a,123-5(x+4)=28

(=)5(x+4)=85

(=) x+4=17

(=) x=13

b,10+2x=2.\(\left(3^2-1\right)\)

(=) 10+2x=16

(=) 2x = 6 

(=)x=3

A, 123-5.(x+4)=38

            5.(x+4)=123-38

           5.(x+4) =85

               (x+4)=85:5

               (x+4)=17

                x      =17-4

                x      =13

B, 10+2x=2(\(3^2-1\))

     10+2x=2.8

     10+2x=16

           2x=16-10

           2x=6

             x=6:2

            x  =3

a, 123 - 5(x + 4) = 38

=>5(x + 4) = 85

=> x + 4 = 17

=> x = 13

vậy_

10 + 2x = 2(3² - 1)

=> 10 + 2x = 2.8

=> 10 + 2x = 16

=> 2x = 6

=> x = 3

vậy_

a: \(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

=>2x+16=6

=>2x=-10

hay x=-5

b: \(\Leftrightarrow3\left(x-1-4x^2+4x\right)+4\left(3x^2+9x-2x-6\right)=-27\)

\(\Leftrightarrow3\left(-4x^2+5x-1\right)+4\left(3x^2+7x-6\right)=-27\)

\(\Leftrightarrow-12x^2+15x-3+12x^2+28x-24=-27\)

=>43x=0

hay x=0

c: \(\Leftrightarrow5\left(2y^2+4y+3y+6\right)-2\left(5y^2-5y-4y+4\right)=75\)

\(\Leftrightarrow10y^2+35y+30-10y^2+18y-8=75\)

=>53y=53

hay y=1

d: \(\Leftrightarrow6x^2+27x+4x+18-\left(6x^2+x+12x+2\right)=x+1-x+6=7\)

\(\Leftrightarrow6x^2+31x+18-6x^2-13x-2=7\)

=>18x+16=7

=>18x=-9

hay x=-1/2

a) |x| + |-5| = |-37|

<=> |x| + 5 = 37

<=> |x| = 37 - 5 = 32

=> x \(\in\) {32 ; -32}

b)|-6| . |x| = |54|

<=> 6 . |x| = 54

|x| = 54 : 6 = 9

=> x \(\in\){9;-9}

c) |x| > 21

Có |x| \(\ge\) 0 > 21

=> |x| \(\in\) { 22 ; 23 ; 24 ; 25 ; ....}

=> x \(\in\) { 22; -22 ; 23; -23; 24; -24; 25; -25; ....}

a, x= -9/2

b, x= 2

kb nha

a) \(\left|x+5\right|-x+23=27\)

\(\Leftrightarrow\left|x+5\right|-x+23-27=0\)

\(\Leftrightarrow\left|x+5\right|-x+\left(23-27\right)=0\)

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

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

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

\(\Leftrightarrow\left|x+5\right|=x+4\)

Xét trường hợp 1:

\(x+5=x+4\)

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

\(\Leftrightarrow x+5-x-4=0\)

\(\Leftrightarrow1=0\left(loai\right)\)

Xét trường hợp 2:

\(x+5=-\left(x+4\right)\)

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

\(\Leftrightarrow x+5+x+4=0\)

\(\Leftrightarrow2x+9=0\)

\(\Leftrightarrow2x=-9\)

\(\Leftrightarrow x=\dfrac{-9}{2}\left(thoa\right)\)

Vậy \(x=\dfrac{-9}{2}\)

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Rightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

\(\Rightarrow x+10=0\) vì \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

\(\Rightarrow x=-10\)

=>\(\frac{x+1}{9}\)+1+\(\frac{x+2}{8}\)+1=\(\frac{x+3}{7}\)+1+\(\frac{x+4}{6}\)+1

=>\(\frac{x+10}{9}\)+ \(\frac{x+10}{8}\)- \(\frac{x+10}{7}\)- \(\frac{x+10}{6}\)= 0

=>(x+10)x(\(\frac{1}{9}\)+ \(\frac{1}{8}\)+ \(\frac{1}{7}\)+\(\frac{1}{6}\))=0

=>x+10=0

=>x= -10

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Vì \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)nên x +10  = 0

x = -10.