Bài 5.

a.

1. \(\overrightarrow{a}.\overrightarrow{b}=1.3+2.\left(-4\right)=3-8=-5\)

2. \(\left(\overrightarrow{a}+\overrightarrow{b}\right)=\left(4;-2\right);\left(\overrightarrow{a}-\overrightarrow{b}\right)=\left(-2;6\right)\) \(\Rightarrow\left(\overrightarrow{a}+\overrightarrow{b}\right)\left(\overrightarrow{a}+\overrightarrow{b}\right)=\left(-8;-12\right)\)

3. \(\left(2\overrightarrow{a}+\overrightarrow{b}\right)=\left(5;0\right);\left(\overrightarrow{a}-2\overrightarrow{b}\right)=\left(-5;10\right)\) \(\Rightarrow\left(2\overrightarrow{a}+\overrightarrow{b}\right)\left(\overrightarrow{a}-2\overrightarrow{b}\right)=\left(-25;0\right)\)

ta có \(x=8+2y\)

\(A=\left(8+2y\right)\left(8+2y+2y-5\right)+y\left(4y-40-10y+10\right)-y\left(8+2y\right)\)

\(=\left(8+2y\right)\left(3+4y\right)-y\left(30+6y\right)-y\left(8+2y\right)\)

\(=\left(3+4y-y\right)\left(8+2y\right)-30y-6y^2=24+6y^2+24y+6y-30y-6y^2=24\)

VT: 2( 5x - 8 ) - 3( 4x - 5 )= 10x - 16 - 12x + 15 = - 2x - 1

VP: 4( 3x - 4 ) + 11 = 12x -16 + 11 = 12x - 5

ta có VT = VP suy ra 

- 2x - 1 = 12x - 5 \(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)

Đáp án là 13121543 nha

Dạng 2

1. \(x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^3-x+3\right)\) = \(2x^2+x-x^3-2x+x^3-x+3=3\)

2. \(x\left(3x^2-x+5\right)-\left(2x^3+3x-16\right)-x\left(x^2-x+2\right)\)

  = \(3x^3-x^2+5x-2x^3-3x+16-x^3+x^2-2x=16\)

Nhân hết vào là ra đáp án đó bạn

\(\int\dfrac{1}{2x+3}dx=\dfrac{1}{2}ln\left|2x+3\right|+C\)

ta có \(f\left(2\right)=\dfrac{1}{2}ln\left|2\times2+3\right|+C=\dfrac{1}{2}ln7+C=1\Leftrightarrow C=1-\dfrac{1}{2}ln7\)

Ta có:

\(Mg+2HCl\rightarrow MgCl_2+H_2\) ; \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Đặt số mol Mg và Al lần lượt là a và b (a,b>0)
theo bài ra ta có hệ 

\(\left\{{}\begin{matrix}24a+27b=5,1\\a+1,5b=\dfrac{5,6}{22,4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\%Mg=\dfrac{0,1\times24}{5,1}=47,06\%\Rightarrow\%Al=100\%-47,06\%=52,94\%\)

Theo PT có \(n_{HCl}=2n_{Mg}+3n_{Al}=2\times0,1+3\times0,1=0,5\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,5\times36,5=18,25\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{18,25}{10\%}=182,5\left(g\right)\)

\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)

\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3\downarrow+3NaCl\)

+ Với NaOH vừa đủ

\(a=m_{Mg\left(OH\right)_2}+m_{Al\left(OH\right)_3}=0,1\times58+0,1\times78=13,6\left(g\right)\)

+ Với NaOH dư có thêm PT

\(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)

\(\Rightarrow a=m_{Mg\left(OH\right)_2}=0,1\times58=5,8\left(g\right)\)