nhỡ sai cũng cho luôn à anh 

Chu vi AMND=36 cm => cạnh AM=DN= 36/4=9cm

=> AB=9+5= 14cm

Tỉ số diện tích AMND/ABCD= AM.MN/AB.BC= 9.9/14.9=9/14

a) x=2

b) x=6

a)\(3.\left(5^x-1\right)-2=70\\3.\left(5^x-1\right)=72\\5^x-1=24\\ 5^x=25=5^2 \\ \Rightarrow x=2 \)

b)\(2^x+2^{x+1}+2^{x+2}=960-2^{x+3}\\2^x+2^{x+1}+2^{x+2}+2^{x+3}=960\\ 2^x\left(1+2+4+8\right)=2^x.15=960\\ 2^x=64=2^6\\ x=6\)

1.You / smoke / you / get yellow teeth

=> If you smoke, you get yellow teeth

2.Children / play outside / they / not get overweight. 

=> If the children not play outside, they not get overweight

3.You / heat ice / it / melt.

=> If you heat ice, it melt

4.I /speak to john / he / get annoyed. 

=> If I speak to John, he get annoyed

5.Lots of people / come/ Jenny / have a party.

=> if lots of people come, Jenny has a party 

6.I / feel good the next day / I / go to bed early.

=> I feel good the next day, if I go to bed early

7.She / buy expensive clothes / she / go shopping.

=> She buy expensive clothes if she go shopping

8.My daughter / pass her exams / she / work hard.

=> My daughter passed her exam because she had worked hard.

9.David / be sick / he / drink milk

=> David'll be sick if he drinks milk.

10.The river / freeze / it / be very cold.

=> The river froze because it was very cold.

11.I / to visit the museums / I / be in a new city.

=> I would to visit the museums if I was in a new city.

12.I / cycle to work / the weather / be fine.

=> I will cycle to work if the weather is fine.

13.My flatmate / clean it really well / she / clean the house.

=> My flatmate cleans it very well when she is cleaning the house.

14.Everybody / be grumpy / it / rain a lot.

=> Everybody will be grumpy if it rains a lot.

1. If you smoke, you'll get yellow tooth.

2. Children should play outside if they don't want to get overweight.

3. If you heat ice, it melts.

4. If I speak to John, he'll be annoyed.

5. Lots of people came to Jenny's house to have a party.

6. If I want to feel good the next day, I should go to bed early.

7. She bought expensive clothes while she was going shopping.

8. My daughter passed her exam because she had worked hard.

9. David'll be sick if he drinks milk.

10. The river froze because it was very cold.

11. I would to visit the museums if I was in a new city.

12. I will cycle to work if the weather is fine.

13. My flatmate cleans it very well when she is cleaning the house.

14. Everybody will be grumpy if it rains a lot.

a, O là tâm đường tròn nội tiếp nên AO là đường trung trực của tam giác ABC. Tam giác ABC cân tại A nên AO cũng là đường phân giác của góc A.

b, Tamm giác ABK và tam giác ADB có: Góc A chung; AKB = ABD vì chắn hai cung bằng nhau AB  và AC. Suy ra tam giác ABK đồng dạng với tam giác ADB. Suy ra\(\frac{AB}{AK}=\frac{AD}{AB}\)suy ra AB²=AD.AK

Câu 2:

a) Điều kiện: \(x\ne-1\)

BPT tương đương:

\(\frac{\left(x+1\right)^2\left(\sqrt{x^2+2x+2}+1\right)}{x^2+2x+1}\ge4+2x\)

\(\Leftrightarrow\sqrt{x^2+2x+2}\ge3+2x\)

\(\Leftrightarrow3+2x< 0\left(h\right)\hept{\begin{cases}3+2x\ge0\\x^2+2x+2\ge9+12x+4x^2\end{cases}}\)

\(\Leftrightarrow x< -\frac{3}{2}\left(h\right)\hept{\begin{cases}x\ge-\frac{3}{2}\\-\frac{7}{3}\le x\le-1\end{cases}}\Leftrightarrow x\le-1\)

Kết hợp ĐK suy ra \(S_a=\left(-\infty;-1\right)\)

b) Hệ tương đương:

\(\hept{\begin{cases}\left(x^2+1\right)=y\left(x+y+2\right)\left(1\right)\\\left(x^2+1\right)\left(x+y-2\right)=5y\left(2\right)\end{cases}}\)

Ta thấy VP(1) = VT (1) = x² + 1 khác 0, vậy thì chia VT(2) và VP(2) cho VT(1) và VP (1), ta được:

\(x+y-2=\frac{5}{x+y+2}\Leftrightarrow\left(x+y\right)^2=9\Leftrightarrow\orbr{\begin{cases}x+y=3\\x+y=-3\end{cases}}\)

+) Nếu \(y=3-x\) thì (1) trở thành:

\(x^2+5x-14=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-7\end{cases}}\Rightarrow\orbr{\begin{cases}\left(x;y\right)=\left(2;1\right)\\\left(x;y\right)=\left(-7;10\right)\end{cases}}\)

+) Nếu \(y=-3-x\) thì (1) trở thành:

\(x^2-x-2=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}\left(x;y\right)=\left(-1;-2\right)\\\left(x;y\right)=\left(2;-5\right)\end{cases}}\)

Vậy \(S_b=\left\{\left(2;1\right);\left(-7;10\right);\left(-1;-2\right);\left(2;-5\right)\right\}.\)

TH 4 bạn nữ hoặc 5 bạn nữ đứng liền nhau: 

Coi nhóm 4 bạn nữ là X, số cách sắp xếp nhóm X là: \(4!\)(cách)

Sắp xếp X, 1 bạn nữ còn lại và 4 bạn nam có:  \(6!\)(cách)

Xếp ngẫu nhiên 9 bạn có: \(9!\)(cách)

Vậy xác suất để không quá 3 bạn nữ đứng liền nhau là: \(\frac{9!-4!.6!}{9!}=\frac{20}{21}\)

a, \(\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}=\sqrt{\frac{6a^2}{24}}=\sqrt{\frac{a^2}{4}}=\left|\frac{a}{2}\right|=\frac{a}{2}\)

do \(a\ge0\)

b, \(\sqrt{13a}.\sqrt{\frac{52}{a}}=\sqrt{\frac{676a}{a}}=\sqrt{676}=26\)

c, \(\sqrt{5a}.\sqrt{45a}-3a=\sqrt{225a^2}-3a=\left|15a\right|-3a\)

\(=15a-3a=12a\)do a > 0 

d, \(=\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\)

\(=\left(3-a\right)^2-\sqrt{36a^2}=\left(3-a\right)^2-\left|6a\right|\)

Với \(a\ge0\Rightarrow\left(3-a\right)^2-6a=a^2-6a+9-6a=a^2-12a+9\)

Với \(a< 0\Rightarrow\left(3-a\right)^2+6a=a^2-6a+9+6a=a^2+9\)

a) Ta có:

b) Ta có:

c) Do a ≥ 0 nên bài toán luôn xác định. Ta có:

 

 

d) Ta có:

Khai phương tích 12.30.40 được:

(A) 1200 ;           (B) 120 ;            (C) 12 ;             (D) 240.

Chọn B

Khai phương tích 12.30.40 (=12.12.10.10) ta được 12.10= 120 (Chọn B)

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