ko ghi đề

\(=25,97+\left(6,54+103,46\right)\)

\(=25,97+110\)

\(=135,97\)

a/135,97

b/1500

c/10

4.x=16

-37 + 14 +26 + 37 

=-37 + 37 + 14 + 36 

=0 + 50

=50

  -24 + 6 + 10 + 24 

=(-24 + 24) +(6+10)

= 0+16

= 16

a, 7\(x\) - \(x\) = 5²¹ : 5¹⁹ + 3.2².7

     6\(x\)    = 5³ + 3.4.7

    6\(x\)    = 125 + 12.7

    6\(x\)  = 125 + 84

    6\(x\) = 209

     \(x\)  = 209 : 6

    \(x\) = \(\dfrac{209}{6}\)

b; 11\(x\) - 7\(x\) + 3⁴ : 3³ = 5⁴ + 2\(x\)

    4\(x\) + 3 = 625 + 2\(x\)

   4\(x\) - 2\(x\) = 625 - 3

   2\(x\)        = 622

     \(x\)        = 622 : 2

    \(x\)        = 311

c; 75 - 5.(\(x-3\))³ = 700

          5.(\(x\) - 3)³ = 700 - 75

         5.(\(x\) - 3)³ = - 625

           (\(x\) - 30)³ = - 625 : 5

           (\(x\) - 30)³ = - 125

           (\(x-3\))³ =  (-5)³

           \(x\) - 3 = - 5 

           \(x\)         = - 5 + 3

            \(x\)       = -2

d, 3.(2\(x\) - 1)² = 75

       (2\(x\) - 1)² = 75 : 3

       (2\(x\) - 1)² = 25

       \(\left[{}\begin{matrix}2x-1=-5\\2x-1=5\end{matrix}\right.\)

       \(\left[{}\begin{matrix}2x=-5+1\\2x=5+1\end{matrix}\right.\)

       \(\left[{}\begin{matrix}2x=-4\\2x=6\end{matrix}\right.\)

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

Bài 1 :

a, 25 + (-42) = - ( 42 - 25 ) = - 17

b, (-17) + 5 + 8 + 17 + (-3 )

= [ (-17) + 17 ] + [ ( 5 + 8 ) + (-3) ]

= 0 + 10

= 10

c, 25. 2² - ( 15-18 ) + ( 12 - 19 + 10 )

= 25 . 4 - (-3) + 3

= 100 - [ (-3) + 3 ]

= 100 - 0

= 100

d, 120 - 35 + 29 - 242

= 85 + 29 - 242

= 114 - 242

= -128

Bài 1 : Tính :

\(a,\text{ }-45+15-17\)

\(=-30-17\)

\(=-47\)

\(b,\text{ }-15\cdot4\text{ : }\left(-12\right)\)

\(=-60\cdot\frac{1}{-12}=5\)

\(c,\text{ }-13\cdot\left(-5\right)+\left(-16\right)\cdot5-\left(-38\right)\text{ : }2-\left(-5^2\right)\)

\(=13\cdot5-16\cdot5+19-25\)

\(=5\left(13-16\right)-6\)

\(=5\cdot\left(-3\right)-6\)

\(=-15-6\)

\(=-21\)

\(d,\text{ }-2020\cdot79+2020\cdot\left(-21\right)\)

\(=2020\left(-79-21\right)\)

\(=2020\cdot\left(-100\right)\)

\(=-202000\)

Bài 2 :                                               Bài giải

\(a,\text{ }x+15=7\)

\(x=7-15\)

\(x=-8 \)

\(b,\text{ }17-2x=23\)

\(2x=17-23\)

\(2x=-6\)

\(x=-6\text{ : }2\)

\(x=-3\)

\(c,\text{ }\left(2x+17\right)^2=169\)

\(\left(2x+17\right)^2=\left(\pm13\right)^2\)

\(2x+17=\pm13\)

\(\Rightarrow\orbr{\begin{cases}2x+17=-13\\2x+17=13\end{cases}}\Rightarrow\orbr{\begin{cases}2x=-30\\2x=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=-15\\x=-2\end{cases}}\)

\(\Rightarrow\text{ }x\in\left\{-15\text{ ; }-2\right\}\)

- Bài 1 và bài 2 a,b dễ bạn tự làm nhé :>

Bài 2: c. ( 2x + 17)² = 169 Ta có: 169 = 13² = (-13)²

TH1: ( 2x + 17 )² = 13² => 2x + 17 = 13 => 2x = 13 - 17 = -4 => x = (-4) : 2 = -2

TH2: ( 2x + 17 )² = (-13)² => 2x + 17 = -13 => 2x = -13 - 17 = -30 => x = (-30) : 2 = -15

Bài 3: a. S = 2 - 5 + 8 - 11+ 14 - 17+......+ 2016 - 2019

=> S = (-3) + (-3) + (-3) + ... + (-3)

- Sau đó bạn nhân -3 với số cặp là ra nhé :)

b. x - 4 là bội của x + 7 => \(\frac{x-4}{x+7}\inℤ\Rightarrow\frac{x+7-11}{x+7}\inℤ\)

\(\Rightarrow\frac{x+7}{x+7}-\frac{11}{x+7}\inℤ\Rightarrow1-\frac{11}{x+7}\inℤ\)

Vì \(1\inℤ\Rightarrow\frac{11}{x+7}\in Z\Rightarrow x+7\inƯ\left(11\right)=\left\{\pm1,\pm11\right\}\)

Ta có bảng sau: 

Ju

Bài 1 câu a:

1/-8 + -5/8

= -1/8 - 5/8

= -6/8

= -3/4

Câu b:

-1/21 + -1/28

= -4/84 - 3/84

= - 7/854

Câu c:

-5/9 + 8/15 + -2/11 + 4/-9 + 7/15

= (8/15+ 7/15) - (5/9 + 4/9) - 2/11

= 1 - 1 - 2/11

= 0 - 2/11

= - 2/11

Câu d:

-25/-75 + 17/34 + 121/- 132

= 1/3 + 1/2 - 11/12

= 4/12 + 6/12 - 11/12

= 10/12 - 11/12

= - 1/12

Bài 2a:

51/23 = 5151/2323 = 515151/232323

\(\frac{5151}{2323}\) = \(\frac{5151:101}{2323:101}\) = \(\frac{51}{23}\) (1)

\(\frac{515151}{232323}\) = \(\frac{515151:10101}{232323:10101}\) = \(\frac{51}{23}\) (2)

Từ (1) và (2) ta có:

\(\frac{51}{23}=\frac{5151}{2323}=\frac{515151}{232323}\) (đpcm)

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