a) \(\left(5\sqrt{2}+2\sqrt{5}\right).\sqrt{5}-\sqrt{250}=5\sqrt{10}+10-5\sqrt{10}=10\)

b) \(\left(2\sqrt{3}-5\sqrt{2}\right).\sqrt{3}-\sqrt{36}=6-5\sqrt{6}-6=-5\sqrt{6}\)

c) \(6\sqrt{\frac{1}{3}}+\frac{9}{\sqrt{3}}-\frac{2}{\sqrt{3}-1}=6\cdot\frac{\sqrt{3}}{3}+\frac{9\sqrt{3}}{3}-\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=2\sqrt{3}+3\sqrt{3}-\frac{2\left(\sqrt{3}+1\right)}{3-1}=5\sqrt{3}-2\sqrt{3}-2=3\sqrt{3}-2\)

d) \(4\sqrt{\frac{1}{2}}-\frac{6}{\sqrt{2}}+\frac{2}{\sqrt{2}+1}=4\cdot\frac{\sqrt{2}}{2}-\frac{6\sqrt{2}}{2}+\frac{2\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)

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

a, \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)

\(=5\sqrt{10}+10-\sqrt{250}\)

\(=5\sqrt{10}+10-5\sqrt{10}\)

\(=10\)

b, \(\left(2\sqrt{3}-5\sqrt{2}\right)\sqrt{3}-\sqrt{36}\)

\(=6-5\sqrt{6}-6\)

\(=-5\sqrt{6}\)

c, \(6\sqrt{\frac{1}{3}}+\frac{9}{\sqrt{3}}-\frac{2}{\sqrt{3}-1}\)

\(=\sqrt{\frac{36}{3}}+\frac{9\sqrt{3}}{3}-\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=\sqrt{12}+3\sqrt{3}-\frac{2\sqrt{3}+2}{2}\)

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

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

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

\(=4\sqrt{3}-1\)

d,\(4\sqrt{\frac{1}{2}}-\frac{6}{\sqrt{2}}+\frac{2}{\sqrt{2}+1}\)

\(=\sqrt{\frac{16}{2}}-\frac{6\sqrt{2}}{2}+\frac{2\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)

\(=\sqrt{8}-3\sqrt{2}+2\sqrt{2}-2\)

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

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

Xét tam giác \(ABD\)vuông tại \(A\):

\(BD^2=AB^2+AD^2\)(định lí Pythagore) 

\(=4^2+10^2=116\)

\(\Rightarrow BD=\sqrt{116}=2\sqrt{29}\left(cm\right)\)

Lấy \(E\)thuộc \(CD\)sao cho \(AE\perp AC\)

Suy ra \(ABDE\)là hình bình hành. 

\(AE=BD=2\sqrt{29}\left(cm\right),DE=AB=4\left(cm\right)\).

Xét tam giác \(AEC\)vuông tại \(A\)đường cao \(AD\):

\(\frac{1}{AD^2}=\frac{1}{AE^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{AC^2}=\frac{1}{AD^2}-\frac{1}{AE^2}=\frac{1}{100}-\frac{1}{116}=\frac{1}{715}\)

\(\Rightarrow AC=\sqrt{715}\left(cm\right)\)

\(AE^2=ED.EC\Leftrightarrow EC=\frac{AE^2}{ED}=\frac{116}{4}=29\left(cm\right)\)suy ra \(DC=25\left(cm\right)\)

Hạ \(BH\perp CD\).

\(BC^2=HC^2+BH^2=21^2+10^2=541\Rightarrow BC=\sqrt{541}\left(cm\right)\)

\(S_{ABCD}=\left(AB+CD\right)\div2\times AD=\frac{4+25}{2}\times10=145\left(cm^2\right)\)

A B C D E P K H Q 1 2 x

a) Xét tam giác ADB vuông tại D có: \(cos\widehat{A}=\frac{AD}{AB}\)

Xét tam giác AEC vuông tại  C có: \(cos\widehat{A}=\frac{AE}{AC}\)

=> \(\frac{AD}{AB}=\frac{AE}{AC}\) => AE.AB = AD.AC

b) Xét tam giác ADE và tam giác ABC

có: \(\widehat{A}\) :chung

  \(\frac{AD}{AB}=\frac{AE}{AC}\) (cmt)

=> tam giác ADE ∽ tam giác ABC (c.g.c)

=> \(\widehat{AED}=\widehat{ACB}\)mà \(\widehat{AED}=\widehat{QEB}\)(đối đỉnh) => \(\widehat{QEB}=\widehat{QCD}\)

Xét tam giác QEB và tam giác QCD

có: \(\widehat{QEB}=\widehat{QCD}\)(cmt); \(\widehat{Q}\) : chung

=> tam giác QEB ∽ tam giác QCD (g.g)

=> \(\frac{QE}{QC}=\frac{QB}{QD}\) => QB. QC = QE . QD

c) CMTT: \(\widehat{BKE}=\widehat{BAC}\); \(\widehat{DKC}=\widehat{BAC}\)

Ta có: \(\widehat{BKE}+\widehat{K_2}=90^0\) (phụ nhau))

   \(\widehat{K_1}+\widehat{DKC}=90^0\) (phụ nhau) 

==> \(\widehat{K_1}=\widehat{K_2}\) => KA là phân giác của \(\widehat{DKE}\)

=> \(\frac{KE}{KD}=\frac{EP}{ED}\)(1)

Gọi Kx là tia đối của tia KD => \(\widehat{DKC}=\widehat{QKx}\) mà \(\widehat{DKC}=\widehat{EKB}\) => \(\widehat{EKQ}=\widehat{QKx}\)

=> KQ là tia phân giác của \(\widehat{EKx}\) => \(\frac{EK}{KD}=\frac{QE}{QD}\)(2)

Từ (1) và (2) => \(\frac{EP}{PD}=\frac{QE}{QD}\) => PD. QE = PE. QD

TC2: a) Ta có : \(cos^2\alpha=1-sin^2\alpha=1-\left(\frac{2}{3}\right)^2=\frac{5}{9}\)

\(\Rightarrow cos\alpha=\sqrt{\frac{5}{9}}=\frac{\sqrt{5}}{3}\)

\(\Rightarrow\hept{\begin{cases}tan\alpha=\frac{sin\alpha}{cos\alpha}=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}=\frac{2}{\sqrt{5}}\\cot\alpha=\frac{cos\alpha}{sin\alpha}=\frac{\frac{\sqrt{5}}{3}}{\frac{2}{3}}=\frac{\sqrt{5}}{2}\end{cases}}\)

b)Ta có :\(A=\frac{sin\alpha-cos\alpha}{sin\alpha+cos\alpha}=\frac{\frac{2}{3}-\frac{\sqrt{5}}{3}}{\frac{2}{3}+\frac{\sqrt{5}}{3}}=\frac{2-\sqrt{5}}{2+\sqrt{5}}=\frac{\left(2-\sqrt{5}\right)^2}{4-5}=4\sqrt{5}-9\)

ĐK: \(x\ge0,x\ne9\).

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

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

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

\(3\sqrt{0,02}-3\sqrt{0,08}+5\sqrt{0,18}\)   

\(=3\cdot\sqrt{0,02}-3\cdot\sqrt{4\cdot0,02}+5\cdot\sqrt{9\cdot0,02}\)   

\(=3\cdot\sqrt{0,02}-3\cdot2\cdot\sqrt{0,02}+5\cdot3\cdot\sqrt{0,02}\)   

\(=3\cdot\sqrt{0,02}-6\cdot\sqrt{0,02}+15\cdot\sqrt{0,02}\)   

\(=12\cdot\sqrt{0,02}\)

6) \(3\sqrt{0,02}-3\sqrt{0,08}+5\sqrt{0,18}\)

\(=3\sqrt{0,02}-3\sqrt{4\cdot0,02}+5\sqrt{9\cdot0,02}\)

\(=3\sqrt{0,02}-6\sqrt{0,02}+15\sqrt{0,02}\)

\(=12\sqrt{0,02}\)

7) \(2\sqrt{\frac{27}{4}}+\sqrt{\frac{48}{9}}+\frac{2}{5}\sqrt{\frac{75}{16}}\)

\(=2\sqrt{3\cdot\frac{9}{4}}+\sqrt{3\cdot\frac{16}{9}}+\frac{2}{5}\sqrt{3\cdot\frac{25}{16}}\)

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

\(=\frac{29}{6}\sqrt{3}=\frac{29\sqrt{3}}{6}\)

8) \(2\sqrt{\frac{16}{5}}-5\sqrt{\frac{9}{125}}-6\sqrt{\frac{121}{45}}\)

\(=2\sqrt{16\cdot\frac{1}{5}}-5\sqrt{\frac{9}{25}\cdot\frac{1}{5}}-6\sqrt{\frac{121}{9}\cdot\frac{1}{5}}\)

\(=8\sqrt{\frac{1}{5}}-3\sqrt{\frac{1}{5}}-22\sqrt{\frac{1}{5}}\)

\(=-17\sqrt{\frac{1}{5}}=\frac{-17\sqrt{5}}{5}\)

Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)có :

\(C\ge\frac{4}{1+\left(a+b\right)^2}\ge\frac{4}{1+1}=2\)

Dấu = khi a=b=1/2