1. On the ocean 

2. Yes

3. do house work , cleaning the floor , cooking meals , washing clothes , feeding his pets .

mn xa lánh em quá hic:((((((

giúp em với mai em nộp bài đó ạ;-;

b) \(\sqrt{29-12\sqrt{5}}+\sqrt{45-20\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}+\sqrt{\left(5-2\sqrt{5}\right)^2}=2\sqrt{5}-3+5-2\sqrt{5}=2\)

h) \(\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4+\sqrt{15}}=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{\left(\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{3}{2}}\right)^2}=\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{3}{2}}\right)=\sqrt{25}+\sqrt{15}-\sqrt{15}-\sqrt{9}=5-3=2\)

h: Ta có: \(\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4+\sqrt{15}}\)

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

=5-3=2

Lời giải:

b. $-138-3(2x^2-1)=-169$

$3(2x^2-1)=-148-(-169)=21$

$2x^2-1=21:3=7$

$2x^2=7+1=8$

$x^2=4=2^2=(-2)^2\Rightarrow x=\pm 2$

c.

$4(-x+5)^2-15=21$

$4(-x+5)^2=21+15=36$

$(-x+5)^2=36:4=9=3^2=(-3)^2$

$\Rightarrow -x+5=3$ hoặc $-x+5=-3$

$\Rightarrow x=2$ hoặc $x=8$

d.

$4(-3x^2-145)=-628$

$-3x^2-145=-628:4=-157$

$3x^2+145=157$

$3x^2=157-145=12$

$x^2=12:3=4=2^2=(-2)^2$

$\Rightarrow x=\pm 2$
e.

$252-3(-x-5)^2=225$

$3(-x-5)^2=252-225=27$

$(-x-5)^2=27:3=9$

$(x+5)^2=3^2=(-3)^2$

$\Rightarrow x+5=3$ hoặc $x+5=-3$

$\Rightarrow x=-2$ hoặc $x=-8$

g.

$-(2x-1)^2-(-4)^2=-41$

$(2x-1)^2+(-4)^2=41$

$(2x-1)^2+16=41$

$(2x-1)^2=41-16=25=5^2=(-5)^2$

$\Rightarrow 2x-1=5$ hoặc $2x-1=-5$

$\Rightarrow x=3$ hoặc $x=-2$

Bài 2: 

b: Ta có: \(B=\dfrac{15-5\sqrt{x}}{x-5\sqrt{x}+6}+\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)

\(=\dfrac{-5\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-2}=1\)

a. \(\widehat{DAB}=\widehat{ABC}=\widehat{BCE}=90^0\)

\(\widehat{ABD}=180^0-\widehat{ABC}-\widehat{EBC}=180^0-60^0-\left(180^0-\widehat{BCE}-\widehat{CEB}\right)=180^0-60^0-\left(180^0-60-\widehat{CEB}\right)=\widehat{CEB}\)\(\Rightarrow\)△ABD∼△CEB (g-g).

\(\Rightarrow\dfrac{AD}{CB}=\dfrac{AB}{CE}\Rightarrow AD.CE=CB.AB\Rightarrow AD.CE=a^2\) không đổi

b. \(\widehat{CAD}=\widehat{BAD}+\widehat{BAC}=60^0+60^0=\widehat{BCE}+\widehat{ACB}=\widehat{ACE}\)

 \(\dfrac{AD}{CB}=\dfrac{AB}{CE}\Rightarrow\dfrac{AD}{AC}=\dfrac{AC}{CE}\)

\(\Rightarrow\)△ACD∼△CEA (c-g-c) 

\(\Rightarrow\left\{{}\begin{matrix}\widehat{ACD}=\widehat{CEA}\\\dfrac{CE}{AC}=\dfrac{EA}{CD}\end{matrix}\right.\)

\(\Rightarrow\)△ACK∼△AEC (g-g).

\(\Rightarrow\dfrac{CK}{EC}=\dfrac{AK}{AC}\Rightarrow\dfrac{CE}{AC}=\dfrac{CK}{AK}\)

\(\Rightarrow\dfrac{AE}{CD}=\dfrac{CK}{AK}\Rightarrow AE.AK=CD.CK\)