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

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

Câu 3: ĐKXĐ: x+1>=0

=>x>=-1

\(\sqrt{x+1}=2\)

=>\(x+1=2^2=4\)

=>x=4-1

=>x=3(nhận)

Câu 2: \(\sqrt8+5\sqrt2-\sqrt{32}+6\sqrt{\frac12}\)

\(=2\sqrt2+5\sqrt2-4\sqrt2+3\sqrt2=7\sqrt2-\sqrt2=6\sqrt2\)

Câu 1: ĐKXĐ: x>=0

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

a: Ta có: \(\overrightarrow{PA}+2\cdot\overrightarrow{PB}=\overrightarrow{0}\)

=>\(\overrightarrow{PA}=-2\cdot\overrightarrow{PB}\)

=>P nằm giữa A và B sao cho AP=2PB

AP+PB=AB

=>AB=2PB+PB=3BP

=>\(BP=\frac13BA;AP=\frac23AB\)

Ta có: \(5\cdot\overrightarrow{AQ}-2\cdot\overrightarrow{AC}=\overrightarrow{0}\)

=>\(5\cdot\overrightarrow{AQ}=2\cdot\overrightarrow{AC}\)

=>\(\overrightarrow{AQ}=\frac25\cdot\overrightarrow{AC}\)

Ta có: \(\overrightarrow{PQ}=\overrightarrow{PA}+\overrightarrow{AQ}\)

\(=-\frac23\cdot\overrightarrow{AB}+\frac25\cdot\overrightarrow{AC}=-2\left(\frac13\cdot\overrightarrow{AB}-\frac15\cdot\overrightarrow{AC}\right)\)

\(=-\frac{2}{15}\left(5\cdot\overrightarrow{AB}-3\cdot\overrightarrow{AC}\right)\) (1)
b: Xét ΔABC có AM là đường trung tuyến

nên \(\overrightarrow{AM}=\frac12\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)

=>\(\overrightarrow{AI}=\frac12\cdot\overrightarrow{AM}=\frac14\cdot\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)

\(\overrightarrow{PI}=\overrightarrow{PA}+\overrightarrow{AI}\)

\(=-\frac23\cdot\overrightarrow{AB}+\frac14\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=-\frac23\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{AC}\)

\(=\frac{-5}{12}\cdot\overrightarrow{AB}+\frac{3}{12}\cdot\overrightarrow{AC}=-\frac{1}{12}\left(5\cdot\overrightarrow{AB}-3\cdot\overrightarrow{AC}\right)\) (2)

Từ (1),(2) suy ra \(\frac{\overrightarrow{PI}}{\overrightarrow{PQ}}=\frac{-1}{12}:\frac{-2}{15}=\frac{1}{12}\cdot\frac{15}{2}=\frac{15}{24}=\frac58\)

=>P,I,Q thẳng hàng

C

D. of