Rin sẽ làm bài 2

a. 4 - 2x = -56

         2x = 4 + 56

         2x = 60

           x = 30

b. -9x + 42 = 66

    -9x         = 24

       x          = 24 : -9

       x          = \(\frac{-24}{9}\)

c. 5 x 1x - 111 = 125

          1x - 111 = 25

          1x          = 136

            x          = 136

d. (2x +1)² = 81

    (2x +1)² = 9²

=> 2x + 1 = 9

     2x       = 10

       x       = 5

Bài 1:

c.. (-159) x 26 - (-26) x 59

 =  159 x (-26) - (-26) x 59

 = (159 - 59) x -26

 = 100 x -26

 = -2600

Rin học cách đây mấy năm r nên ko nhớ câu a, b làm thế nào mong bạn thông cảm cho Rin

pls help me mk đang cần vội :(

Bài 1:

\(a,=6x^2+19x-7-6x^3-4x^2+7x=-6x^3+2x^2+26x-7\\ b,B=26\cdot\left(63^2+63\cdot37+37^2\right):26+63\cdot37\\ =63^2+63\cdot37+37^2+63\cdot37\\ =\left(63+37\right)^2=100^2=10000\)

Bài 2:

\(a,=x\left(y^2-25\right)=x\left(y-5\right)\left(y+5\right)\\ b,=\left(x-y\right)\left(x+2\right)\\ c,=\left(x-3\right)\left(x^2-4\right)=\left(x-2\right)\left(x-3\right)\left(x+2\right)\)

a: \(x^2-2x+3\)

\(=x^2-2x+1+2=\left(x-1\right)^2+2\ge2\forall x\)

=>\(A=\frac{37}{x^2-2x+3}\le\frac{37}{2}\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

b: \(x^2-5x+10\)

\(=x^2-5x+\frac{25}{4}+\frac{15}{4}\)

\(=\left(x-\frac52\right)^2+\frac{15}{4}\ge\frac{15}{4}\forall x\)

=>\(\frac{26}{x^2-5x+10}\le26:\frac{15}{4}=26\cdot\frac{4}{15}=\frac{104}{15}\forall x\)

=>\(B=-\frac{26}{x^2-5x+10}\ge-\frac{104}{15}\forall x\)

Dấu '=' xảy ra khi \(x-\frac52=0\)

=>\(x=\frac52\)

c: \(x^2-x+6\)

\(=x^2-x+\frac14+\frac{23}{4}\)

\(=\left(x-\frac12\right)^2+\frac{23}{4}\ge\frac{23}{4}\forall x\)

=>\(\frac{2023}{x^2-x+6}\le2023:\frac{23}{4}=2023\cdot\frac{4}{23}=\frac{8092}{23}\forall x\)

=>\(C=-\frac{2023}{x^2-x+6}\ge-\frac{8092}{23}\forall x\)

Dấu '=' xảy ra khi \(x-\frac12=0\)

=>\(x=\frac12\)

d: \(x^2+x+5\)

\(=x^2+x+\frac14+\frac{19}{4}\)

\(=\left(x+\frac12\right)^2+\frac{19}{4}\ge\frac{19}{4}\forall x\)

=>\(D=\frac{0.75}{x^2+x+5}\le\frac34:\frac{19}{4}=\frac{3}{19}\forall x\)

Dấu '=' xảy ra khi \(x+\frac12=0\)

=>\(x=-\frac12\)

e: \(2x^2-x+37=2\left(x^2-\frac12x+\frac{37}{2}\right)\)

\(=2\left(x^2-2\cdot x\cdot\frac14+\frac{1}{16}+\frac{295}{16}\right)=2\left(x-\frac14\right)^2+\frac{295}{8}\ge\frac{295}{8}\forall x\)

=>\(\frac{13}{2x^2-x+37}\le13:\frac{295}{8}=\frac{104}{295}\forall x\)

Dấu '=' xảy ra khi \(x-\frac14=0\)

=>\(x=\frac14\)

f: \(3x^2-x+19\)

\(=3\left(x^2-\frac13x+\frac{19}{3}\right)\)

\(=3\left(x^2-2\cdot x\cdot\frac16+\frac{1}{36}+\frac{227}{36}\right)=3\left(x-\frac16\right)^2+\frac{227}{12}\ge\frac{227}{12}\forall x\)

=>\(\frac{61}{3x^2-x+19}\le61:\frac{227}{12}=61\cdot\frac{12}{227}=\frac{732}{227}\forall x\)

=>\(-\frac{61}{3x^2-x+19}\ge-\frac{732}{227}\forall x\)

Dấu '=' xảy ra khi x-1/6=0

=>x=1/6

7+2x=-37-(-26)-x

7+2x=-11-x

7+2x+x=-11

7+3x=-11

3x=-11-7

3x=-18 

x=-18:3

x=-6

-7 - 2x = 962 

-2x = 969

x = -484,5

7 - 2x = 962 

-2x = 969

x = -484,5

Bài 2: 

c: \(=x^2\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

2x - 15 = -11 - ( -16 )

2x - 15 = -11 + 16 

2x - 15 = 5

2x         = 5 + 15 

2x         = 20

   x        = 20 : 1 

   x        = 10 

Vậy x = 10

7 - 2x = -37 - ( -26 )

7 - 2x = -37 + 26

7 - 2x = -11

      2x = 7 - ( - 11 )

      2x = 7+ 11

      2x = 18

        x = 18 : 2 

        x = 9

Vậy x = 9 

| x + 7 | = 9 

TH1 :

x + 7 = 9

=> x  = 9 - 7 

=> x  = 2    

TH2 :

x - 7 = -9

=> x  = -9 - 7

=> x  = - 9 + ( -7 )

=> x   = -16 

Vậy x = 2 và x = -16

Câu cuối bn tự làm nha! ^-^

2x-15=-11-(-16)

2x-15= -11+16

2x-15=5

2x=5+15

2x=20

x=20:2

x=10

-7-2x = 37-(-26)

Suy ra -7-2x=11

Suy ra -2x=11-7

Suy ra -2x=4

Suy ra 2x= -4

Suy ra x = -4 :2 =-2