bài 1. 

\(cos88^0< sin7^0< sin29^0< cos58^0< cos50^0< sin64^0\)

b.\(cos38^0< sin56^0< cos31^0< sin61^0\)'

c.\(cot70^0< tan28^0< tan33^0< cot55^0< cot40^0\)

\(a,4x-\sqrt{x^2-4x+4}\)

\(4x-\sqrt{\left(x-2\right)^2}\)

\(4x-\left|x-2\right|\)

\(4x-x-2=3x-2\)

\(b,3x+\sqrt{x^2+6x+9}\)

\(3x+\sqrt{\left(x+3\right)^2}\)

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

\(3x-x-3=2x-3\)

\(c,\frac{\sqrt{x^2+4x+4}}{x+2}\)

\(\frac{\sqrt{\left(x+2\right)^2}}{x+2}\)

\(\frac{\left|x+2\right|}{x+2}\)

\(TH1:x< -2\)

\(\frac{-x-2}{x+2}\)

\(=-1\)

\(TH2:x>-2\)

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

a. ta có : \(4x-\sqrt{x^2-4x+4}=4x-\sqrt{\left(x-2\right)^2}=4x-x+2=3x+2\)

b.\(3x+\sqrt{x^2+6x+9}=3x+\sqrt{\left(x+3\right)^2}=3x-x-3=2x-3\)

c.\(\frac{\sqrt{x^2+4x+4}}{x+2}=\frac{\sqrt{\left(x+2\right)^2}}{x+2}=\frac{\left|x+2\right|}{x+2}\)

\(d,\)để căn thức \(\sqrt{x^2+2x+3}\)có nghĩa thì \(x^2+2x+3\ge0\)

\(x^2+2x+3\ge0\)

\(\left(x+1\right)^2+2\ge2>0\)(luôn đúng)

vậy căn thức có nghĩa với \(\forall\)gt của x

...............................................................................

..........................................................................................

...........................................................................tgbvn JGKGITJNNFJFJNFJBFÒNBFOHRJ;FFJh' IIIor   ỉie

a, Với x > 0 ; \(x\ne1\)

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

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

b, Ta có : A = 1/3 => \(\frac{\sqrt{x}-1}{\sqrt{x}}=\frac{1}{3}\Rightarrow3\sqrt{x}-3=\sqrt{x}\Leftrightarrow2\sqrt{x}-3=0\Leftrightarrow x=\frac{9}{4}\)

a, Thay x = 4 => \(\sqrt{x}=2\)vào A ta được 

\(A=\frac{4}{2+1}=\frac{4}{3}\)

b, Với x >= 0 ; \(x\ne1\)

\(B=\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{3\sqrt{x}+3}{1-\sqrt{x}}+\frac{3+5\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+3\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+3+5\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{x-\sqrt{x}+3\left(x+4\sqrt{x}+3\right)+3+5\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{x+4\sqrt{x}+3+3x+12\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{4x+16\sqrt{x}+12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{4\sqrt{x}+4}{\sqrt{x}-1}\)( đpcm )

a) Với \(x=4\):

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

b) \(B=\left(\frac{3\sqrt{x}+6}{x-9}-\frac{2}{\sqrt{x}-3}\right)\div\frac{1}{\sqrt{x}+3}\)

\(=\frac{3\sqrt{x}+6-2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\left(\sqrt{x}+3\right)\)

\(=\frac{\sqrt{x}}{\sqrt{x}-3}\)

c) \(P=A.B=\frac{\sqrt{x}-3}{x-\sqrt{x}+1}.\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)

Có \(x-\sqrt{x}+1=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}>0,\sqrt{x}\ge0\)

nên \(P\ge0\).

Do đó \(\left|P\right|=P\).

a. khi x=4 ta có \(A=\frac{\sqrt{4}-3}{4-\sqrt{4}+1}=\frac{2-3}{4-2+1}=-\frac{1}{3}\)

b. \(B=\left(\frac{3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2}{\sqrt{x}-3}\right)\times\left(\sqrt{x}+3\right)\)

\(B=\frac{3\sqrt{x}+6}{\sqrt{x}-3}-\frac{2\left(\sqrt{x}+3\right)}{\sqrt{x}-3}=\frac{\sqrt{x}}{\sqrt{x}-3}\)

ta có \(P=AB=\frac{\sqrt{x}-3}{x-\sqrt{x}+1}.\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{\sqrt{x}}{x-\sqrt{x}+1}\ge0\forall x\Rightarrow P=\left|P\right|\)

?????????????????????????

khi \(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{2}-1\)

ta có \(A=\frac{2\left(\sqrt{2}-1\right)+1}{\sqrt{2}-1}=\frac{2\sqrt{2}-1}{\sqrt{2}-1}=\left(2+\sqrt{2}+1\right)=3+\sqrt{2}\)

\(B=\frac{x-3\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\frac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}}\)

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

Để \(\left|P\right|>P\Rightarrow P< 0\Leftrightarrow\sqrt{x}-2< 0\Leftrightarrow0\le x< 4\)