a) \(\Rightarrow5\left(x-10\right)=10\)

\(\Rightarrow x-10=2\Rightarrow x=12\)

b) \(\Rightarrow3\left(70-x\right)+5=92\)

\(\Rightarrow3\left(70-x\right)=87\)

\(\Rightarrow70-x=29\Rightarrow x=41\)

c) \(\Rightarrow230+\left[16+\left(x-5\right)\right]=315\)

\(\Rightarrow11+x=85\Rightarrow x=74\)

d) \(\Rightarrow707:\left(2^x-5+74\right)=7\)

\(\Rightarrow2^x-5+74=101\Rightarrow2^x-5=27\)

\(\Rightarrow2^x=32\Rightarrow x=5\)

Lời giải chi tiết: 

5-1=4     4-1=3    3-1=2   2+3=5     

5-2=3     4-2=2    3-2=1   3+2=5

5-3=2     4-3=1    2-1=1   5-2=3

5-4=1                               5-3=2

a) A = 2 + 2² + 2³ + ... + 2¹⁰⁰

⇒ 2A = 2² + 2³ + 2⁴ + ... + 2¹⁰¹

⇒ A = 2A - A

= (2² + 2³ + 2⁴ + ... + 2¹⁰¹) - (2 + 2² + 2³ + ... + 2¹⁰⁰)

= 2¹⁰¹ - 2

b) B = 1 + 5 + 5² + ... + 5¹⁵⁰

⇒ 5B = 5 + 5² + 5³ + ... + 5¹⁵¹

⇒ 4B = 5B - B

= (5 + 5² + 5³ + ... + 5¹⁵¹) - (1 + 5 + 5² + ... + 5¹⁵⁰)

= 5¹⁵¹ - 1

⇒ B = (5¹⁵¹ - 1) : 4

a) `(2x+5)^3-(2x-5)^3-(120x^2+49)`

`=(2x+5-2x+5)[(2x+5)^2+(2x+5)(2x-5)+(2x-5)^2]-(120x^2+49)`

`=10(12x^2+25)-(120x^2+49)`

`=120x^2+250-120x^2-49`

`=201`

b) `(4-5x)^2-(3+5x)^2=(4-5x+3+5x)(4-5x-3-5x)=7.(-10x+1)=-70x+7`

Lời giải:
a. 

$(2x+5)^3-(2x-5)^3-(120x^2+49)$

$=[(2x+5)-(2x-5)][(2x+5)^2+(2x+5)(2x-5)+(2x-5)^2]-(120x^2+49)$

$=10(4x^2+20x+25+4x^2-25+4x^2-20x+25)-(120x^2+49)$

$=10(12x^2+25)-(120x^2+49)=250-49=201$

b.

$(4-5x)^2-(3+5x)^2=[(4-5x)+(3+5x)][(4-5x)-(3+5x)]$

$=7(1-10x)$

1: =-15+10-35+28=-12

3: =20-12-8+12=12

2) -3(4 - 7) + 5(-3 + 2)

= -3.4 + 3.7 - 5.3 + 5.2

= -12 + 21 -15 + 10

= 31 - 27

= 4

4) -5(2 - 7) + 4(2 - 5)

= -5.2 + 5.7 + 4.2 - 4.5

= -10 + 35 + 8 - 20

= 38 - 30

= 8

Nhìn nó rối ghê á

1: =-15+10-35+28=-32

2 câu c,d làm tương tự  

5:

a: \(3^{2n}=\left(3^2\right)^n=9^n\)

\(\left(2^{3n}\right)=\left(2^3\right)^n=8^n\)

=>\(3^{2n}>2^{3n}\)

b: \(199^{20}=\left(199^4\right)^5=1568239201^5\)

\(2003^{15}=\left(2003^3\right)^5=8036054027^5\)

mà \(1568239201< 8036054027\)

nên \(199^{20}< 2003^{15}\)

4: \(100< 5^{2x-1}< 5^6\)

mà \(25< 100< 125\)

nên \(125< 5^{2x-1}< 5^6\)

=>3<2x-1<6

=>4<2x<7

=>2<x<7/2

mà x nguyên

nên x=3

a: Ta có: \(x^2-4-\left(x+2\right)^2\)

\(=x^2-4-x^2-4x-4\)

=-4x-8

b: Ta có: \(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)

\(=x^2-4-x^2+2x+3\)

=2x-1

c: ta có: \(\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)\)

\(=\left(x-2\right)\left(x+2-x-5\right)\)

\(=-3x+6\)

d: Ta có: \(\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)

\(=\left(6x+1-6x+1\right)^2\)

=4

e: ta có: \(7a\left(3a-5\right)+\left(2a-3\right)\left(4a+1\right)-\left(6a-2\right)^2\)

\(=21a^2-35a+8a^2+2a-12a-3-\left(36a^2-24a+4\right)\)

\(=29a^2-45a-3-36a^2+24a-4\)

\(=-7a^2-21a-7\)

g: ta có: \(\left(5y-3\right)\left(5y+3\right)-\left(5y-4\right)^2\)

\(=25y^2-9-25y^2+40y-16\)

=40y-25

h: Ta có: \(\left(3x+1\right)^3-\left(1-2x\right)^3\)

\(=27x^3+27x^2+9x+1-1+6x-12x^2+8x^3\)

\(=35x^3+15x^2+15x\)

i: Ta có: \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)

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

\(=16x^2\)

Bài 1:

a, 3\(\dfrac{2}{5}\) - \(\dfrac{1}{2}\) 

= \(\dfrac{17}{5}\) - \(\dfrac{1}{2}\) 

= \(\dfrac{34}{10}\) - \(\dfrac{5}{10}\)

= \(\dfrac{29}{10}\)

b, \(\dfrac{4}{5}\) + \(\dfrac{1}{5}\) x \(\dfrac{3}{4}\)

= \(\dfrac{4\times4}{5\times4}\) + \(\dfrac{1\times3}{5\times4}\)

= \(\dfrac{16}{20}\) + \(\dfrac{3}{20}\)

= \(\dfrac{19}{20}\)

c, 4\(\dfrac{4}{9}\) : 2\(\dfrac{2}{3}\) + 3\(\dfrac{1}{6}\)

= \(\dfrac{40}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{19}{6}\)

= \(\dfrac{5}{3}\) + \(\dfrac{19}{6}\)

= \(\dfrac{10}{6}\) + \(\dfrac{19}{6}\)

= \(\dfrac{29}{6}\)

Bài 2:

   3\(\dfrac{2}{5}\) + 2\(\dfrac{1}{5}\) 

= \(\dfrac{17}{5}\) + \(\dfrac{11}{5}\)

= \(\dfrac{28}{5}\)

b, 7\(\dfrac{1}{6}\) : 5\(\dfrac{2}{3}\)

= \(\dfrac{43}{6}\) : \(\dfrac{17}{3}\)

= \(\dfrac{43}{34}\)