AB^2+BC^2}}=7,2\left(cm\right)\)

a) Xét ΔHBA có ^AHB = 90⁰ ( BH ⊥ AC ) => ΔHBA vuông tại H

Khi đó ta có : \(\sin A=\frac{BH}{AB}=\frac{4}{5};\cos A=\frac{AH}{AB}=\frac{\sqrt{AB^2-BH^2}}{AB}=\frac{3}{5};\tan A=\frac{BH}{AH}=\frac{BH}{\sqrt{AB^2-BH^2}}=\frac{4}{3}\)

(bonus cho cotgA) : \(\cot A=\frac{AH}{BH}=\frac{\sqrt{AB^2-BH^2}}{BH}=\frac{3}{4}\)

b) Vì ΔHBA vuông tại H (cmt) => ^A + ^ABH = 90⁰

Khi đó : \(\sin\widehat{ABH}=\cos A=\frac{3}{5};\cos\widehat{ABH}=\sin A=\frac{4}{5};\tan\widehat{ABH}=\cot A=\frac{3}{4};\cot\widehat{ABH}=\tan A=\frac{4}{3}\)

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

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

\(\left|x^2+3\right|=\left|x+3\right|\)

\(\orbr{\begin{cases}x^2+3=x+3\\x^2+3=-x-3\end{cases}\orbr{\begin{cases}x^2=x\\x^2=-x-6\end{cases}}}\)

\(\orbr{\begin{cases}x=1\\x^2+x+6=0\end{cases}}\)

\(\orbr{\begin{cases}x=1\\x^2+x+\frac{1}{2}+\frac{11}{2}=0\end{cases}}\)

\(\orbr{\begin{cases}x=1\left(TM\right)\\\left(x+\frac{1}{4}\right)^2+\frac{11}{2}=0\left(ktm\right)\end{cases}}\)

\(b,\sqrt{3x+2}^2=\sqrt{x}^2\)

\(\left|3x+2\right|=\left|x\right|\)

\(\orbr{\begin{cases}3x+2=x\\3x+2=-x\end{cases}\orbr{\begin{cases}2x+2=0\\4x+2=0\end{cases}\orbr{\begin{cases}x=-1\left(TM\right)\\x=-\frac{1}{2}\left(TM\right)\end{cases}}}}\)

\(c,\sqrt{3-x}^2=\sqrt{x-3}^2\)

\(\left|3-x\right|=\left|x-3\right|\)

\(\orbr{\begin{cases}3-x=x-3\\3-x=3-x\end{cases}}\)

\(\orbr{\begin{cases}x=3\left(TM\right)\\3=3-0x\left(KTM\right)\end{cases}}\)

dễ vậy tự làm đi

\(1=\sqrt{1}\)   

\(\sqrt{1}< \sqrt{2}\)   

\(\Rightarrow1< \sqrt{2}\) 

\(2=1+1\)   

\(1+1< \sqrt{2}+1\)   

\(\Rightarrow2< \sqrt{2}+1\)   

\(7=\sqrt{49}\)   

\(5\sqrt{2}=\sqrt{50}\)   

\(\sqrt{49}< \sqrt{50}\Rightarrow7< 5\sqrt{2}\)

\(7=\sqrt{49}\)   

\(\sqrt{49}>\sqrt{47}\)   

\(\Rightarrow7>\sqrt{47}\)   

\(1=2-1=\sqrt{4}-1\)   

\(\sqrt{4}-1>\sqrt{3}-1\)   

\(\Rightarrow1>\sqrt{3}-1\)   

\(2\sqrt{31}=\sqrt{124}\)   

\(10=\sqrt{100}\)   

\(\sqrt{124}>\sqrt{100}\Rightarrow2\sqrt{31}>10\)