1: (2x-1)*x>0

TH1: \(\left\{{}\begin{matrix}2x-1>0\\x>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>\dfrac{1}{2}\\x>0\end{matrix}\right.\Leftrightarrow x>\dfrac{1}{2}\)

TH2: \(\left\{{}\begin{matrix}2x-1< 0\\x< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x< 0\end{matrix}\right.\)

=>x<0

2: 
ĐKXĐ: x<>1

\(\dfrac{x+3}{x-1}< 0\)

TH1: \(\left\{{}\begin{matrix}x+3>0\\x-1< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>-3\\x< 1\end{matrix}\right.\)

=>-3<x<1

TH2: \(\left\{{}\begin{matrix}x+3< 0\\x-1>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< -3\\x>1\end{matrix}\right.\)

=>Loại

c: 

ĐKXĐ: x<>0

\(\dfrac{x^2-2}{5x}< 0\)

TH1: \(\left\{{}\begin{matrix}x^2-2< 0\\5x>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2< 2\\x>0\end{matrix}\right.\Leftrightarrow0< x< \sqrt{2}\)

TH2: \(\left\{{}\begin{matrix}x^2-2>0\\5x< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2>2\\x< 0\end{matrix}\right.\Leftrightarrow-\sqrt{2}< x< 0\)

d: (x-3)(x+7)>0

TH1: \(\left\{{}\begin{matrix}x-3>0\\x+7>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>3\\x>-7\end{matrix}\right.\Leftrightarrow x>3\)

TH2: \(\left\{{}\begin{matrix}x-3< 0\\x+7< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 3\\x< -7\end{matrix}\right.\)

=>x<-7

\(x^2=x^4\)

=>\(x^2\left(1-x^2\right)=0\)

=>\(\left[{}\begin{matrix}x^2=0\\1-x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

\(x^2=x^4\\ x^4-x^2=0\\ x^2\left(x^2-1\right)=0\)

TH1: `x^2=0`

`=> x=0`

TH2: `x^2-1=0`

`=>x^2=1^2`

`=>x=1` hoặc `x=-1` 

Tuổi con là (28-8):(3-1)=20:2=10(tuổi)

Tuổi mẹ là 3x10+8=38(tuổi)

2 lần tuổi conlà:

28 - 8 = 20 (tuổi) 

Tuổi con là:

20 : 2 = 10 (tuổi)

Tuổi mẹ là: 

28 + 10 = 38 (tuổi)

ĐS: ... 

Đặt A=1+4+7+...+58

Số số hạng là \(\dfrac{58-1}{3}+1=\dfrac{57}{3}+1=20\left(số\right)\)

Tổng của dãy số là \(\left(58+1\right)\cdot\dfrac{20}{2}=59\cdot10=590\)

\(\dfrac{18\cdot123+9\cdot4567\cdot2+3\cdot5310\cdot6}{1+4+7+...+58-410}\)

\(=\dfrac{18\left(123+4567+5310\right)}{590-410}=\dfrac{18\cdot10000}{180}\)

=10000:10=1000

3,62 x 10 = 36,2 

3,62 x 100 = 362 

3,62 x 1000 = 3620

3,62 : 10 = 0,362 

3,62 : 100 = 0,0362 

3,62 : 1000 = 0,00362 

\(31\cdot121^{39}\cdot6+17\cdot11^{60}\cdot18\)

\(=31\cdot11^{78}\cdot6+17\cdot11^{60}\cdot18\)

\(=11\left(31\cdot11^{77}\cdot6+17\cdot11^{59}\cdot18\right)⋮11\)

\(31\cdot121^{39}\cdot6+17\cdot11^{60}\cdot18\\ =31\cdot\left(11^2\right)^{39}\cdot6+17\cdot11^{60}\cdot18\\ =31\cdot11^{78}\cdot6+17\cdot11^{60}\cdot18\\ 11\cdot\left(31\cdot11^{77}\cdot6+17\cdot11^{59}\cdot18\right)⋮11\)

\(0< a< \dfrac{\Omega}{2}\)

=>\(sina>0\)

=>\(sina=\sqrt{1-cos^2a}=\dfrac{4}{5}\)

\(\dfrac{3}{2}\Omega< b< 2\Omega\)

=>\(sinb< 0\)

=>\(sinb=-\sqrt{1-\left(\dfrac{12}{13}\right)^2}=-\dfrac{5}{13}\)

\(tana=\dfrac{sina}{cosa}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)

\(tanb=\dfrac{sinb}{cosb}=\dfrac{-5}{13}:\dfrac{12}{13}=-\dfrac{5}{12}\)

\(tan\left(a+b\right)=\dfrac{tana+tanb}{1-tana\cdot tanb}\)

\(=\dfrac{\dfrac{4}{3}+\dfrac{-5}{12}}{1-\dfrac{4}{3}\cdot\dfrac{-5}{12}}=\dfrac{11}{12}:\left(1+\dfrac{20}{36}\right)=\dfrac{11}{12}:\dfrac{14}{9}\)

\(=\dfrac{11}{12}\cdot\dfrac{9}{14}=\dfrac{11\cdot3}{4\cdot14}=\dfrac{33}{56}\)

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