a. ta có \(BC=\sqrt{AB^2+AC^2}=\sqrt{5^2+12^2}=13cm\)

b. ta có \(sinB=\frac{AC}{BC}=\frac{12}{13},cosB=\frac{BA}{BC}=\frac{5}{13},tanB=\frac{AC}{AC}=\frac{12}{5},cotB=\frac{1}{tanB}=\frac{5}{12}\)

ta có \(sinC=\frac{5}{13}\Rightarrow C\simeq23^0\)

ta có a nhọn nên sin, cos ,tan và cotg của a đều là các số dương

nên ta có :

\(cosa=\sqrt{1-sin^2a}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}\)

\(tana=\frac{sina}{cosa}=\frac{2}{\sqrt{5}},cotga=\frac{1}{tana}=\frac{\sqrt{5}}{2}\)

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

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...........................................................................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 )