3²⁰⁰ = 3^2.100= ( 3²)¹⁰⁰

2³⁰⁰ = 2^3.100 = (2³)¹⁰⁰  

Vì 3² > 2³ nên (3²)¹⁰⁰ > ( 2³)¹⁰⁰ hay 3²⁰⁰> 2³⁰⁰ 

1) \(3^{200}=\left(3^2\right)^{100}=9^{100}\)

\(2^{300}=\left(2^3\right)^{100}=8^{100}\)

Do 9^100 > 8^100 => 3^200 > 2^300

2) 4^x+3 - 3.4^x+1= 13.4¹¹

4^x+1.4² - 3.4^x+1= 13.4¹¹

4^x+1 ( 4² - 3) = 13.4¹¹

4^x+1 . 13 = 13. 4¹¹

4^x+1 = 4¹¹

=> x + 1 = 11

=> x= 10

Ta có 51/2.52/2...100/2

       = 1.2.3....100/1.2...50.2.2...2 (nhân cả tử và mẫu với 1.2.3...50)

      = 1.2.3...100/(1.2)(2.2)(3.2)...(50.2)

      = 1.2.3...100/2.4.6...100

      = 1.3.5...99 => đpcm nhớ giữ lời hứa đấy

a)x-7  = 0 

x=0+7=7

b, ( x - 3 ) . ( x^2 + 3 ) = 0

-> x -3=0  hoặc x^2+3 =0 

+ Nếu x -3 =0 

-> x=3 

+ Nếu x^2+3 =0 

 -> x^2 =-3 ( loại) 

Vậy x=3 

Bài2

6x + 3 chia hết cho x 

 Ta có x chia hết cho x

-> 6x chia hết cho x 

Mà 6x+3 chia hết cho x 

-> (6x+3)-6x chia hết cho x 

-> 3 chia hết cho x

......

Bạn tự làm 

Câu b tương tự

1. 

x - 7 = 0 => x = 7

( x - 3 ) ( x² + 3 ) = 0

=> \(\orbr{\begin{cases}x-3=0\\x^2+3=0\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x^2=-3\end{cases}}\)

Bình phương một số \(\ge\)0 => x² \(\ne\)-3

=> x = 3

2. a) 6x + 3 chia hết cho x

=> 3 chia hết cho x

=> x thuộc Ư(3) = { -3 ; -1 ; 1 ; 3 }

b) 4x + 4 chia hết cho 2x - 1

=> 2(2x - 1) + 6 chia hết cho 2x - 1

=> 4x - 2 + 6 chia hết cho 2x - 1

=> 6 chia hết cho 2x - 1

=> 2x - 1 thuộc Ư(6) = { -6 ; -3 ; -2 ; -1 ; 1 ; 2 ; 3 ; 6 }

Vì x thuộc Z => x thuộc { -1 ; 0 ; 1 ; 2 }

Bài 1:

a, \(\frac{1}{-16}-\frac{3}{45}=\frac{-1}{16}-\frac{1}{15}\)

\(=\frac{-15}{240}-\frac{16}{240}\)

\(=\frac{-31}{240}\)

b, \(=\frac{-10}{12}-\frac{-12}{12}\)

\(=\frac{2}{12}=\frac{1}{6}\)

c, \(=\frac{-30}{6}-\frac{1}{6}\)

\(=\frac{-31}{6}\)

Bài 2:

a, \(x=-\frac{1}{2}-\frac{3}{4}\)

\(x=-\frac{1}{4}\)

b,   \(\frac{1}{2}+x=-\frac{11}{2}\)

\(x=-\frac{11}{2}-\frac{1}{2}\)

\(x=-6\)

Bạn nhớ k đúng và chọn câu trả lời này nhé!!!! Mình giải đúng và chính xác hết ^_^

a)x-5=7

=> x=9

b)2.x+6=24

=>2x=18

=>x=9

Tìm x :

a) \(x-5=49:7\)

\(\Leftrightarrow x-5=7\)

\(\Leftrightarrow x=7+5\)

\(\Leftrightarrow x=12\)

Vậy : \(x=12\)

b) \(2x+6=24\)

\(\Leftrightarrow2x=24-6=18\)

\(\Leftrightarrow x=18:2\)

\(\Leftrightarrow x=9\)

Vậy : \(x=9\)

c) \(\frac{1}{3}:x+\frac{1}{2}=5\)

\(\Leftrightarrow\frac{1}{3}:x=5-\frac{1}{2}=\frac{9}{5}\)

\(\Leftrightarrow x=\frac{1}{3}:\frac{9}{5}\)

\(\Leftrightarrow x=\frac{5}{27}\)

Vậy : \(x=\frac{5}{27}\)

d) \(\frac{1}{6}.x-\frac{1}{3}=2\)

\(\Leftrightarrow\frac{1}{6}.x=2-\frac{1}{3}=\frac{5}{3}\)

\(\Leftrightarrow x=\frac{5}{3}:\frac{1}{6}\)

\(\Leftrightarrow x=10\)

Vậy : \(x=10\)

e) \(\frac{x}{27}=\frac{3}{x}\)

\(\Leftrightarrow x.x=27.3\)

\(\Leftrightarrow x^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=9\end{matrix}\right.\) mà \(x\in N\)

\(\Rightarrow x=9\)

Vậy : \(x=9\)

g) \(1200:24-\left(17-x\right)=36\)

\(\Leftrightarrow50-17+x=36\)

\(\Leftrightarrow33+x=36\)

\(\Leftrightarrow x=36-33\)

\(\Leftrightarrow x=3\)

Vậy : \(x=3\)

h) \(674-\left(12+x\right)=427\)

\(\Leftrightarrow12+x=674-427=247\)

\(\Leftrightarrow x=247-12\)

\(\Leftrightarrow x=230\)

Vậy : \(x=230\)

k) \(36.\left(x-9\right)=900\)

\(\Leftrightarrow x-9=900:36\)

\(\Leftrightarrow x-9=25\)

\(\Leftrightarrow x=25+9\)

\(\Leftrightarrow x=34\)

m) \(1,2:x+3,8:x=2,5\)

\(\Leftrightarrow\left(1,2-3,8\right):x=2,5\)

\(\Leftrightarrow-2,6:x=2,5\)

\(\Leftrightarrow x=\frac{-2,6}{2,5}=-\frac{26}{25}\)

Vậy : \(x=-\frac{26}{25}\)

n) \(\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right).x=\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right).x=\frac{1}{3}\)

\(\Leftrightarrow\left[\frac{1}{2}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\right].x=\frac{1}{3}\)

\(\Leftrightarrow\left[\frac{1}{2}.\left(1-\frac{1}{10}\right)\right].x=\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{2}.\frac{9}{10}.x=\frac{1}{3}\)

\(\Leftrightarrow\frac{9}{20}.x=\frac{1}{3}\)

\(\Leftrightarrow x=\frac{1}{3}:\frac{9}{20}\)

\(\Leftrightarrow x=\frac{20}{27}\)

Vậy : \(x=\frac{20}{27}\)

1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)

  \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)

 \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)

 \(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)

\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)

\(x=\frac{45}{4}+\frac{9}{4}\)

\(x=\frac{27}{2}\)

Bước cưối 58/21 minh man viết nhầm nên sai 

a,3x-10=2x+13

\(\Rightarrow\)3x-2x=10+13

\(\Rightarrow\)x=23

b,x+12=-5-x

\(\Rightarrow\)x+x=-12-5

\(\Rightarrow\)2x=-17

\(\Rightarrow\)x=-8,5

c,x+5=10-x

\(\Rightarrow\)x+x=-5+10

\(\Rightarrow\)2x=5

\(\Rightarrow\)x=2,5

e,12-x=x+1

\(\Rightarrow\)-x-x=-12+1

\(\Rightarrow\)-2x=-11

\(\Rightarrow\)x=5,5

f,14+4x=3x+20

\(\Rightarrow\)4x-3x=-14+20

\(\Rightarrow\)x=6

g,2.(x-1)+3(x-2)=x-4

\(\Rightarrow\)2x-2+3x-6=x-4

\(\Rightarrow\)2x-2+3x-6-x+4=0

\(\Rightarrow\)4x-4=0

\(\Rightarrow\)4x=4

\(\Rightarrow\)x=1

h,3(4-x)-2(x-1)=x+20

\(\Rightarrow\)12-3x-2x+2-x-20=0

\(\Rightarrow\)-6x-6=0

\(\Rightarrow\)-6x=6

\(\Rightarrow\)x=-1

i,4(2x+7)-3(3x-2)=24

\(\Rightarrow\)8x+28-9x+6=24

\(\Rightarrow\)-x+34=24

\(\Rightarrow\)-x=-10

\(\Rightarrow\)x=10

k,3(x-2)+2x=10

\(\Rightarrow\)3x-6+2x=10

\(\Rightarrow\)5x-6=10

\(\Rightarrow\)5x=16

\(\Rightarrow\)x=3,2

Phần d, tớ không biết làm!!!!

Nhưng bn cũng đã giúp tớ rùi . Thank kiu bạn

\(\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)\left(1-\frac{3}{7}\right)...\left(1-\frac{10}{7}\right)\)

\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right)...\left(1-\frac{10}{7}\right)\)

\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-1\right)...\left(1-\frac{10}{7}\right)\)

\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...0...\left(1-\frac{10}{7}\right)\)

= 0

Tìm \(x\) câu a:

\(\frac13.x\) + \(\frac25.\left(x+1\right)\) = 0

\(\frac{5}{15}x\) + \(\frac{6}{15}x\) + \(\frac25\) = 0

\(\frac{11}{15}x\) = - \(\frac25\)

\(x=-\frac25:\frac{11}{15}\)

\(x\) = - \(\frac25\times\frac{15}{11}\)

\(x\) = - \(\frac{6}{11}\)

Vậy \(x=-\frac{6}{11}\)

Tìm \(x\) câu b:

\(x\) x 25% = 0,5

\(x\times0,25\) = 0,5

\(x=0,5:0,25\)

\(x=2\)

Vậy \(x=2\)