a) \(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)^2=\dfrac{4}{25}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{22}{15}\\x=-\dfrac{2}{15}\end{matrix}\right.\)

b) \(\Rightarrow\left(1-\dfrac{1}{4}x\right)^2=\dfrac{121}{49}\)

\(\Rightarrow\left[{}\begin{matrix}1-\dfrac{1}{4}x=\dfrac{11}{7}\\1-\dfrac{1}{4}x=-\dfrac{11}{7}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{16}{7}\\x=\dfrac{72}{7}\end{matrix}\right.\)

\(\dfrac{x+1}{124}+1+\dfrac{x+2}{123}+1=\dfrac{x+3}{122}+1+\dfrac{x+4}{121}+1\)

\(\Leftrightarrow\dfrac{x+125}{124}+\dfrac{x+125}{123}=\dfrac{x+125}{122}+\dfrac{x+125}{121}\)

\(\Leftrightarrow\left(x+125\right)\left(\dfrac{1}{124}+\dfrac{1}{123}-\dfrac{1}{122}-\dfrac{1}{121}\ne0\right)=0\Leftrightarrow x=-125\)

<=>\(\dfrac{x+1}{124}+\dfrac{x+2}{123}-\dfrac{x+3}{122}-\dfrac{x+4}{121}=0\)

<=>\(\left(\dfrac{x+1}{124}+1\right)+\left(\dfrac{x+2}{123}+1\right)-\left(\dfrac{x+3}{122}+1\right)-\left(\dfrac{x+4}{121}+1\right)=0\)

<=>\(\dfrac{x+125}{124}+\dfrac{x+125}{123}-\dfrac{x+125}{122}-\dfrac{x+125}{121}=0\)

<=>\(\left(x+125\right)\left(\dfrac{1}{124}+\dfrac{1}{123}-\dfrac{1}{122}-\dfrac{1}{121}\right)=0\)

<=>x+125=0

<=>x=-125

1. So sánh

a) \(25^{50}\) và \(2^{300}\)

\(25^{50}=25^{1.50}=\left(25^1\right)^{50}=25^{50}\)

\(2^{300}=2^{6.50}=\left(2^6\right)^{50}=64^{50}\)

Vì \(25< 64\) nên \(25^{50}< 64^{50}\)

Vậy \(25^{50}< 2^{300}\)

b) \(625^{15}\) và \(12^{45}\)

\(625^{15}=625^{1.15}=\left(625^1\right)^{15}=625^{15}\)

\(12^{45}=12^{3.15}=\left(12^3\right)^{15}=1728^{15}\)

Vì \(625< 1728\) nên \(625^{15}< 1728^{15}\)

Vậy \(625^{15}< 12^{45}\)

1.So sánh

a)\(25^{50}\) và \(2^{300}\)

Ta có : \(2^{300}=\left(2^6\right)^{50}=64^{50}\)

Vì \(25^{50}< 64^{50}\) nên \(25^{50}< 2^{300}\)

b)\(625^{15}\) và \(12^{45}\)

Ta có : \(12^{45}=\left(12^3\right)^{15}=1728^{15}\)

Vì \(625^{15}< 1728^{15}\) nên \(625^{15}< 12^{45}\)

\(B=\left(11^2-1^2\right)×...×\left(11^2-11^2\right)×...×\left(11^2-29^2\right)\)

\(B=0\)

a) 4.(3.X-4)-2=30

   4.(3.X-4)    =30+2

  4.(3.X-4)     =32

      3.X-4      =32:4

     3.X-4       =8

     3.X          =8+4

     3.X          =12

       X           =12:3

        X           =4

vậy X  =4

bạn ơi hình như đề b sai đó 

bạn xem lại nha

chúc bạn học tốt nhé

KHÔNG SAI ĐÂU DẠNG BÀI NÓ THẾ MƠI SANG NAY MINH MOI THI MA

cau a . 4.( 3.x-4)-2=30

           4.(3.x-4)    =30+2

           4.(3.x-4)    =32

              (3.x-4)    =32:4

              (3.x-4)    =8

               3.x        =8+4

               3.x        =12

                  x        =12:3

                  x        =4

suy ra x = 4 

còn đề b mình ko biết làm nhưng mình nghĩ đề b bị sai , bạn xem lại đi nhé chúc bạn thi tốt và nhận nhiều lời mời kết bạn nhé hi hi

a,

\(\left(x+\frac{1}{3}\right)^2=\frac{1}{4}\)

⇒ \(\left[{}\begin{matrix}x+\frac{1}{3}=\frac{1}{2}\\x+\frac{1}{3}=-\frac{1}{2}\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=\frac{3}{6}-\frac{2}{6}=\frac{1}{6}\\x=-\frac{3}{6}-\frac{2}{6}=-\frac{5}{6}\end{matrix}\right.\)

Vậy.....

b, \(\left(2x+3\right)^2=\frac{9}{121}\)

⇒ \(\left[{}\begin{matrix}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}2x=\frac{3}{11}-\frac{33}{11}=-\frac{30}{11}\\2x=-\frac{3}{11}-\frac{33}{11}=-\frac{36}{11}\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=-\frac{15}{11}\\x=-\frac{12}{11}\end{matrix}\right.\)

c, \(\left(3x-1\right)^3=-\frac{8}{27}\)

⇒ \(3x-1=-\frac{2}{3}\)

⇒ \(3x=-\frac{2}{3}+1=\frac{1}{3}\)

⇒ \(x=\frac{1}{9}\)

d, \(4^x+4^{x+2}=112\)

⇒ \(4^x.\left(1+16\right)=112\)

=> \(4^x.17=112\)

Chỗ này kì nè :33 bạn xem lại đềcoi

giúp mình với các bạn ơi

a,(x + \(\frac{1}{3})^2\) = \(\frac{1}{4}\)

\(\Rightarrow(x+\frac{1}{3})^2=(\frac{1}{2})^2\)

nên \(x+\frac{1}{3}=\frac{1}{2}\) hoặc \(x+\frac{1}{3}=-\frac{1}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{3}=\frac{1}{2}\\x+\frac{1}{3}=\frac{-1}{2}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{6}\\x=\frac{-5}{6}\end{matrix}\right.\)

Vậy x\(\in\left\{\frac{1}{6},\frac{-5}{6}\right\}\)

`a, (4x+5) : 3 -121:11=14`

`=> (4x+5) : 3 - 11=14`

`=> (4x+5) : 3 =14+11`

`=> (4x+5) : 3 = 25`

`=> 4x+5=25xx3`

`=>4x+5= 75`

`=> 4x=75-5`

`=>4x=70`

`=>x= 70/4`

`=>x= 35/2`

`b, 2+4+6+...+x=2450`

Số lượng của dãy là :

`(x-2)/2 + 1= (x-2)/2 +2/2= x/2`

Tổng số lượng là :

\(\dfrac{\left(x+2\right)\cdot\dfrac{x}{2}}{2}=\dfrac{\dfrac{x^2}{2}+\dfrac{2x}{2}}{2}=\dfrac{x\left(x+2\right)}{4}\)

\(\dfrac{x\left(x+2\right)}{4}=2450\)

\(\Rightarrow x\left(x+2\right)=2450\cdot4\\ \Rightarrow x\left(x+2\right)=9800\)

`=>x=98`

`c,` `1` nhân `32` sao?

`d, 2x+3x=1505`

`=> (2+3)x=1505`

`=> 5x=1505`

`=> x= 1505:5`

`=>x= 301`

\(\left(1\dfrac{3}{4}-\dfrac{4}{6}\right):\left(1\dfrac{1}{5}+2\dfrac{2}{5}+\dfrac{1}{5}\right)< x< 1\dfrac{1}{5}.1\dfrac{1}{4}+3\dfrac{2}{11}:2\dfrac{3}{121}\)

\(a,\left(2x+5\right):3-121:11=4\)

\(\left(2x+5\right):3-11=4\)

\(\left(2x+5\right):3=4+11\)

\(\left(2x+5\right):3=15\)

\(2x+5=15.3\)

\(2x+5=45\)

\(2x=45-5\)

\(2x=40\)

\(x=40:2\)

\(x=20\)

Vậy \(x=20\)

\(b,\left(4x+1\right)^3=125\)

\(\left(4x+1\right)^3=5^3\)

\(\Rightarrow4x+1=5\)

\(4x=5-1\)

\(4x=4\)

\(x=4:4\)

\(x=1\)

Vậy \(x=1\)