Hỏi đáp Tìm x

ta có 

\(x^2-12x+36+5x\left(x-6\right)=0\Leftrightarrow\left(x-6\right)^2+5x\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(6x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=6\\x=1\end{cases}}\)

Answer:

\(x^2-12x+36+5x\left(x-6\right)=0\)

\(\Rightarrow x^2-12x+36+5x^2-30x=0\)

\(\Rightarrow6x^2-42x+36=0\)

\(\Rightarrow6x^2-36x-6x+36=0\)

\(\Rightarrow6x\left(x-6\right)-6\left(x-6\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x-6=0\\6x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=6\\x=1\end{cases}}}\)

x-(-12)+(-8)+(-11)

X x 19 = 4066 x 2

x - 4536 = 2378

x= 2378 + 4536

x = 6914

ko bt ko bt laf ko bt OK chx

5 × y = 4075

      y  = 4075 : 5

      y = 815

68921-a=345:5

68921-a=69

a=68921-69

a=68852 

       HT

Help gấp!!!!!!! 

Tìm a nhé giúp với

x=400

x=200

ở trên nhầm bn nhé

thảo nào thấy sai bấm 3 lần

11

kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

(7/3+7/2).x+25/6=22/7

x=-43/245

nha bạn chúc bạn học tốt nha 

\(\left(\frac{7}{3}+\frac{7}{2}\right).x+\frac{25}{6}=\frac{22}{7}\)

\(\frac{35}{6}.x+\frac{25}{6}=\frac{22}{7}\)

\(\frac{35}{6}.x=\frac{22}{7}-\frac{25}{6}\)

\(\frac{35}{6}.x=-\frac{43}{42}\)

\(x=-\frac{43}{42}:\frac{35}{6}\)

\(x=-\frac{43}{245}\)

\(\left(\frac{7}{3}+\frac{7}{2}\right)x+\frac{25}{6}=\frac{22}{7}\)

\(\Leftrightarrow\frac{35x}{6}+\frac{25}{6}=\frac{22}{7}\)

\(\Leftrightarrow\frac{35x+25}{6}=\frac{22}{7}\)

\(\Leftrightarrow35x+25=\frac{22.6}{7}\)

\(\Leftrightarrow35x+25=\frac{132}{7}\)

\(\Leftrightarrow x=\frac{\frac{132}{7}-25}{35}\)

\(\Leftrightarrow x=-\frac{43}{245}\)

Vậy \(x=-\frac{43}{245}\)

a, \(\left(4x-1\right)^2-\left(4x-5\right)\left(4x+5\right)=3\)

\(\Leftrightarrow16x^2-8x+1-16x^2+25=3\Leftrightarrow-8x+23=0\Leftrightarrow x=\frac{23}{8}\)

b, \(\left(x+2\right)^3-\left(x^3+8\right)=0\Leftrightarrow\left(x+2\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=0\)

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

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

\(\Leftrightarrow6x\left(x+2\right)=0\Leftrightarrow x=-2;x=0\)

a/ (4x-1)²-(4x-5)(4x+5)=3

     x = 23/8

b/(x+2)³-(x³+8)=0

   x = -2 

  x = 0

Số các số hạng là: ( X-1 ) :1+1= X ( số hạng)TA có:Tổng vế trái: ( X+1). X:2=> ( X+1) .X:2= 78=> ( X+1 ) . X= 78.2=>( X+1 ) . X=156Có 156= 2^2.3.13 =>156=( 2^2.3).13=12.13Vậy x=12 Đáp án đấy nha bn. Hok tốt!!!!

7 x X + 7 + 15= 4 x X+ 25                                                                                                                                                                                    7 x X + 22      =   4 x X +25

7 x X - 4 x X   = 25 - 22

X x (7-4)         =   3                                                                                      

X x 3               = 3 

X                     =3: 3 

X                     =   1

7 x (\(x\)+ 1) +15 = 4 x \(x\)+25 nha :))))

x = \(\frac{32}{7}\): 21

x = \(\frac{32}{147}\)

\(x=\frac{32}{7}:21\)

\(x=\frac{32}{147}\)

5 hoặc -5

x + 1/3 = 2/9 + 1/2

x + 1/3 = 13/18

         x = 13/18 - 1/3 = 7/18

x  + \(\frac{1}{3}\)=\(\frac{2}{9}\)+ \(\frac{1}{2}\)

\(\Rightarrow\)x  + \(\frac{1}{3}\)=\(\frac{13}{18}\)

\(\Rightarrow\)x = \(\frac{13}{18}\)- \(\frac{1}{3}\)

\(\Rightarrow\)x = \(\frac{7}{18}\)

Vậy x = \(\frac{7}{18}\)

a) Ta có: \(x^2-2x+1=6y^2-2x+2\)\(\Rightarrow x^2-1=6y^2\Rightarrow6y^2=\left(x-1\right)\left(x+1\right)⋮2\)( vì \(6y^2⋮2\))

Mặt khác: \(x-1+x+1=2x⋮2\Rightarrow\left(x-1\right);\left(x+1\right)\)cùng chẵn hoặc cùng lẻ.

=> (x+1),(x-1) là 2 số chẵn liên tiếp

=> (x+1)(x-1)\(⋮\)8 => \(6y^2⋮8\Rightarrow3y^2⋮4\Rightarrow y^2⋮4\Rightarrow y⋮2\)

=> y=2 

Thay vào =>x=5