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

Bài 2.8
Cửa hàng lãi:\(\left(1000000:800000\right)-100\%=25\%\left(giavon\right)\)
Bài 2.9
Số tiền phải bán để lãi 25% giá vốn:\(740000+\left(740000\times25\%\right)==925000\left(đồng\right)\)

a) \(\Rightarrow\left|\dfrac{3}{4}+x\right|=0\Rightarrow\dfrac{3}{4}+x=0\Rightarrow x=-\dfrac{3}{4}\)

b) \(\Rightarrow x+0,4=\dfrac{4}{9}:\dfrac{2}{3}=\dfrac{2}{3}\Rightarrow x=\dfrac{2}{3}-0,4=\dfrac{4}{15}\)

`A=1/(x+sqrtx)+(2sqrtx)/(x-1)-1/(x-sqrtx)`

`=(sqrtx-1+2x-sqrtx-1)/(sqrtx(x-1))`

`=(2x-2)/(sqrtx(x-1))`

`=2/sqrtx`

`b)A=1`

`<=>2/sqrtx=1`

`<=>sqrtx=2`

`<=>x=4(tm)`

\(\left(3\sqrt{7}\right)^2=63>28=\left(\sqrt{28}\right)^2\) hoặc \(3\sqrt{7}>2\sqrt{7}=\sqrt{28}\)

C1: $\sqrt{28}=\sqrt{4.7}=2\sqrt 7$

Ta có: $3>2$

$\Leftrightarrow 3\sqrt 7>3\sqrt 7$ hay $3\sqrt 7>\sqrt{28}$

C2: $3\sqrt{7}=\sqrt{63}$

Ta có: $63>28$

$\Leftrightarrow\sqrt{63}>\sqrt{28}$ hay $3\sqrt 7>\sqrt{28}$

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

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