22x+1 - 15 = 17
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a)
\(4\left(x-1\right)-\left(3x+2\right)=16\)
=> 4x - 4 - 3x - 2 = 16
=> x - 6 = 16
=> x = 22
Vậy .......
b)
\(7\left(x-5\right)+5\left(15-x\right)=130\)
=> 7x - 35 + 75 - 5x = 130
=> 2x + 40 = 130
=> 2x = 90
=> x = 45
Vậy........
c)
\(30\left(x-2\right)-6\left(x-5\right)-22x=100\)
=> 30x - 60 - 6x + 30 - 22x = 100
=> 2x - 30 = 100
=> 2x = 130
=> x = 65
Vậy.....
d) \(x-17+x=x-23\)
=> 2x - 17 = x - 23
=> x = -6
Vậy .......
5 x - 1 + 2 6 - 7 x - 1 4 = 2 2 x + 1 7 - 5 ⇔ 5 x - 3 6 - 7 x - 1 4 = 4 x + 2 7 - 5
⇔ 14(5x – 3) – 21(7x – 1) = 12(4x + 2) – 5.84
⇔ 70x – 42 – 147x + 21 = 48x + 24 – 420
⇔ 70x – 147x – 48x = 24 – 420 + 42 – 21
⇔ -125x = -375
⇔ x = 3
Phương trình có nghiệm x = 3
Đặt \(A=2^{2x}+2^{2x+1}+...+2^{2x+1918}\)
=>\(2\cdot A=2^{2x+1}+2^{2x+2}+...+2^{2x+1919}\)
=>\(A=2^{2x+1919}-2^{2x}\)
Theo đề, ta có; \(2^{2x+1919}-2^{2x}=2^{1923}-2^4\)
=>\(2^{2x}\cdot\left(2^{2019}-1\right)=2^4\left(2^{2019}-1\right)\)
=>2x=4
=>x=2
\(-\dfrac{15}{23}:\dfrac{22x}{7}=-\dfrac{14x}{11}:\left(13+\dfrac{4}{5}\right)\)
=>\(-\dfrac{15}{23}\cdot\dfrac{7}{22x}=\dfrac{-14x}{11}:\dfrac{69}{5}\)
=>\(-\dfrac{105}{23\cdot22x}=\dfrac{-70x}{11\cdot69}\)
=>\(\dfrac{-3}{2x}=\dfrac{-2x}{3}\)
=>\(4x^2=9\)
=>\(x^2=\dfrac{9}{4}\)
=>\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(\dfrac{-15}{23}:\dfrac{22x}{7}=\dfrac{-14x}{11}:\left(13+\dfrac{4}{5}\right)\)
\(\Leftrightarrow-\dfrac{15}{23}\cdot\dfrac{7}{22x}=\dfrac{-14x}{11}:\dfrac{69}{5}\)
=>\(\dfrac{-15\cdot7}{23\cdot22x}=\dfrac{-14x}{11}\cdot\dfrac{5}{69}\)
=>\(\dfrac{5\cdot3\cdot7}{23\cdot2\cdot11x}=\dfrac{2\cdot7x}{11}\cdot\dfrac{5}{3\cdot23}\)
=>\(\dfrac{3}{2x}=\dfrac{2x}{3}\)
=>\(4x^2=9\)
=>\(x^2=\dfrac{9}{4}\)
=>\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(2^{2x+1}-15=17\\ =>2^{2x+1}=32=2^5\\ =>2x+1=5\\ =>2x=4\\ =>x=2\)
22x + 1 - 15 = 17
22x+1 = 17 + 15
22x+1 = 32
22x+1 = 25
2x + 1 = 5
2x = 5 - 1
2x = 4
x = 4 : 2
x = 2
Vậy x = 2.