Bạn thêm góc alpha nhọn nhé, vì chương trình lớp 9 mới học góc nhọn còn chẳng hạn như góc alpha tù thì cos alpha<0.

a. Ta có: 

\(\left(\sin\alpha+\cos\alpha\right)^2=\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha=1+2\frac{1}{4}=\frac{3}{2}\)

\(\Rightarrow\sin\alpha+\cos\alpha=\sqrt{\frac{3}{2}}\)

b. \(\sin\alpha+\cos\alpha\le\sqrt{2\left(\sin^2\alpha+\cos^2\alpha\right)}=\sqrt{2}\)

Đẳng thức xảy ra \(\Leftrightarrow\sin\alpha=\cos\alpha\Leftrightarrow\sin\alpha=\sin\left(90-\alpha\right)\Leftrightarrow\alpha=90-\alpha\Leftrightarrow\alpha=45\)

(Bạn thêm kí hiệu độ nữa nhé)

Kết luận:

Làm đông đá 

BẢO

ta có : 

\(B=\frac{\sqrt{x}\left(1-x\right)^2}{1+x}:\left[\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}-1}+\sqrt{x}\right).\left(\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}-\sqrt{x}\right)\right]\)

\(=\frac{\sqrt{x}\left(1-x\right)^2}{1+x}:\left[\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2\right]=\frac{\sqrt{x}\left(1-x\right)^2}{1+x}:\left[\left(1-x\right)^2\right]==\frac{\sqrt{x}}{1+x}\)

ta có :

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

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

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

\(-4\sqrt{\frac{\sqrt{3}-1}{2+\sqrt{3}}}\)

\(-4\sqrt{\frac{\left(\sqrt{3}-1\right)\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}\)

\(=-4\sqrt{\frac{2\sqrt{3}-2-3+\sqrt{3}}{4-3}}\)

\(=-\left(4\sqrt{3\sqrt{3}-5}\right)\)

\(=-\sqrt{48\sqrt{3}-80}\)

là sao bạn

ròi bạn sẽ biết 

\(\sqrt{x-3}+\sqrt{y-5}+\sqrt{z-1}=20-\frac{4}{\sqrt{x-3}}-\frac{9}{\sqrt{y-5}}-\frac{25}{\sqrt{z-1}}\)(ĐK: \(x>3,y>5,z>1\))

\(\Leftrightarrow\sqrt{x-3}+\frac{4}{\sqrt{x-3}}+\sqrt{y-5}+\frac{9}{\sqrt{y-5}}+\sqrt{z-1}+\frac{25}{\sqrt{z-1}}=20\)

Ta có: 

\(\sqrt{x-3}+\frac{4}{\sqrt{x-3}}\ge2\sqrt{\sqrt{x-3}.\frac{4}{\sqrt{x-3}}}=4\)

\(\sqrt{y-5}+\frac{9}{\sqrt{y-5}}\ge2\sqrt{\sqrt{y-5}.\frac{9}{\sqrt{y-5}}}=6\)

\(\sqrt{z-1}+\frac{25}{\sqrt{z-1}}\ge2\sqrt{\sqrt{z-1}.\frac{25}{\sqrt{z-1}}}=10\)

Do đó \(\sqrt{x-3}+\frac{4}{\sqrt{x-3}}+\sqrt{y-5}+\frac{9}{\sqrt{y-5}}+\sqrt{z-1}+\frac{25}{\sqrt{z-1}}\ge20\)

Dấu \(=\)xảy ra khi \(\hept{\begin{cases}\sqrt{x-3}=\frac{4}{\sqrt{x-3}}\\\sqrt{y-5}=\frac{9}{\sqrt{y-5}}\\\sqrt{z-1}=\frac{25}{\sqrt{z-1}}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=7\\y=14\\z=26\end{cases}}\)(thỏa mãn)

nham = 1300

\(\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)

\(=\frac{\left(\sqrt{x}+1\right)\left(2-\sqrt{x}\right)+2\sqrt{x}\left(2-\sqrt{x}\right)+2+5\sqrt{x}}{4-x}\)

\(=\frac{2\sqrt{x}+2-x-\sqrt{x}+4\sqrt{x}-2x+2+5\sqrt{x}}{4-x}\)

\(=\frac{10\sqrt{x}-3x+4}{4-x}\)

p/s : phần a do \(\sqrt{x}+2>0\)nên chỉ cần xét ước dương thôi => trình bày như mình cũng được mà thừa quá, ngại sửa, nhưng như nào cũng đúng nhé 

a,ĐK x>=0 \(D=\frac{6}{\sqrt{x}+2}\Rightarrow\sqrt{x}+2\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

b, ĐK : x>=0 \(E=\frac{\sqrt{x}-3}{\sqrt{x}+4}=\frac{\sqrt{x}+4-7}{\sqrt{x}+4}=1-\frac{7}{\sqrt{x}+4}\)

\(\Rightarrow\sqrt{x}+4\inƯ\left(7\right)=\left\{1;7\right\}\)

c, ĐK : x> = 0 \(F=\frac{2\sqrt{x}-4}{\sqrt{x}+5}=\frac{2\left(\sqrt{x}+5\right)-14}{\sqrt{x}+5}=2-\frac{14}{\sqrt{x}+5}\)

\(\Rightarrow\sqrt{x}+5\inƯ\left(14\right)=\left\{1;2;7;14\right\}\)