1 was arrested before the garden by the boy yesterday

2 his cats taken care of carefully by Jack everyday

3 Has the country been protected from Covid-19 successfully

4 is reported that the man is running out of money

is reported to be running out of money\

5 has been said that the thief got out of the prison

has been said to have got out of the prison

6 was thought that Mary had turned down the job

Mary was thought to have  turned down the job

\(1,\\ a,=\dfrac{\left(3+2\sqrt{3}\right)\sqrt{3}}{3}+\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{1}\\ =\dfrac{3\sqrt{3}+6}{3}+\sqrt{2}=\sqrt{3}+1+\sqrt{2}\\ b,=\left(\dfrac{\sqrt{5}+\sqrt{2}}{3}-\dfrac{\sqrt{5}-\sqrt{2}}{3}+1\right)\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\\ =\dfrac{\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}+3}{3}\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\\ =\dfrac{2\sqrt{2}+3}{3\left(3+2\sqrt{2}\right)}=\dfrac{1}{3}\)

\(2,\\ A=2x+\sqrt{\left(x-3\right)^2}=2x+\left|x-3\right|\\ =2\left(-5\right)+\left|-5-3\right|=-10+8=-2\\ B=\dfrac{\sqrt{\left(2x+1\right)^2}}{\left(x-4\right)\left(x+4\right)}\left(x-4\right)^2=\dfrac{\left|2x+1\right|\left(x-4\right)}{x+4}\\ B=\dfrac{17\cdot4}{12}=\dfrac{17}{3}\)

2:

a: BC=căn 15^2+20^2=25cm

AH=15*20/25=12cm

góc ADH=góc AEH=góc DAE=90 độ

=>ADHE là hình chữ nhật

=>DE=AH=12cm

b: ΔAHB vuông tại H có HD vuông góc AB

nên AD*AB=AH^2

ΔAHC vuông tại H có HE vuông góc AC

nên AE*AC=AH^2

=>AD*AB=AE*AC

c: góc IAC+góc AED

=góc ICA+góc AHD

=góc ACB+góc ABC=90 độ

=>AI vuông góc ED

4:

a: góc BDH=góc BEH=góc DBE=90 độ

=>BDHE là hình chữ nhật

b: BDHE là hình chữ nhật

=>góc BED=góc BHD=góc A

Xét ΔBED và ΔBAC có 

góc BED=góc A

góc EBD chung

=>ΔBED đồng dạng với ΔBAC
=>BE/BA=BD/BC

=>BE*BC=BA*BD

c: góc MBC+góc BED

=góc C+góc BHD

=góc C+góc A=90 độ

=>BM vuông góc ED

Em tách ra từng bài nhé !

Bài 3:

\(a,=\sqrt[3]{\left(x-1\right)^3}-\sqrt[3]{\left(5x+1\right)^3}=x-1-5x-1=-4x-2\\ b,=6a-6a+20a=20a\)

Bài 2:

\(a,=2\sqrt[3]{6}+3\sqrt[3]{5}-4\sqrt[3]{6}-2\sqrt[3]{5}=\sqrt[3]{5}-2\sqrt[3]{6}\\ b,=\sqrt[3]{8}-4\sqrt[3]{27}+2\sqrt[3]{64}=2-12+16=6\\ c,=\sqrt[3]{64}+\sqrt[3]{48}+\sqrt[3]{36}-\sqrt[3]{48}-\sqrt[3]{36}-\sqrt[3]{27}=4-3=1\\ d,=\sqrt[3]{162\left(-2\right)\cdot\dfrac{2}{3}}=\sqrt[3]{-216}=-6\)