a,sửa đề : đk x khác -2;  2 

 \(x^2+x-2+5x-10=12+x^2-4\)

\(\Leftrightarrow6x-20=0\Leftrightarrow x=\dfrac{10}{3}\left(tm\right)\)

b, \(3x-12+5+5x=105\Leftrightarrow8x=112\Leftrightarrow x=14\)

c, \(3x^2+14x-49=-\left(x^2+2x-15\right)\)

\(\Leftrightarrow4x^2+16x-34=0\Leftrightarrow x=\dfrac{-4\pm5\sqrt{2}}{2}\)

a. ko hỉu đề lắm :v

b.\(\dfrac{x-4}{5}+\dfrac{1+x}{3}=7\)

\(\Leftrightarrow\dfrac{3\left(x-4\right)+5\left(1+x\right)}{15}=\dfrac{105}{15}\)

\(\Leftrightarrow3\left(x-4\right)+5\left(1+x\right)=105\)

\(\Leftrightarrow3x-12+5+5x-105=0\)

\(\Leftrightarrow8x-112=0\)

\(\Leftrightarrow8x=112\)

\(\Leftrightarrow x=14\)

c.\(\left(3x-7\right)\left(x+7\right)=\left(5+x\right)\left(3-x\right)\)

\(\Leftrightarrow3x^2+21x-7x-49=15-5x+3x-x^2\)

\(\Leftrightarrow4x^2+16x-64=0\)

Nghiệm xấu lắm bạn

1. a

\(\dfrac{8}{5}-\dfrac{5}{6}\cdot\dfrac{3}{4}\)

\(=\dfrac{8}{5}-\dfrac{5\cdot3}{3\cdot2\cdot4}\)

\(=\dfrac{8}{5}-\dfrac{5}{8}=\dfrac{39}{40}\)

1.b

\(=\dfrac{7}{8}+\dfrac{5}{6}\cdot\dfrac{3}{2}\)

\(=\dfrac{7}{8}+\dfrac{5\cdot3}{3\cdot2\cdot2}\)

\(=\dfrac{7}{8}+\dfrac{5}{4}=\dfrac{17}{8}\)

2.a

\(\dfrac{4}{5}+x=\dfrac{11}{10}\)

\(x=\dfrac{11}{10}-\dfrac{4}{5}=\dfrac{3}{10}\)

2.b

\(x-\dfrac{3}{4}=\dfrac{5}{7}\)

\(x=\dfrac{5}{7}+\dfrac{3}{4}=\dfrac{41}{28}\)

a) y + 3/7 x 21/9 = 5

      y + 3/7 x 7/3 = 5

      y + 1             = 5

            y             = 4

b) y  x 3/5 - 1/2 = 4/5

    y x 3/5          = 4/5 + 1/2

    y x 3/5          = 13/10

    y                   = 13/6

c) 9/8 - y + 2/5 = 1/4

            y - 2/5  = 9 /8 - 1/4

            y - 2/5  = 7/8

                   y   = 51/40 

d) y - 6/5 : 2/3 = 5/6

    y - 9/5         = 5/6

          y           = 79/30

e) y : 7/8 + 3/4 = 7/4

    y : 7/8          = 1

          y            = 7/8

\(y+\frac{3}{7}\times\frac{21}{9}=5\)

\(y+\frac{3}{7}\times\frac{7}{3}=5\)

\(y+1=5\)

\(\Rightarrow y=4\)

Mình ko ghi lại đề , bạn ghi ra xong rồi suy ra như mình nha .

1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)

\(=>A=-12x+16\)

2) \(=>B=8x^3+27-8x^3+2=29\)

3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)

4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)

5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)

\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)

\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)

6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)

k cho mik nha , 

a, Xét : x-4 = 0 => x= 4

            2x+1 = 0 => x= \(\frac{1}{2}\)

            x+3 = 0 => x = -3

            x + 9 = 0 => x = -9

Khi đó ta có bảng xét dấu : 

=> có 5 trường hợp:

TH1 : \(x\le-9\)

TH2 : \(-9\le x< -3\)

TH3 : \(-3\le x< \frac{1}{2}\)

TH4 : \(\frac{1}{2}\le x< 4\)

Do đó :

TH1 : \(x\le-9\)

Ta có :  /x-4/ = -(x-4) = 4 - x

            /2x+1/ = -(2x+1) = -2x -1

           /x+3/   = -(x + 3 ) = -x - 3

          /x-9/ = -(x-9) = -x + 9                  Thay vào đề bài ta có:

                                               3.(4-x) + 2x-1 +5(-x - 3) -x-9 = 5

                                    => 12 - 3x + 2x - 1 + -5x - 15 - x - 9 = 5

                                    =>(12 - 1 - 15 -9 ) +(-3x +2x -5x -x) = 5

                                   => -13 - 7x                                        = 5

                                             7x                                =     -13 - 5

                                                 7x =      -18

                                              x = \(\frac{-18}{7}\)( Ko TM)

Tương tự với 4 trường hợp còn lại.

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-13}{12}\)

Em nên gõ công thức trực quan để đề bài rõ ràng nhé