a: Xét ΔADM và ΔCBN có 

AD=CB

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

AM=CN

Do đó: ΔADM=ΔCBN

Suy ra: DM=BN

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

1.A

2.B

3.B

4.B

5.C

6.B

7.B

8.B

9.D

10.B

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: \(0\le x\le4\)

Câu 1:

a: Ta có: \(2\sqrt{48}-\sqrt{300}+3\sqrt{75}\)

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

\(=13\sqrt{3}\)

b: Ta có: \(\sqrt{16x}+2\sqrt{9x}-3\sqrt{4x}\)

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

\(=4\sqrt{x}\)

c: Ta có: \(3\sqrt{98}+4\sqrt{8}-7\sqrt{32}\)

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

\(=\sqrt{2}\)

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

lx

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

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

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

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

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

13

c,\(\left(x+1\right)^2-\left(x-1\right)^2=12\)

\(< =>x^2+2x+1-x^2+2x-1=12\)

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

d,\(3\left(x-1\right)< x+5\)

\(< =>3x-3< x+5\)

\(< =>3x-x< 5+3=8\)

\(< =>x< \frac{8}{2}=4\)