Trắc nghiệm
Câu 1:B

Câu 2:C

Câu 3:A

Câu 4:C

Câu 5:A 

Câu 6:B

Câu 7:D

Câu 8:D

Bài 8:

a: \(A=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)

\(=7\cdot\dfrac{6}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)

b: \(B=2\left(\dfrac{1}{15}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{21}+...+\dfrac{1}{87}-\dfrac{1}{90}\right)\)

\(=2\left(\dfrac{1}{15}-\dfrac{1}{90}\right)\)

\(=2\cdot\dfrac{5}{90}=\dfrac{10}{90}=\dfrac{1}{9}\)

c: \(C=3\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)

\(=3\cdot\dfrac{24}{200}=\dfrac{72}{200}=\dfrac{9}{25}\)

a) \(\Leftrightarrow x^2=\sqrt{4}\)

\(\Leftrightarrow x^2=2\Leftrightarrow x=\pm2\)

b) \(\Leftrightarrow\sqrt{\left(\dfrac{1}{2}x+1\right)^2}=9\)

\(\Leftrightarrow\left|\dfrac{1}{2}x+1\right|=9\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+1=9\\\dfrac{1}{2}x+1=-9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=-16\end{matrix}\right.\)

c) \(\Leftrightarrow\sqrt{2x}-4\sqrt{2x}+16\sqrt{2x}=52\left(đk:x\ge0\right)\)

\(\Leftrightarrow13\sqrt{2x}=52\Leftrightarrow\sqrt{2x}=4\Leftrightarrow2x=16\Leftrightarrow x=8\left(tm\right)\)

f: Ta có: \(\sqrt{\dfrac{50-25x}{4}}-8\sqrt{2-x}+\sqrt{18-9x}=-10\)

\(\Leftrightarrow\sqrt{2-x}\cdot\dfrac{5}{2}-8\sqrt{2-x}+3\sqrt{2-x}=-10\)

\(\Leftrightarrow\sqrt{2-x}=4\)

\(\Leftrightarrow2-x=16\)

hay x=-14

9NaOH+3AlCl3- 3Al(OH)3+9NaCl

c: Ta có: \(\left(x-3\right)^3-\left(x^3-27\right)+9\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-9x^2+27x+9x^2+18x+9=15\)

\(\Leftrightarrow45x=6\)

hay \(x=\dfrac{2}{15}\)

d: Ta có: \(x\left(x-5\right)\left(x+5\right)-\left(x^3+8\right)=3\)

\(\Leftrightarrow x^3-25x-x^3-8=3\)

\(\Leftrightarrow-25x=11\)

hay \(x=-\dfrac{11}{25}\)

Đổi: 300m2 = 3000000cm2

Diện tích 1 viên gạch là:

50 x 50 = 2500 (cm2)

Vậy số viên gạch có là:

3000000 : 2500 = 1200 (viên)

Đ/S: 1200 viên