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

\(a,=x^2+x+4x+4=\left(x+1\right)\left(x+4\right)\\ b,=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\\ c,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ d,=3\left(x^2-2x+5x-10\right)=3\left(x-2\right)\left(x+5\right)\\ e,=-3x^2+6x-x+2=\left(x-2\right)\left(1-3x\right)\\ f,=x^2-x-6x+6=\left(x-1\right)\left(x-6\right)\\ h,=4\left(x^2-3x-6x+18\right)=4\left(x-3\right)\left(x-6\right)\\ i,=3\left(3x^2-3x-8x+5\right)=3\left(x-1\right)\left(3x-8\right)\\ k,=-\left(2x^2+x+4x+2\right)=-\left(2x+1\right)\left(x+2\right)\\ l,=x^2-2xy-5xy+10y^2=\left(x-2y\right)\left(x-5y\right)\\ m,=x^2-xy-2xy+2y^2=\left(x-y\right)\left(x-2y\right)\\ n,=x^2+xy-3xy-3y^2=\left(x+y\right)\left(x-3y\right)\)

Bào quan riboxom trong chất tế bào có chức năng gì? 

Bài 5: 

d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y-z}{2+3-4}=\dfrac{-20}{1}=-20\)

Do đó: x=-40; y=-60; z=-80

1.Yes, they do

2..Yes, it is

3.People buy fruits and flowers from the market and decorate their house

4.People visit their family and friends

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

Câu A

A

a) ĐKXĐ: \(\left\{{}\begin{matrix}2x+3\ne0\\2x+1\ne0\\\left(2x+3\right)\left(2x+1\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{3}{2}\\x\ne-\dfrac{1}{2}\\\left(2x+3\right)\left(2x+1\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{3}{2}\\x\ne-\dfrac{1}{2}\end{matrix}\right.\)

b) \(\Rightarrow P=\dfrac{2\left(2x+1\right)+3\left(2x+3\right)-6x-5}{\left(2x+3\right)\left(2x+1\right)}\)

\(\Rightarrow P=\dfrac{4x+2+6x+9-6x-5}{\left(2x+3\right)\left(2x+1\right)}\)

\(\Rightarrow P=\dfrac{4x+6}{\left(2x+3\right)\left(2x+1\right)}\)

\(\Rightarrow P=\dfrac{2\left(2x+3\right)}{\left(2x+3\right)\left(2x+1\right)}\)

\(\Rightarrow P=\dfrac{2}{2x+1}\)

c) \(P=-1\Rightarrow\dfrac{2}{2x+1}=-1\\ \Rightarrow2=-2x-1\\ \Rightarrow2x=-3\\ \Rightarrow x=-\dfrac{3}{2}\)