a) x - 12 - 2x - 32 = -11

-x - 44 = -11

-x  = 33

x = -33 

a/ (x-12)-(2x+31)=-6-5

    x-12-2x-31=-11

    -x=(-11)+43

-x=32

=> x=-32

2: x=3

3: x=18

1) \(x=51\)

2) \(x=3\)

3) \(x=18\)

4) \(x=2\)

5) \(x=7\)

a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)

\(\Rightarrow3x-6x^2+6x+14=29\)

\(\Rightarrow-6x^2+9x-15=0\)

\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)

\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)

Vậy \(S=\varnothing\)

a. \(2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

*) \(-7+2x=-37-\left(-26\right)\)

⇔ \(-7+2x=-11\)

⇔ \(2x=\left(-11\right)-\left(-7\right)\)

⇔ \(2x=-4\)

⇔ \(x=\left(-4\right):2\)

⇔ \(x=-2\)

\(\text{(3x +9).(11-x ) =0}\)

⇔\(\left[{}\begin{matrix}3x+9=0\\11=x=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}3x=-9\\x=11\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=-3\\x=11\end{matrix}\right.\)

Vậy x= -3 và x= 11

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

3x^2-x = 0

x(3x-1) = 0

TH1: x = 0

TH2: 3x-1 = 0

x= 1/3

b) Ta có: 2x(x – 5) – x(3 + 2x) = 26

      ⇔ 2$x^2$ – 10x – 3x – 2$x^2$ =26

      ⇔ - 13x = 26

      ⇔ x = - 2

\(a,3x-31=-40\Rightarrow3x=-9\Rightarrow x=-3\)

\(b,-3x+37=\left(-4\right)^2\Rightarrow-3x=-21\Rightarrow x=7\)

\(c,\left|2x+7\right|=5\)

\(\Rightarrow\left\{{}\begin{matrix}2x+7=5\Rightarrow x=-1\\2x+7=-5\Rightarrow x=-6\end{matrix}\right.\)

\(d,-x+21=15+2x\Rightarrow3x=6\Rightarrow x=2\)

a) Ta có: 3x-31=-40

\(\Leftrightarrow3x=-9\)

hay x=-3

Vậy: x=-3

b) Ta có: \(-3x+37=\left(-4\right)^2\)

\(\Leftrightarrow-3x+37=16\)

\(\Leftrightarrow-3x=16-37=-21\)

hay x=7

Vậy: x=7

Bài 2: 

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

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

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