\(P=\left(x+5\right)\left(ax^2+bx+25\right)\) (Sửa =25 thành +25)

\(Q=x^3+125=x^3+5^3=\left(x+5\right)\left(x^2-5x+25\right)\) (Sửa =25 thành +25)

Để \(P=Q\Rightarrow\left(ax^2+bx+25\right)=\left(x^2-5x+25\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=-5\end{matrix}\right.\)

\(3A=3^2+3^3+...+3^{101}\)

=>\(A=\dfrac{3^{101}-3}{2}\)

`a)25/(x+1)-1 1/6=-1/3-0,5`

`=>25/(x+1)=-1/3-1/2+1+1/6`

`=>25/(x+1)=1/3`

`=>75=x+1`

`=>x=74`

Vậy `x=74`

`b)(2x+25 3/5)^2-9/25=0`

`=>(2x+128/5)=9/25`

`**2x+128/5=3/5`

`=>2x=-125/5=-25`

`=>x=-25/2`

`**2x+128/5=-3/5`

`=>2x=-131/5`

`=>x=-131/10`

Giải:

a) \(\dfrac{25}{x+1}-1\dfrac{1}{6}=\dfrac{-1}{3}-0,5\) 

              \(\dfrac{25}{x+1}=\dfrac{-5}{6}+\dfrac{7}{6}\) 

              \(\dfrac{25}{x+1}=\dfrac{1}{3}\) 

\(\Rightarrow1.\left(x+1\right)=25.3\)  

\(\Rightarrow x+1=75\) 

\(\Rightarrow x=75-1\) 

\(\Rightarrow x=74\) 

b) \(\left(2x+25\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\) 

              \(\left(2x+\dfrac{128}{5}\right)^2=0+\dfrac{9}{25}\) 

             \(\left(2x+\dfrac{128}{5}\right)^2=\dfrac{9}{25}\) 

\(\Rightarrow\left[{}\begin{matrix}\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\\\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{128}{5}=\dfrac{3}{5}\\2x+\dfrac{128}{5}=\dfrac{-3}{5}\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-25}{2}\\x=\dfrac{-131}{10}\end{matrix}\right.\) 

Chúc bạn học tốt!

a) \(2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)

\(x+\frac{2}{5}=\frac{-7}{20}\)

\(x=\frac{-13}{20}\)

Vậy \(x=\frac{-13}{20}\)

b)\(x-1\frac{1}{8}-\frac{2}{3}x-\frac{5}{6}x=75\%\)

  \(\left(x-\frac{2}{3}x-\frac{5}{6}x\right)-\frac{9}{8}=\frac{3}{4}\)

\(\frac{-1}{2}x-\frac{9}{8}=\frac{3}{4}\)

\(\frac{-1}{2}x=\frac{15}{8}\)

\(x=\frac{-15}{4}\)

Vậy \(x=\frac{-15}{4}\)

\(a,2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)=\frac{11}{20}-\frac{5}{4}\)

\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)

\(x+\frac{2}{5}=\frac{-7}{10}:2\)

\(x+\frac{2}{5}=\frac{-7}{20}\)

\(x=\frac{-7}{20}-\frac{2}{5}\)

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

a) x - 1/2 = 3/5

x = 3/5 + 1/2

x = 11/10

b) x - 1/2 = -2/3

x = -2/3 + 1/2

x = -1/6

c) 2/5 - x = 0,25

x = 2/5 - 0,25

x = 2/5 - 1/4

x = 3/20

Mình làm câu khó thôi nhé.

2x chia hết cho 3

=>(2x+x-x) chia hết cho 3

=>(3x-x) chia hết cho 3

3x chia hết cho 3=>x chia hết cho 3

=>x thuộc B(3)={0;3;6;...}

Vậy x thuộc {0;3;6;...ư}

Bổ sung: vì 0<x<40=>x thuộc {3;6;9;...:39}

a) Các ước của -12: 1;-1;2;-2;3;-3;4;-4;6;-6;12;-12

Vậy x\(\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

b) (-300):20+5.(3x-1)=25

    (-15) + 5.(3x-1)     = 25

               5.(3x-1)      = 25-(-15)

               5.(3x-1)      = 40

                   3x-1        = 40:5                   

                   3x-1        = 8

                   3x           = 8+1

                   3x           = 9

                     x           = 9:3

                     x           = 3

Vậy x=3

c) 2x \(⋮\) 3

=>(2x+x-x) \(⋮\) 3

=>(3x-x) \(⋮\) 3

3x \(⋮\) 3=>x \(⋮\) 3

=>x \(\in\) B(3)={0;3;6;...}

Vậy x \(\in\) {0;3;6;...}

TICK VÀ LIKE NHA. CẢM ƠN

3/25 x ( 15/7 - 2/7 ) + 3/7 x 1/25 

= 3/25 x 13/7 + 3/7 x 1/25 

= (3 x 13/7 + 3/7 ) x 1/25 

= 42/7 x 1/25 

= 6 x 1/25 

= 6/25

\(\dfrac{3}{25}\times\dfrac{15}{7}+\dfrac{3}{7}\times\dfrac{1}{25}-\dfrac{2}{7}\times\dfrac{3}{25}\)

\(=\dfrac{3}{25}\times\left(\dfrac{15}{7}-\dfrac{2}{7}\right)+\dfrac{3}{7}\times\dfrac{1}{25}\)

\(=\dfrac{3}{25}\times\dfrac{13}{7}+\dfrac{3}{7}\times\dfrac{1}{25}\)

\(=\dfrac{3\times13}{25\times7}+\dfrac{3\times1}{7\times25}\)

\(=\dfrac{39}{175}+\dfrac{3}{175}\)

\(=\dfrac{39+3}{175}\)

\(=\dfrac{42}{175}\)

\(=\dfrac{6}{25}\)

A = \(\dfrac{3}{25}\) \(\times\) \(\dfrac{15}{7}\) + \(\dfrac{3}{25}\) \(\times\) \(\dfrac{1}{7}\) - \(\dfrac{2}{7}\times\) \(\dfrac{3}{25}\)

A = \(\dfrac{3}{25}\) \(\times\)( \(\dfrac{15}{7}\) + \(\dfrac{1}{7}-\dfrac{2}{7}\))

A = \(\dfrac{3}{25}\) \(\times\) \(\dfrac{14}{7}\)

A = \(\dfrac{6}{25}\)