\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)

\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)

\(\Leftrightarrow-3x=-\frac{13}{4}\)

\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)

\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)

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

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

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

\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)

\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)

\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)

\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)

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

d,e,f Tương tự

Bài 1 :

a) 7^2x-1 = 343

=> 7^2x-1 = 7³

=> 2x - 1 = 3 => 2x = 4 => x = 2

b) (7x - 11)³ = 2⁵.3² + 200

=> (7x - 11)³ = 32.9 + 200

=> (7x - 11)³ = 488

xem kĩ lại đề này :vvv

c) 174 - (2x - 1)² = 5³

=> (2x - 1)² = 174 - 5³

=> (2x - 1)² = 174 - 125 = 49

=> (2x - 1)² = (\(\pm\)7)²

=> \(\orbr{\begin{cases}2x-1=7\\2x-1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)

Mà x \(\in\)N nên x = 4( thỏa mãn điều kiện)

Bài 2 :

a) x⁵ = 32 => x⁵ = 2⁵ => x = 2

b) (x + 2)³ = 27

=> (x + 2)³ = 3³

=> x + 2 = 3 => x = 3 - 2 = 1

c) (x - 1)⁴ = 16

=> (x - 1)⁴ = 2⁴

=> x - 1 = 2 => x = 3 ( vì đề bài cho x thuộc N nên thỏa mãn)

d) (x - 1)⁸ = (x - 1)⁶

=> (x - 1)⁸ - (x - 1)⁶ = 0

=> (x - 1)⁶ [(x - 1)² - 1] = 0

=> \(\orbr{\begin{cases}\left(x-1\right)^6=0\\\left(x-1\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=\left(\pm1\right)^2\end{cases}}\)

+) x - 1 = 1 => x = 2 ( tm)

+) x - 1 = -1 => x = 0 ( tm)

Vậy x = 1,x = 2,x = 0

a. 37-7.(x+1)=40-8.3

37 - 7(x+1) = 16 

7(x+1) = 21

x+1 = 3

x=2

b. (x+5).3=11=10+(3+4).4

(sao có 2 dấu bằng, dấu nào là dấu cộng do viết nhầm ???)

c. 25-5.3+4.10=100-5.x

50 = 100 - 5x 

5x = 50

x=10

d. 36+2.(x-7)=12+8.(3+5)-36

36 + 2(x-7) = 40

2(x-7) = 4

x-7 =2

x=9

e. 3.(x+7)-9=11.5-5-8

3(x+7) - 9 = 42 

3(x+7) = 51

x+7 = 17 

x=10

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

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

\(\Leftrightarrow\frac{1}{2}:x=\frac{3}{15}-\frac{5}{15}\)

\(\Leftrightarrow\frac{1}{2}:x=\frac{-2}{15}\)

\(\Leftrightarrow x=\frac{1}{2}:\frac{-15}{2}=\frac{-15}{4}\)

\(b,\frac{1}{3}x+\frac{2}{5}\left[x+1\right]=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)

\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\Leftrightarrow x=\frac{-2}{5}:\frac{11}{15}=\frac{-2}{5}\cdot\frac{15}{11}=\frac{-2}{1}\cdot\frac{3}{11}=\frac{-6}{11}\)

\(c,\frac{11}{12}-\left[\frac{2}{5}+x\right]=\frac{2}{3}\)

\(\Leftrightarrow\frac{11}{12}-\frac{2}{3}=\frac{2}{5}+x\)

\(\Leftrightarrow\frac{11}{12}-\frac{8}{12}=\frac{2}{5}+x\)

\(\Leftrightarrow\frac{1}{4}=\frac{2}{5}+x\)

\(\Leftrightarrow\frac{1}{4}-\frac{2}{5}=x\Leftrightarrow x=-\frac{3}{20}\)

a: \(A=\dfrac{5}{4}\cdot\dfrac{11}{3}\cdot\dfrac{-1}{11}=\dfrac{-5}{12}=\dfrac{-25}{60}=\dfrac{-50}{120}\)

b: \(B=\dfrac{3}{4}\cdot\dfrac{1}{12}\cdot\dfrac{2}{3}=\dfrac{1}{24}=\dfrac{5}{120}\)

c: \(C=\dfrac{5}{4}\cdot\dfrac{1}{15}\cdot\dfrac{2}{5}=\dfrac{2}{60}=\dfrac{1}{30}=\dfrac{4}{120}\)

\(D=-3\cdot\dfrac{-7}{12}\cdot\dfrac{1}{-7}=-\dfrac{1}{4}=\dfrac{-30}{120}\)

Vì -50<-30<4<5

nên A<D<B<C

a: \(A=\dfrac{5}{4}\cdot\dfrac{11}{3}\cdot\dfrac{-1}{11}=\dfrac{-5}{12}=\dfrac{-25}{60}=\dfrac{-50}{120}\)

b: \(B=\dfrac{3}{4}\cdot\dfrac{1}{12}\cdot\dfrac{2}{3}=\dfrac{1}{24}=\dfrac{5}{120}\)

c: \(C=\dfrac{5}{4}\cdot\dfrac{1}{15}\cdot\dfrac{2}{5}=\dfrac{2}{60}=\dfrac{1}{30}=\dfrac{4}{120}\)

\(D=-3\cdot\dfrac{-7}{12}\cdot\dfrac{1}{-7}=-\dfrac{1}{4}=\dfrac{-30}{120}\)

Vì -50<-30<4<5

nên A<D<B<C