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\(2^{x-1}-12=20\\ \Rightarrow2^{x-1}=32\\ \Rightarrow2^{x-1}=2^5\\ \Rightarrow x-1=5\\ \Rightarrow x=6\)
\(\Leftrightarrow-2x+3\cdot\left\{12-2\left[3x-20-2x-4x\right]+1\right\}=45\)
\(\Leftrightarrow-2x+3\cdot\left\{12-2\cdot\left(-3x-20\right)+1\right\}=45\)
=>-2x+3(11+6x+40)=45
=>-2x+153+18x=45
=>16x=45-153=-108
hay x=-6,75
Ta có :-2x + 3x . { 12-2.[ 3.x-(20+2.x) - 4.x ] + 1 } = 45
<=>-2.x +3.x. [ 12-2 .( 3.x - 20 - 2.x - 4.x )+1 ) = 45
<=> -2.x + 3.x . [ 12- 2 ( -3.x -20 ) + 1 ] = 45
<=> -2.x + 3.x . ( 12 + 6.x + 40 + 1 ) = 45
<=>-2.x + 3.x . ( 53 + 6.x ) = 45
<=> -2.x + 159.x + 18.x2 = 45
<=> 157 .x + 18.x2 = 45
<=> 157.x + 18.x2 - 45 = 0
<=> 18.x2 - 5.x + 162.x - 45 = 0
<=> x. ( 18.x-5 ) + 9 .( 18.x-5) = 0
<=> ( 18.x-5 ) . ( x + 9 ) = 0
<=> \(\orbr{\begin{cases}18.x-5=0\\x+9=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{5}{18}\\x=-9\end{cases}}\)
Vậy x = { \(\frac{5}{18};-9\)}
\(\Rightarrow2^{x-1}=32=2^5\\ \Rightarrow x-1=5\\ \Rightarrow x=6\)
Sửa đề: \(2x+\dfrac{11}{6}+\dfrac{23}{12}+\dfrac{39}{20}+\dfrac{59}{30}+\dfrac{83}{42}+\dfrac{111}{56}=12\)
\(\Leftrightarrow2x+\left(2+2+2+2+2+2\right)-\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\right)=12\)
\(\Leftrightarrow2x-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)
\(\Leftrightarrow2x-\dfrac{3}{8}=0\)
hay x=3/16
= 100-2x+(-20+74-12-2x)
= 100-2x+(42-2x)
= 100-2x+42-2x
= 142-2x
k mk nha
2x + 12 = 20
2x = 20 - 12
2x = 18
x = 18 : 2
x = 9
=> x =9