Bài 6: Phép trừ và phép chia. Luyện tập 1. Luyện tập 2

Bài 1:

a, (-313) + 176 - 87 + (-76)

= - (313 + 87) + (176 - 76)

= - 400 + 100  = -300

b, 17. (-56) + 17. (-30) + 17. (-4)

= -17. (56+30+4)

= -17. 90 = -1530

c, (-2) + (-5) + (-8) +...+ (-182)

= (-182 -2) : 2 x 61

= -5612

d, 1 - 3 + 5 - 7 + 9 - 11 + ... + 97 - 99

= (1-3) + (5-7) + (9-11) +...+ (97 - 99)

= -2 - 2 - 2 +... - 2

= -2 x 25 = -50

Bài 2:

\(a,3\left(3x-1\right)+2\left(1-5x\right)-4\left(2-3x\right)\\ =9x-3+2-10x-8+12x\\ =\left(9x-10x+12x\right)+\left(2-3-8\right)\\ =11x-9\\ ---\\ b,-3\left(x-1\right)-4\left(4-2x\right)-2\left(x+5\right)\\ =-3x+3-16+8x-2x-10\\ =\left(-3x+8x-2x\right)+\left(3-16-10\right)\\ =3x-23\)

\(\left(57-725\right)-\left(605-53\right)\)

\(=57-725-605+53\)

\(=\left(57+53\right)-\left(725+605\right)\)

\(=110-1330=-1220\)

=(57+53)−(725+605)

\(71\cdot64+32\cdot\left(-7\right)-13\cdot32\)

\(=32\cdot2\cdot71+32\cdot\left(-7\right)+32\cdot\left(-13\right)\)

\(=32\left(142-7-13\right)\)

\(=32\cdot122=3904\)

A = 4 + 4² + 4³ + ... + 4⁹⁰

4A = 4² + 4³ + 4⁴ + ... + 4⁹¹

3A = 4A - A

= (4² + 4³ + 4⁴ + ... + 4⁹¹) - (4 + 4² + 4³ + ... + 4⁹⁰)

= 4⁹¹ - 4

A = (4⁹¹ - 4)/3

a: \(4x^3+12=120\)

=>\(4x^3=108\)

=>\(x^3=27=3^3\)

=>x=3

b: \(\left(x-4\right)^2=64\)

=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)

c: (x+1)^3-2=5^2

=>\(\left(x+1\right)^3=25+2=27\)

=>x+1=3

=>x=2

d: 136-(x+5)^2=100

=>(x+5)^2=36

=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)

e: \(4^x=16\)

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

=>x=2

f: \(7^x\cdot3-147=0\)

=>\(3\cdot7^x=147\)

=>\(7^x=49\)

=>x=2

g: \(2^{x+3}-15=17\)

=>\(2^{x+3}=32\)

=>x+3=5

=>x=2

h: \(5^{2x-4}\cdot4=10^2\)

=>\(5^{2x-4}=\dfrac{100}{4}=25\)

=>2x-4=2

=>2x=6

=>x=3

i: (32-4x)(7-x)=0

=>(4x-32)(x-7)=0

=>4(x-8)*(x-7)=0

=>(x-8)(x-7)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

k: (8-x)(10-2x)=0

=>(x-8)(x-5)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)

m: \(3^x+3^{x+1}=108\)

=>\(3^x+3^x\cdot3=108\)

=>\(4\cdot3^x=108\)

=>\(3^x=27\)

=>x=3

n: \(5^{x+2}+5^{x+1}=750\)

=>\(5^x\cdot25+5^x\cdot5=750\)

=>\(5^x\cdot30=750\)

=>\(5^x=25\)

=>x=2

`(x + 1)^3 - 1 = 7`

`(x + 1)^3 = 7 + 1`

`(x + 1)^3 = 8`

`(x + 1)^3 = 2^3`

`=> x + 1 = 2`

`x = 2 - 1`

`x = 1`

`5^6 : 5^3 +3^2 : 3^3 - 3^4 . 3`

`= 5^(6-3)+  (3^2)/(3^3) - 3^(4+1)`

`= 5^3 + 9/27 - 3^5`

`= 125 + 1/3 -243`

`=-353/3`

\(13\cdot15-196:14-16:8-2022^0\)

\(=13\cdot15-14^2:14-2+1\)

\(=195-14+1\)

\(=181+1\)

\(=182\)

\(13.15-196:14+16:8-2022^0\)

\(=195-14+2-1\)

\(=181+2-1\)

\(=183-1=182\)

13.15 -196:14+16 :8-2022⁰.

=13.15-196:14+16:8-1.

=195-14+3-1.

=181+3-1.

=184-1.

=183.

\(B4\\ \overline{aaa}=100a+10a+a=111a=3a.37⋮37\)

Vậy \(\overline{aaa}\) luôn là bội của 37 

Bài 3:

\(Ư\left(24\right)=\left\{1;2;3;4;6;8;12;24\right\}\)

Vậy hai số cần tìm là 24 và 12