Bài 1:

a; (\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\)) x \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)

     \(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) : \(\dfrac{3}{4}\)

      \(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) x \(\dfrac{4}{3}\)

    \(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) =  \(\dfrac{1}{3}\)

      \(\dfrac{1}{4}x\) = \(\dfrac{1}{3}\) + \(\dfrac{1}{8}\)

       \(\dfrac{1}{4}\) \(x\)=  \(\dfrac{8}{24}\) + \(\dfrac{11}{24}\)

          \(\dfrac{1}{4}x=\dfrac{11}{24}\)

           \(x=\dfrac{11}{24}:\dfrac{1}{4}\)

           \(x=\dfrac{11}{24}\times4\)

           \(x=\dfrac{11}{6}\) 

b; \(\dfrac{12}{5}:x\) = \(\dfrac{14}{3}\) x \(\dfrac{4}{7}\)

     \(\dfrac{12}{5}\) : \(x\) = \(\dfrac{8}{3}\)

            \(x\) = \(\dfrac{12}{5}\) : \(\dfrac{8}{3}\)

            \(x\) = \(\dfrac{12}{5}\) x \(\dfrac{3}{8}\)

             \(x\) = \(\dfrac{9}{10}\)

c; \(x+\dfrac{2}{3}\) = 8:4 - 1

   \(x+\dfrac{2}{3}\) = 2 - 1

   \(x+\dfrac{2}{3}\) = 1

  \(x=1-\dfrac{2}{3}\)

  \(x\) = \(\dfrac{1}{3}\)

a; 

c; \(\dfrac{1}{10}\) + \(\dfrac{4}{20}\) + \(\dfrac{9}{30}\)+\(\dfrac{16}{40}+\dfrac{25}{50}+\dfrac{36}{60}+\dfrac{49}{70}+\dfrac{64}{80}+\dfrac{81}{90}\)

= \(\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}\)

= \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)

=  \(\dfrac{\left(1+9\right)+\left(2+8\right)+\left(3+7\right)+\left(4+6\right)+5}{10}\)

= \(\dfrac{10+10+10+10+5}{10}\)

= \(\dfrac{\left(10+10+10+10\right)+5}{10}\)

= \(\dfrac{10\times4+5}{10}\)

= \(\dfrac{45}{10}\)

= \(\dfrac{9}{2}\)

mình xin lỗi!!!a=x nha!!!

a)\(\left(5x+1\right)^2=\frac{36}{49}\\ \left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\\ \Rightarrow\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{matrix}\right.\)

vậy...

2.

a) \(\left(5x+1\right)^2=\frac{36}{49}\)

⇒ \(5x+1=\pm\frac{6}{7}\)

⇒ \(\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}5x=\frac{6}{7}-1=-\frac{1}{7}\\5x=\left(-\frac{6}{7}\right)-1=-\frac{13}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-\frac{1}{7}\right):5\\x=\left(-\frac{13}{7}\right):5\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{1}{35};-\frac{13}{35}\right\}.\)

Chúc bạn học tốt!

9: =>x-3=2

=>x=5

10: =>x+1/2=1/5 hoặc x+1/2=-1/5

=>x=-7/10 hoặc x=-3/10

12:

a: =>x^2=900

=>x=30 hoặc x=-30

b: =>x=1/18*27=3/2

7: =>|x-0,4|=1,1

=>x-0,4=1,1 hoặc x-0,4=-1,1

=>x=1,5 hoặc x=-0,7

5.

P = ( x - 1 )( x + 2 )( x + 3 )( x + 6 ) < sửa rồi nhé :v >

= [ ( x - 1 )( x + 6 ) ][ ( x + 2 )( x + 3 ) ]

= ( x² + 5x - 6 )( x² + 5x + 6 ) (1)

Đặt t = x² + 5x 

(1) = ( t - 6 )( t + 6 )

     = t² - 36 ≥ -36 ∀ t

Dấu "=" xảy ra khi t = 0

=> x² + 5x = 0

=> x( x + 5 ) = 0

=> x = 0 hoặc x = -5

=> MinP = -36 <=> x = 0 hoặc x = -5

6.

a) ( x² + x )² + 4( x² + x ) = 12

Đặt t = x² + x

pt <=> t² + 4t = 12

     <=> t² + 4t - 12 = 0

     <=> t² - 2t + 6t - 12 = 0

     <=> t( t - 2 ) + 6( t - 2 ) = 0

     <=> ( t - 2 )( t + 6 ) = 0

     <=> ( x² + x - 2 )( x² + x + 6 ) = 0

     <=> x² + x - 2 = 0 hoặc x² + x + 6 = 0

+) x² + x - 2 = 0

=> x² - x + 2x - 2 = 0

=> x( x - 1 ) + 2( x - 1 ) = 0

=> ( x - 1 )( x + 2 ) = 0

=> x = 1 hoặc x = -2

+) x² + x + 6 = ( x² + x + 1/4 ) + 23/4 = ( x + 1/2 )² + 23/4 ≥ 23/4 > 0 ∀ x

=> x ∈ { -2 ; 1 }

b) x² - 12x + 36 = 81

<=> ( x - 6 )² = ( ±9 )²

<=> x - 6 = 9 hoặc x - 6 = -9

<=> x = 15 hoặc x = -3

1: \(\left(x^2+1\right)^2-4x\left(1-x^2\right)\)

\(=x^4+2x^2+1-4x+4x^3\)

\(=x^4+4x^3+2x^2-4x+1\)

\(=\left(x^2+2x-1\right)^2\)

2: \(\left(x^2-8\right)^2+36\)

\(=x^4-16x^2+64+36\)

\(=x^4-16x^2+100\)

\(=x^4+20x^2+100-36x^2\)

\(=\left(x^2+10\right)^2-\left(6x\right)^2\)

\(=\left(x^2+6x+10\right)\left(x^2-6x+10\right)\)

3: \(x^4+4=x^4+4x^2+4-4x^2\)

\(=\left(x^2+2\right)^2-4x^2\)

\(=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)

4: \(x^4+64=x^4+16x^2+64-16x^2\)

\(=\left(x^2+8\right)^2-16x^2\)

\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)

5: \(64x^4+1=64x^4+16x^2+1-16x^2\)

\(=\left(8x^2+1\right)^2-16x^2\)

\(=\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)\)

1).( 27,56 x 35 ) + ( 27,56 x 67 ) - ( 27,56 x 2)

= (964 + 1846,52) - 55,12

=2810,52 - 55,12

= 2755,4

2).( 4x 35 ) x ( 25 x 5 ) x 2

= ( 140 x 125 ) x2

= 17500 x 2

=35000

4). 3/10

5). 1188

6).  61/6

1)2756   2)35000   3)0   4)3/10   5)10/1/6

1,2755,4

2,35000

3,0

4,\(\frac{3}{10}\)

5,1188

6,\(\frac{61}{6}\)

â) \(\sqrt{x+9}=7\\ \Rightarrow x+9=49\\ \Rightarrow x=40\)

b) \(\sqrt{x-4}=4-x\\ \Rightarrow x-4=16-8x+x^2\\ \Rightarrow x^2-9x+20=0\\ \Rightarrow\left(x-4\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

c) \(\sqrt{x^2-12x+36}=81\\ \Rightarrow x-6=81\\ \Rightarrow x=87\)

a: Ta có: \(\sqrt{x+9}=7\)

\(\Leftrightarrow x+9=49\)

hay x=40

b: Ta có: \(\sqrt{x-4}=4-x\)

\(\Leftrightarrow\left(x-4\right)^2-\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=5\left(loại\right)\end{matrix}\right.\)

c: Ta có: \(\sqrt{x^2-12x+36}=81\)

\(\Leftrightarrow\left|x-6\right|=81\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=81\\x-6=-81\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=87\\x=-75\end{matrix}\right.\)