1 do you do

2 had done - went

3 went - had read

4 will attend

5 hadn't worn

6 to be

7 weren't sleeping - were playing

8 to be

9 had lived - moved

10 locking

11 had work - retired 

12 told - had learned

13 won't call

14 had met

do you do

had done-went

went- had read

will attend

hadn't worn

to be

\(\dfrac{x}{27}=\dfrac{2}{9}-\dfrac{1}{3}\Rightarrow\dfrac{x}{27}=-\dfrac{1}{9}\Rightarrow\dfrac{x}{27}=\dfrac{-3}{27}\Rightarrow x=27\)

\(\dfrac{x}{27}=\dfrac{2}{9}-\dfrac{1}{3}=-\dfrac{1}{9}\Rightarrow x=-\dfrac{1}{9}.27=-3\).

1 were - would you play

2 weren't studying - would have 

3 had taken - wouldn't have got

4 would you go - could

5 will you give - is

6 recycle - won't be

7 had heard - wouldn't have gone

8 would you buy - had

9 don't hurry - will miss

10 had phoned - would have given

11 were - wouldn't eat

12 will go - rains

13 had known - would have sent

14 won't feel - swims

15 hadn't freezed - would have gone

a Xét tứ giác AEHF có \(\hat{AEH}=\hat{AFH}=\hat{FAE}=90^0\)

nên AEHF là hình chữ nhật

b: AEHF là hình chữ nhật

=>AH cắt EF tại trung điểm của mỗi đường

mà O là trung điểm của AH

nên O là trung điểm của EF

=>E,O,F thẳng hàng

c: Hình chữ nhật AEHF trở thành hình vuông khi AH là phân giác của góc EAF

=>AH là phân giác của góc BAC

Xét ΔABC có

AH là đường cao

AH là đường phân giác

Do đó: ΔABC cân tại A

=>AB=AC

d: AEHF là hình vuông

=>ΔABC vuông cân tại A

mà AH là đường cao

nên H là trung điểm của BC và BC=2AH

H là trung điểm của BC

=>BC=2CH=2*3=6(cm)

=>AH=6/2=3(cm)

AEHF là hình vuông

=>AH=EF

=>EF=3(cm)

Diện tích hình vuông AEHF là:

\(S_{AEHF}=\frac12\cdot AH\cdot EF=\frac12\cdot3\cdot3=\frac92\left(\operatorname{cm}^2\right)\)

Bài 2: 

a) Ta có: \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\dfrac{6}{2-\sqrt{10}}-\dfrac{20}{\sqrt{10}}\)

\(=\dfrac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}+\dfrac{6\cdot\left(\sqrt{10}+2\right)}{\left(\sqrt{10}-2\right)\left(\sqrt{10}+2\right)}-\dfrac{\sqrt{10}\cdot2\sqrt{10}}{\sqrt{10}}\)

\(=\sqrt{10}+\sqrt{10}-2-2\sqrt{10}\)

=-2

b) Ta có: \(\left(\dfrac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\dfrac{4}{1+\sqrt{5}}+4\right)\)

\(=\left(\sqrt{5}-1-2\right)\left(\sqrt{5}-1+4\right)\)

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

=5-9=-4

c) Ta có: \(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\dfrac{\sqrt{5}+1}{\sqrt{5}-1}\)

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

\(=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\dfrac{6+2\sqrt{5}}{4}\)

\(=\dfrac{16}{2}-\dfrac{6+2\sqrt{5}}{4}\)

\(=\dfrac{32-6-2\sqrt{5}}{4}\)

\(=\dfrac{26-2\sqrt{5}}{4}\)

\(=\dfrac{13-\sqrt{5}}{2}\)

ĐKXĐ: y>=3

TA có: \(A_{y}^3+C_{y}^{y-2}=14y\)

=>\(\frac{y!}{\left(y-3\right)!}+\frac{y!}{\left(y-y+2\right)!\cdot\left(y-2\right)!}=14y\)

=>\(\frac{y\left(y-1\right)\left(y-2\right)}{1}+\frac{y\left(y-1\right)}{2}=14y\)

=>\(\frac{2y\left(y-1\right)\left(y-2\right)+y\left(y-1\right)}{2}=14y\)

=>y(y-1)[2(y-2)+1]=28y

=>y(y-1)(2y-3)-28y=0

=>y[(y-1)(2y-3)-28]=0

=>(y-1)(2y-3)-28=0

=>\(2y^2-5y+3-28=0\)

=>\(2y^2-5y-25=0\)

=>\(2y^2-10y+5y-25=0\)

=>(y-5)(2y+5)=0

=>y=5(nhận) hoặc y=-5/2(loại)

\(M=A_{y+1}^4+3\cdot C_{y}^3\)

\(=A_{5+1}^4+3\cdot C_5^3=A_6^4+3\cdot C_5^3\)

\(=\frac{6!}{\left(6-4\right)!}+3\cdot\frac{5!}{\left(5-3\right)!\cdot3!}=\frac{6!}{2!}+3\cdot\frac{5\cdot4}{2!}=6\cdot5\cdot4\cdot3+3\cdot5\cdot2\)

=30*12+15*2

=360+30

=390

=>Chọn B

a: Số cần tìm là -4:2/3=-6

b: Số cần tìm là -4/5:2/3=-6/5