Lời giải:
a.

$(x-15).27=0$

$x-15=0:27=0$

$x=15+0=15$

b.

$23(42-x)=0$

$42-x=0$

$x=42$

c.

$(9x+2).3=60$

$9x+2=60:3=20$

$9x=18$

$x=2$
d.

$71+(26-3x):5=75$

$(26-3x):5=75-71=4$

$26-3x=4.5=20$

$3x=26-20=6$

$x=6:2=3$

1b) có thiếu ko cau?

`1a)5^3` và `3^5`

`5^3=125`

`3^5=243`

Vì `243>125` nên `3^5>5^3`

__

`c)3^24` và `27^7`

`27^7=(3^3)^7=3^21`

Vì `3^24>3^31` nên `3^24>27^7`

`2a)x^3=216`

`=>x^3=6^3`

`=>x=6`

__

`b)3^x+15=18`

`=>3^x=18-15`

`=>3^x=3`

`=>x=1`

a) x=1
b) x=2

\(a,\Rightarrow x=30-18=12\\ b,\Rightarrow x+6=45:5=9\\ \Rightarrow x=9-6=3\\ c,\Rightarrow38-3x=4^2=16\\ \Rightarrow3x=38-16=22\\ \Rightarrow x=\dfrac{22}{3}\)

a)

(x - 50) : 45 + 240 = 300

(x - 50) : 45 = 60

x – 50 = 2700

x = 2750

b)

7200 : [200 + (33 600 : x) - 500] = 4

200 + (33 600 : x) – 500 = 1800

33 600 : x = 2100

x = 16

\(a,3\cdot x-15=x+35\)

\(\Rightarrow3x-x=35+15\)

\(\Rightarrow 2x=50\)

\(\Rightarrow x = 50:2\)

\(\Rightarrow x= 25\)

\(b,(8x-16)(x-5)=0\)

\(+, TH1: 8x-16=0\)

\(\Rightarrow8x=16\)

\(\Rightarrow x = 16:8\)

\(\Rightarrow x=2\)

\(+,TH2: x-5=0\)

\(\Rightarrow x =5\)

\(c,x(x+1)=2+4+6+8+10+...+2500\)  \(^{\left(1\right)}\)

Đặt \(A=2+4+6+8+10+...+2500\)

Số các số hạng của \(A\) là: \(\left(2500-2\right):2+1=1250\left(số\right)\)

Tổng \(A\) bằng: \(\left(2500+2\right)\cdot1250:2=1563750\)

Thay \(A=1563750\) vào \(^{\left(1\right)}\), ta được:

\(x\left(x+1\right)=1563750\)

\(\Rightarrow x\left(x+1\right)=1250\cdot1251\)

\(\Rightarrow x =1250\)

#\(Toru\)

( x - 50 ) : 45 + 240 = 300

( x - 50 ) : 45 = 300 - 240

( x - 50 ) : 45 = 60

 x - 50  = 60 x 45

 x - 50  = 2700

        x  = 2700 + 50

        x = 2750 

\(\Leftrightarrow\left(x-50\right)\div45=300-240\)

\(\Leftrightarrow\left(x-50\right)\div45=60\)

\(\Leftrightarrow x-50=60\times45\)

\(\Leftrightarrow x-50=2700\)

\(\Leftrightarrow x=2700+50\)

\(\Leftrightarrow x=2750\)

Vậy x = 2750