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 .......

a) 4(x-1)-(3x+2)=16 b)7(x-5)+5(15-x)=130

<=>4x-4-3x-2=16 <=>7x-35+75-5x=130

<=>x-6=16 <=>2x+40=130

<=>x=22 <=>2x=90

<=>x=45

c) 30(x- 2 ) - 6( x - 5 ) - 22x = 100 d) x-17+x=x-23

<=>30x-60-6x+30-22x=100 <=>2x-x=-23+17

<=>2x-30=100 <=>x=-6

<=>2x=130

<=>x=65

a) 4(x-1)-(3x+2)=16

<=>4x-4-3x-2=16

<=>x-6=16

<=>x=22

c) 30(x- 2 ) - 6( x - 5 ) - 22x = 100

<=>30x-60-6x+30-22x=100

<=>2x-30=100

<=>2x=130

<=>x=65

b)7(x-5)+5(15-x)=130

<=>7x-35+75-5x=130

<=>2x+40=130

<=>2x=90

<=>x=45

d) x-17+x=x-23

<=>2x-x=-23+17

<=>x=-6

<=>2x=130

<=>x=65

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

\(-\dfrac{15}{23}:\dfrac{22x}{7}=-\dfrac{14x}{11}:\left(13+\dfrac{4}{5}\right)\)

=>\(-\dfrac{15}{23}\cdot\dfrac{7}{22x}=\dfrac{-14x}{11}:\dfrac{69}{5}\)

=>\(-\dfrac{105}{23\cdot22x}=\dfrac{-70x}{11\cdot69}\)

=>\(\dfrac{-3}{2x}=\dfrac{-2x}{3}\)

=>\(4x^2=9\)

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

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

\(\dfrac{-15}{23}:\dfrac{22x}{7}=\dfrac{-14x}{11}:\left(13+\dfrac{4}{5}\right)\)

\(\Leftrightarrow-\dfrac{15}{23}\cdot\dfrac{7}{22x}=\dfrac{-14x}{11}:\dfrac{69}{5}\)

=>\(\dfrac{-15\cdot7}{23\cdot22x}=\dfrac{-14x}{11}\cdot\dfrac{5}{69}\)

=>\(\dfrac{5\cdot3\cdot7}{23\cdot2\cdot11x}=\dfrac{2\cdot7x}{11}\cdot\dfrac{5}{3\cdot23}\)

=>\(\dfrac{3}{2x}=\dfrac{2x}{3}\)

=>\(4x^2=9\)

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

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

Vậy bất phương trình có tập nghiệm