a: \(x+40\%\cdot x=5\)

=>\(x\left(1+0,4\right)=5\)

=>1,4x=5

=>\(x=\dfrac{5}{1,4}=\dfrac{50}{14}=\dfrac{25}{7}\)

b: \(1,2x-80\%x=\dfrac{1}{4}\)

=>1,2x-0,8x=0,25

=>0,4x=0,25

=>\(x=\dfrac{0.25}{0.4}=\dfrac{25}{40}=\dfrac{5}{8}\)

d: \(x\cdot x-\dfrac{1}{9}=\dfrac{1}{3}\)

=>\(x^2=\dfrac{1}{9}+\dfrac{1}{3}=\dfrac{4}{9}\)

=>\(\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

2x-31=4646:46

=>2x-31=101

=>2x=31+101=132

=>\(x=\dfrac{132}{2}=66\)

12+40:(17-x)=52

=>40:(17-x)=40

=>17-x=40:40=1

=>x=17-1=16

40:( 17-x)=40

17-x=40: 40

17-x=1

x=17-1

x=16

6: \(\widehat{mOn}=\widehat{mOy}+\widehat{nOy}\)

\(=\dfrac{1}{2}\left(\widehat{xOy}+\widehat{zOy}\right)\)

\(=\dfrac{1}{2}\cdot180^0=90^0\)

5:

a: tia Oc nằm giữa hai tia Oa và Ob

=>\(\widehat{aOc}+\widehat{bOc}=\widehat{aOb}\)

=>\(\widehat{bOc}=100^0-40^0=60^0\)

b: Od là phân giác của góc cOb

=>\(\widehat{cOd}=\dfrac{\widehat{cOb}}{2}=\dfrac{60^0}{2}=30^0\)

Bài 7:

a: \(\left[0,\left(30\right)+0,\left(60\right)\right]x=10\)

=>\(\left(\dfrac{10}{33}+\dfrac{20}{33}\right)\cdot x=10\)

=>\(\dfrac{30}{33}\cdot x=10\)

=>\(x\cdot\dfrac{10}{11}=10\)

=>\(x=10:\dfrac{10}{11}=11\)

b: \(0,\left(12\right):1,\left(6\right)=x:0,\left(4\right)\)

=>\(x:\dfrac{4}{9}=\dfrac{4}{33}:\dfrac{5}{3}\)

=>\(x:\dfrac{4}{9}=\dfrac{4}{33}\cdot\dfrac{3}{5}=\dfrac{4}{11\cdot5}=\dfrac{4}{55}\)

=>\(x=\dfrac{4}{55}:\dfrac{4}{9}=\dfrac{9}{55}\)

\(3n+22⋮2n+3\)

=>\(6n+44⋮2n+3\)

=>\(6n+9+35⋮2n+3\)

=>\(35⋮2n+3\)

mà 2n+3>=3(Vì n là số tự nhiên)

nên \(2n+3\in\left\{5;7;35\right\}\)

=>\(n\in\left\{1;2;16\right\}\)

a: \(1,\left(6\right)+\left(\dfrac{-2}{7}\right)-\left(-1,2\right)\)

\(=\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}\)

\(=\dfrac{175}{105}-\dfrac{30}{105}+\dfrac{126}{105}=\dfrac{271}{105}\)

b: \(0,\left(3\right)-\dfrac{-5}{6}+\dfrac{3}{4}=\dfrac{1}{3}+\dfrac{5}{6}+\dfrac{3}{4}\)

\(=\dfrac{4}{12}+\dfrac{10}{12}+\dfrac{9}{12}=\dfrac{23}{12}\)

c: \(0,\left(3\right)-1,\left(3\right)+\dfrac{2}{7}=\dfrac{1}{3}-\dfrac{4}{3}+\dfrac{2}{7}=-1+\dfrac{2}{7}=-\dfrac{5}{7}\)

d: \(-0,8\left(3\right)-\left(\dfrac{-3}{8}+\dfrac{1}{10}\right)\)

\(=-\dfrac{5}{6}+\dfrac{3}{8}-\dfrac{1}{10}\)

\(=-\dfrac{100}{120}+\dfrac{45}{120}-\dfrac{12}{120}=\dfrac{-67}{120}\)

\(16\left(143-x\right)-2\left(143-x\right)=1414\)

=>\(14\left(143-x\right)=1414\)

=>143-x=1414:14=101

=>x=143-101=42

16 x (143 - \(x\)) - 2 x (143 - \(x\)) = 1414

        (143 - \(x\)) x (16 - 2) = 1414

       (143 - \(x\)) x 14 = 1414

        143 - \(x\)          = 1414 : 14

        143 - \(x\)         = 101

                 \(x\)         = 143 - 101

                 \(x\)         = 42

              Vậy \(x=42\) 

\(A=\dfrac{7}{19\cdot31}+\dfrac{5}{19\cdot43}+\dfrac{3}{23\cdot43}+\dfrac{11}{23\cdot57}+\dfrac{19}{38\cdot57}\)

\(=2\left(\dfrac{7}{31\cdot38}+\dfrac{5}{38\cdot43}+\dfrac{3}{43\cdot46}+\dfrac{11}{46\cdot57}+\dfrac{19}{57\cdot76}\right)\)

\(=2\left(\dfrac{1}{31}-\dfrac{1}{38}+\dfrac{1}{38}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{57}+\dfrac{1}{57}-\dfrac{1}{76}\right)\)

\(=2\left(\dfrac{1}{31}-\dfrac{1}{76}\right)=2\cdot\dfrac{45}{31\cdot76}=\dfrac{45}{1178}\)

Bài 5:

a: \(3,\left(15\right)=3+\dfrac{15}{99}=3+\dfrac{5}{33}=\dfrac{3\cdot33+5}{33}=\dfrac{104}{33}\)

b: \(0,2\left(07\right)=0,2+0,0\left(07\right)=\dfrac{41}{198}\)

c: \(0,1\left(37\right)=0,1+0,0\left(37\right)=\dfrac{1}{10}+\dfrac{37}{990}=\dfrac{68}{495}\)

d: \(0,20\left(23\right)=0,20+0,00\left(23\right)=0,2+\dfrac{23}{9900}=\dfrac{2003}{9900}\)