1) \(A=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}:\dfrac{\sqrt{x}-1}{5}\)

        \(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{5}{\sqrt{x}-1}\) \(=\dfrac{5}{x+\sqrt{x}+1}\)

2) Ta thấy \(x+\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}+1\right)+1>1\forall x\)

\(\Rightarrow A< 5\)

a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)

\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{1}\)

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

b: Để A<0 thì x-1<0

hay x<1

Kết hợp ĐKXĐ, ta được: 0<x<1

với \(x\ge0\) ta có :\(D=\dfrac{3\sqrt{x+7}}{\sqrt{x}+2}=\dfrac{3\left(\sqrt{x}+2\right)+1}{\sqrt{x}+2}=3+\dfrac{1}{\sqrt{x}+2}\)

D lớn nhất \(\Leftrightarrow\sqrt{x}+2\) nhỏ nhất:

Mà:\(\sqrt{x}+2\ge2\)

vậy:\(\max\limits_D=3+\dfrac{1}{2}=\dfrac{7}{2}\Leftrightarrow x=0\)

\(M=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\left(đk:x\ge0,x\ne9\right)\)

Để \(M=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}< 0\) thì 

\(\sqrt{x}-3< 0\) ( do \(\sqrt{x}+3\ge3>0\))

\(\Leftrightarrow\sqrt{x}< 3\Leftrightarrow0\le x< 9\)

Mà \(x\in Z\)

\(\Rightarrow x\in\left\{0;1;2;3;4;5;6;7;8\right\}\)

a: Xét (O) có

ΔADB nội tiếp

AB là đường kính

Do đó: ΔADB vuông tại D

=>AD⊥BC tại D

Xét tứ giác AHDC có \(\hat{AHC}=\hat{ADC}=90^0\)

nên AHDC là tứ giác nội tiếp

b: AHDC nội tiếp

=>\(\hat{AHD}+\hat{ACD}=180^0\)

mà \(\hat{AHD}+\hat{MHD}=180^0\) (hai góc kề bù)

nên \(\hat{MHD}=\hat{ACD}=\hat{ACB}\)

Xét ΔOAC vuông tại A có AH là đường cao

nên \(OH\cdot OC=OA^2\)

=>\(OH\cdot OC=OB^2\)

=>\(\frac{OH}{OB}=\frac{OB}{OC}\)

Xét ΔOHB và ΔOBC có

\(\frac{OH}{OB}=\frac{OB}{OC}\)

góc HOB chung

Do đó: ΔOHB~ΔOBC

=>\(\hat{OHB}=\hat{OBC}=\hat{ABC}\)

mà \(\hat{OHB}+\hat{MHB}=\hat{OHM}=90^0\) và \(\hat{ABC}+\hat{ACB}=90^0\) (ΔABC vuông tại A)

nên \(\hat{MHB}=\hat{ACB}\)

=>\(\hat{MHB}=\hat{DHM}\)

=>HM là phân giác của góc DHB

\(b,B=\dfrac{x-4+2\sqrt{x}+6-3\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ B=\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\\ c,M=B:A=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}+3}{x-\sqrt{x}+2}=\dfrac{\sqrt{x}+1}{x-\sqrt{x}+2}\\ M=\dfrac{x-\sqrt{x}+2-x+2\sqrt{x}-1}{x-\sqrt{x}+2}\\ M=1-\dfrac{x-2\sqrt{x}+1}{x-\sqrt{x}+2}=1-\dfrac{\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+2}\)

Ta có \(\left(\sqrt{x}-1\right)^2\ge0;x-\sqrt{x}+2=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\)

Do đó \(\dfrac{\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+2}\ge0\)

\(\Leftrightarrow M=1-\dfrac{\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+2}\le1-0=1\)

Vậy \(M_{max}=1\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(tm\right)\)

a: Thay \(x=3+2\sqrt{2}\) vào A, ta được:

\(A=\dfrac{3+2\sqrt{2}-\sqrt{2}-1+2}{\sqrt{2}+1+3}=\dfrac{4+\sqrt{2}}{4+\sqrt{2}}=1\)

1.

a. Ta có: \(AB^2+AC^2=6^2+8^2=36+64=100\)

\(BC^2=10^2=100\)

 \(\Rightarrow AB^2+AC^2=BC^2\) \(\Rightarrow\Delta\)ABC vuông tại A

b. \(\Delta\)ABC vuông tại A, đường cao AH. Ta có:

AB.AC = AH.BC

hay 6.8 = AH.10

=> AH = \(\dfrac{6.8}{10}=4.8\)

Bài 1:

\(cos45^0=cos\left(2\cdot22^030p\right)=2\cdot cos^2\alpha-1\)

=>\(2\cdot cos^2\alpha=1+\frac{\sqrt2}{2}=\frac{2+\sqrt2}{2}\)

=>\(cos^2\alpha=\frac{2+\sqrt2}{4}\)

=>\(\sin^2a=1-\frac{2+\sqrt2}{4}=\frac{4-2-\sqrt2}{4}=\frac{2-\sqrt2}{4}\)

\(cos\alpha=\sqrt{\frac{2+\sqrt2}{4}}=\frac{\sqrt{2+\sqrt2}}{2}\)

\(\sin a=\sqrt{\frac{\left(2-\sqrt2\right)}{4}}=\frac{\sqrt{2-\sqrt2}}{2}\)

\(\tan\alpha=\frac{\sin\alpha}{cos\alpha}=\frac{\sqrt{2-\sqrt2}}{2}:\frac{\sqrt{2+\sqrt2}}{2}=\sqrt{\frac{2-\sqrt2}{2+\sqrt2}}=\sqrt{\frac{\sqrt2-1}{\sqrt2+1}}\)

\(=\sqrt{\frac{\left(\sqrt2-1\right)^2}{\left(\sqrt2-1\right)\left(\sqrt2+1\right)}}=\sqrt{\left(\sqrt2-1\right)^2}=\sqrt2-1\)

\(\cot\alpha=\frac{1}{\tan\alpha}=\frac{1}{\sqrt2-1}=\sqrt2+1\)

b: \(cos30^0=cos\left(2\cdot15^0\right)=2\cdot cos^215^0-1\)

=>\(2\cdot cos^215^0-1=\frac{\sqrt3}{2}\)

=>\(2\cdot cos^215^0=\frac{2+\sqrt3}{2}\)

=>\(cos^215^0=\frac{2+\sqrt3}{4}\)

=>\(cos15^0=\sqrt{\frac{2+\sqrt3}{4}}=\frac{\sqrt{2+\sqrt3}}{2}=\frac{\sqrt{4+2\sqrt3}}{2\sqrt2}=\frac{\sqrt3+1}{2\sqrt2}=\frac{\sqrt6+\sqrt2}{4}\)

Ta có: \(cos^215^0+\sin^215^0=1\)

=>\(\sin^215^0=1-\frac{2+\sqrt3}{4}=\frac{2-\sqrt3}{4}=\frac{\left(4-2\sqrt3\right)}{8}=\frac{\left(\sqrt3-1\right)^2}{\left(2\sqrt2\right)^2}\)

=>\(\sin15^0=\frac{\sqrt3-1}{2\sqrt2}=\frac{\sqrt6-\sqrt2}{4}\)

\(\tan15^0=\frac{\sin15^0}{cos15^0}=\frac{\sqrt6-\sqrt2}{4}:\frac{\sqrt6+\sqrt2}{4}=\frac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2}\)

\(=\frac{\sqrt3-1}{\sqrt3+1}=\frac{\left(\sqrt3-1\right)^2}{2}=\frac{4-2\sqrt3}{2}=2-\sqrt3\)

\(\cot15^0=\frac{1}{\tan15^0}=\frac{1}{2-\sqrt3}=2+\sqrt3\)

Bài 2:

\(cos^2a-cos^2b\)

\(=1-\sin^2a-\left(1-\sin^2b\right)=1-\sin^2a-1+\sin^2b\)

\(=\sin^2b-\sin^2a\)