a: Xét ΔADM và ΔCBN có 

AD=CB

\(\widehat{DAM}=\widehat{BCN}\)

AM=CN

Do đó: ΔADM=ΔCBN

Suy ra: DM=BN

41. have been used
42. making
43. driving
44. be finished
45. to post
46. mending
47. is being sung
48. work
49. were blown
50. do
51. will
52. were reading
53. has bought
54. listening
55. stopped
56. is spoken
57. to practice

1. A

2. D

3. B

4. B

5. C

6. B

7. B

8. B

9. D

10. B

1A 2D 3B 4B 5C 6B 7B 8B 9D 10B

Câu 1

a)=\(8\sqrt{3}-10\sqrt{3}+15\sqrt{3}=13\sqrt{3}\)

b)=\(4\sqrt{x}+6\sqrt{x}-6\sqrt{x}=4\sqrt{x}\)

c)=\(21\sqrt{2}+8\sqrt{2}-28\sqrt{2}=\sqrt{2}\)

d)\(\Rightarrow\)\(8\sqrt{2\sqrt{3}}-\sqrt{5\sqrt{3}}-4\sqrt{5\sqrt{3}}\)

\(\Rightarrow\)\(8\sqrt{2\sqrt{3}}-5\sqrt{5\sqrt{3}}\)

câu 2

a)\(\Rightarrow4x=64\)\(\Rightarrow x=16\)

b)\(\Rightarrow9x\le36\)\(\Rightarrow x\le4\)

Câu 2: 

a: Ta có: \(\sqrt{4x}=8\)

\(\Leftrightarrow4x=64\)

hay x=16

b: Ta có: \(\sqrt{9x}\le6\)

\(\Leftrightarrow9x\le36\)

\(\Leftrightarrow x\le4\)

Kết hợp ĐKXĐ, ta được: \(0\le x\le4\)

a: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7\cdot b^2k^2+3\cdot bk\cdot b}{11\cdot b^2k^2-8\cdot b^2}=\dfrac{b^2k\left(7k+3\right)}{b^2\left(11k^2-8\right)}=\dfrac{k\left(7k+3\right)}{11k^2-8}\)

\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7\cdot d^2k^2+3\cdot dk\cdot d}{11\cdot d^2k^2-8d^2}=\dfrac{k\left(7k+3\right)}{11k^2-8}\)

Do đó: \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)

c: \(\dfrac{3a+2c}{3b+2d}=\dfrac{3bk+2dk}{3b+2d}=k\)

\(\dfrac{a}{b}=\dfrac{bk}{b}=k\)

Do đó: \(\dfrac{a}{b}=\dfrac{3a+2c}{3b+2d}\)

lỗi

Em cần giúp câu nào hả em? Em nên chụp 1-2 ý cho 1 lần hỏi nhá, như thế mọi người sẽ dễ dàng giúp em hơn

13

a, \(3x-4=-x+8\)

\(< =>3x+x=8+4\)

\(< =>4x=12\)

\(< =>x=\frac{12}{4}=3\)

b, \(\frac{2x+1}{6}+\frac{x-7}{12}=10\)

\(< =>\frac{2\left(2x+1\right)}{12}+\frac{x-7}{12}=\frac{120}{12}\)

\(< =>4x+2+x-7=120\)

\(< =>5x=120+5=125\)

\(< =>x=\frac{125}{5}=\frac{5^3}{5}=5^2=25\)