a) \(2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)

\(x+\frac{2}{5}=\frac{-7}{20}\)

\(x=\frac{-13}{20}\)

Vậy \(x=\frac{-13}{20}\)

b)\(x-1\frac{1}{8}-\frac{2}{3}x-\frac{5}{6}x=75\%\)

  \(\left(x-\frac{2}{3}x-\frac{5}{6}x\right)-\frac{9}{8}=\frac{3}{4}\)

\(\frac{-1}{2}x-\frac{9}{8}=\frac{3}{4}\)

\(\frac{-1}{2}x=\frac{15}{8}\)

\(x=\frac{-15}{4}\)

Vậy \(x=\frac{-15}{4}\)

2.x² = 18

 x² = 18 : 2

x² = 9

x² = 3²

\(\Rightarrow\)x = 3

( 2⁰ + 2¹ + 2² ) . x - 3 = 18

=> ( 1 + 2 + 4 ) . x = 21

=> 7 . x = 21

=> x = 21 : 7 = 3

2 . x² = 18

=> x² = 9

=> x² = 3² = ( -3 )²

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

a) Ta có 

1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x + 2 = 5

rút gọn ta được : 1 - 1/x+2 = 5

<=> x + 2 - 1 / x+ 2

<=> x + 1 / x + 2 = 5

<=> x+ 1 = 5x + 10

<=> - 4x = 9

<=> x = -9/4

B) / x +4/ = 2^0 + 1 ^ 2013

=> /x + 4/ = 1 + 1

=> / x + 4 / = 2

TH 1 :  x+ 4 = 2

 => x = 2 - 4 = -2

TH2 : x + 4 = -2

=> x = -2 - 4 = -6

=> x = { - 2 , -6 }

\((x-3)^{3}=3^{0}+3^{1}+2^{2}.5\)

\(=>(x-3)^{3}=1+3+4.5\)

\(=>(x-3)^{3}=24\)

Vì không có cơ số nào có số mũ là \(3\) mà là \(24\)

Vậy \(x\in\left\{\varnothing\right\}\)

a) \(\left(x+7\right).\left(x-9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0-7\\x=0+9\end{cases}}\Rightarrow\orbr{\begin{cases}x=-7\left(TM\right)\\x=9\left(TM\right)\end{cases}}\)

Vậy \(x\in\left\{-7;9\right\}.\)

b) \(\left(x+2\right).\left(x^2+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x^2+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0-2\\x^2=0-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\left(TM\right)\\x^2=-1\left(vôlí,loại\right)\end{cases}}\)

Vậy \(x=-2.\)

Chúc bạn học tốt!

Câu đầu:\(\left(x+7\right).\left(x-9\right)=0\)

Vì tích trên bằng 0 nên 1 trong 2 vế phải bằng 0.

TH1: \(x+7=0\)

\(\Rightarrow x=-7\)

TH2: \(x-9=0\Rightarrow x=9\)

Vậy \(x\in\left\{-7;9\right\}\)

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

=> \(\orbr{\begin{cases}\left(x+1\right)^2=0\\\left(x-2\right)^2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

b) \(\left(x-9\right)^5\left(x-5\right)^8=0\)

=> \(\orbr{\begin{cases}\left(x-9\right)^5=0\\\left(x-5\right)^8=0\end{cases}}\)

=> \(\orbr{\begin{cases}x-9=0\\x-5=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=9\\x=5\end{cases}}\)

\(2^x+2^{x+2}=32.\left(2^2+1\right)\)

\(\Rightarrow2^x+2^{x+2}=32.5\)

\(\Rightarrow2^x+2^{x+2}=160\)

\(\Rightarrow2^x\left(1+4\right)=160\)

\(\Rightarrow2^x.5=160\)

\(\Rightarrow2^x=160:5=32\)

\(\Rightarrow2^x=2^5\)

\(\Rightarrow x=5\)

Vậy x = 5