a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)

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

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

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

a) 150- x = -9

x = 150 -(-9)

x=150+9

x=159

b)4 (x-3)=48

x-3=48÷4

x-3=12

x=12+3

x=15

Bài 7:

a)(x-15)-75=0

⇔x-15=75

⇔x=90

b)575-(6x+70)=445

⇔6x+70=130

⇔6x=60

⇔x=10

c)x-105:21=15

⇔x-5=15

⇔x=20

d)(x-105):21=15

⇔x-105=315

⇔x=420

x=15 và x=-9 bạn nhé

\(a,\Leftrightarrow3\left(x+3\right)=0\Leftrightarrow x=-3\\ b,\Leftrightarrow\left(x^2-2\right)\left(6x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=2\\6x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\\x=-\dfrac{1}{6}\end{matrix}\right.\\ c,\Leftrightarrow\left(x-2013\right)\left(4x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2013\\x=\dfrac{1}{4}\end{matrix}\right.\\ d,\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

a) x+7=-12

        x=(-12)-7

        x=-19

b)x-15=-21

        x=(-21)+15

        x=-6

c)13-x=20

        x=13-20

        x=-7

d)17-(2+x)=3

      x=17-3

      x=14

      x=14-2

      x=12

a,x+7=-12

=>x= -12-7

=>x= -19

b,x-15= -21

=>x= -21+15

=>x= -6

c,13-x=20

=>x=13-20

=>x= -7

d, 17-(2+x)=3

=>2+x=17-3

=>2+x=14

=>x=14-2

=>x=12

a) \(x+7=-12\) 

    \(x=-12-7\) 

    \(x=-19\) 

b) \(x-15=-21\)  

    \(x=-21+15\) 

    \(x=-6\)  

c) \(13-x=20\) 

            \(x=13-20\)  

            \(x=-7\) 

d) \(17-\left(2+x\right)=3\)  

             \(2+x=17-3\)  

             \(2+x=14\)   

                   \(x=14-2\)  

                   \(x=12\)

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

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

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

a) Ta có: 12-5x=37

\(\Leftrightarrow5x=-25\)

hay x=-5

Vậy: x=-5

b) Ta có: 7-3|x-2|=-11

\(\Leftrightarrow3\left|x-2\right|=18\)

\(\Leftrightarrow\left|x-2\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{8;-4\right\}\)

c) Ta có: \(x+\dfrac{2}{8}=-\dfrac{15}{4}\)

\(\Leftrightarrow x=\dfrac{-15}{4}-\dfrac{2}{8}=\dfrac{-15}{4}-\dfrac{1}{4}\)

hay x=-4

Vậy: x=-4

a, \(\Leftrightarrow5x=12-37=-25\)

\(\Leftrightarrow x=-\dfrac{25}{5}=-5\)

Vậy ...

b, \(\Leftrightarrow3\left|x-2\right|=7+11=18\)

\(\Leftrightarrow\left|x-2\right|=\dfrac{18}{3}=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)

Vậy ...

c, \(\Leftrightarrow x=-\dfrac{15}{4}-\dfrac{2}{8}=-4\)

Vậy ..

a)-5; b)8 hoặc -4;c)-4 nhé. Bn tham khảo nhé

\(a.b=15\) ⇒ \(a=\dfrac{15}{b}\)

Thay vào \(a+b=18\)

⇒ \(\dfrac{15}{b}+b=18\)

⇒ \(\dfrac{b^2+15}{b}=18\)

⇒ \(b^2-2.b.9+18=3\)

⇒ \(\left(b-9\right)^2=3\)

Còn lại tự lm

Bài 1:

a)-54

b)-8

Bài 2:

a)(x-14):5=4¹⁵:4¹³

⇔(x-14):5=4²

⇔(x-14):5=16

⇔x-14=80

⇔x=94

b)7x-15x=15-175

⇔-8x=-160

⇔x=20