a, \(\Leftrightarrow2x^2=72\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow x=\pm6\)

Vậy ...

\(b,\Leftrightarrow\dfrac{3}{5}x-0,75=2\dfrac{4}{5}.\dfrac{3}{7}=\dfrac{6}{5}\)

\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{6}{5}+0,75=\dfrac{39}{20}\)

\(\Leftrightarrow x=\dfrac{39}{20}:\dfrac{3}{5}=\dfrac{13}{4}\)

Vậy ...

\(c,\Leftrightarrow2x=1\dfrac{5}{6}.\dfrac{6}{11}-\dfrac{3}{10}=\dfrac{7}{10}\)

\(\Leftrightarrow x=\dfrac{7}{10}:2=\dfrac{7}{20}\)

Vậy ...

\(d,\Leftrightarrow\dfrac{1}{x-7\dfrac{1}{3}}=1.5:2\dfrac{1}{4}=\dfrac{2}{3}\)

\(\Leftrightarrow x-7\dfrac{1}{3}=\dfrac{3}{2}\)

\(\Leftrightarrow x=\dfrac{3}{2}+7\dfrac{1}{3}=\dfrac{53}{6}\)

Vậy ...

a) 2x² - 72 = 0

\(\Rightarrow\) 2x² = 72

\(\Rightarrow\) x² = 36 = 6² = (- 6)²

\(\Rightarrow\) x = 6 hoặc x = - 6

Vậy x = 6 hoặc x = - 6

b) (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(2\dfrac{4}{5}\)

\(\Rightarrow\)  (\(\dfrac{3}{5}\)x - 0,75) : \(\dfrac{3}{7}\) = \(\dfrac{14}{5}\)

\(\Rightarrow\)  \(\dfrac{3}{5}\)x - \(\dfrac{3}{4}\) = \(\dfrac{6}{5}\)

\(\Rightarrow\) \(\dfrac{3}{5}\)x = \(\dfrac{39}{20}\)

\(\Rightarrow\) x = \(\dfrac{13}{4}\)

Vậy x = \(\dfrac{13}{4}\)

1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)

\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)

\(\left|4-2x\right|=\dfrac{1}{3}.3\)

\(\left|4-2x\right|=1\)

=>\(4-2x=\pm1\)

+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)

\(2x=4-1\) \(2x=4-\left(-1\right)\)

\(2x=3\) \(2x=4+1\)

\(x=3:2\) \(2x=5\)

\(x=1,5\) \(x=5:2\)

Vậy x=1,5 \(x=2,5\)

Vậy x=2,5

2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)

\(9:\left|x+\left(-1\right)\right|=-3\)

\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)

\(\left|x+\left(-1\right)\right|=-3\)

=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )

Vậy x = \(\varnothing\)

a: \(4x^3+12=120\)

=>\(4x^3=108\)

=>\(x^3=27=3^3\)

=>x=3

b: \(\left(x-4\right)^2=64\)

=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)

c: (x+1)^3-2=5^2

=>\(\left(x+1\right)^3=25+2=27\)

=>x+1=3

=>x=2

d: 136-(x+5)^2=100

=>(x+5)^2=36

=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)

e: \(4^x=16\)

=>\(4^x=4^2\)

=>x=2

f: \(7^x\cdot3-147=0\)

=>\(3\cdot7^x=147\)

=>\(7^x=49\)

=>x=2

g: \(2^{x+3}-15=17\)

=>\(2^{x+3}=32\)

=>x+3=5

=>x=2

h: \(5^{2x-4}\cdot4=10^2\)

=>\(5^{2x-4}=\dfrac{100}{4}=25\)

=>2x-4=2

=>2x=6

=>x=3

i: (32-4x)(7-x)=0

=>(4x-32)(x-7)=0

=>4(x-8)*(x-7)=0

=>(x-8)(x-7)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

k: (8-x)(10-2x)=0

=>(x-8)(x-5)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)

m: \(3^x+3^{x+1}=108\)

=>\(3^x+3^x\cdot3=108\)

=>\(4\cdot3^x=108\)

=>\(3^x=27\)

=>x=3

n: \(5^{x+2}+5^{x+1}=750\)

=>\(5^x\cdot25+5^x\cdot5=750\)

=>\(5^x\cdot30=750\)

=>\(5^x=25\)

=>x=2

Lời giải:

1. $(x+2)-2=0$

$x+2=2$

$x=0$

2.

$(x+3)+1=7$

$x+3=7-1=6$

$x=6-3=3$

3.

$(3x-4)+4=12$

$3x-4+4=12$

$3x=12$

$x=12:3=4$

4.

$(5x+4)-1=13$

$5x+4=13+1=14$

$5x=14-4=10$

$x=10:5=2$

5.

$(4x-8)-3=5$

$4x-8=5+3=8$

$4x=8+8=16$

$x=16:4=4$

6.

$3+(x-5)=7$

$x-5=7-3=4$

$x=4+5=9$

7.

$8-(2x-4)=2$

$2x-4=8-2=6$

$2x=6+4=10$

$x=10:2=5$

8.

$7+(5x+2)=14$

$5x+2=14-7=7$

$5x=7-2=5$

$x=5:5=1$

9.

$5-(3x-11)=1$

$3x-11=5-1=4$

$3x=11+4=15$

$x=15:3=5$

10.

$16-(8x+2)=6$

$8x+2=16-6=10$

$8x=10-2=8$

$x=8:8=1$

a. 5 - 3(x + 4) = -1

⇔ 5 - 3x - 12 = -1

⇔ 3x = -1 - 5 + 12

⇔ 3x = 6

⇔ x = 2

\(d,2x^2-3=5\)

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=\pm2\)

\(e,x\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)

MẤY DÒNG NÀO BẠN THẤY KO CẦN THIẾT THÌ LƯỢC BỎ NHA!!!

a) \(2\left(x-5\right)-3\left(x+6\right)=4\left(x-7\right)\)

   \(2x-10-3x-18=4x-28\)

   \(2x-3x-4x-10-18=-28\)

   \(-5x-28=-28\)

   \(-5x=-28+28=0\)

    \(x=\frac{0}{-5}=0\)

b) \(3\left(x-1\right)-2\left(x+5\right)=2\left(x-3\right)\)

  \(3x-3-2x-10=2x-6\)

   \(3x-2x-2x-3-10=-6\)

   \(-x-13=-6\)

   \(-x=-6+13=7\)

    \(x=-7\)

c) ​​\(5\left(1-x\right)-6\left(1+x\right)=7\left(3-x\right)\)

    ​​\(5-5x-6-6x=21-7x\)

    \(-5x-6x+7x+5-6=21\)

    \(-4x-1=21\)

    \(-4x=22\)

     \(x=\frac{22}{-4}=\frac{-11}{2}\)

d) \(2x+5-3\left(3x+7\right)=6\left(1-x\right)+8\)

    \(2x+5-9x-21=6-6x+8\)

    \(2x-9x+6x+5-21=6+8\)

    ​\(-x-16=14\)

    \(-x=14+16=30\)

    \(x=-30\)

e) \(x-2+3\left(x-4\right)=5\left(x-6\right)+7\)

   \(x-2+3x-12=5x-30+7\)

   \(x+3x-5x-2-12=-30+7\)

  \(-x-14=-23\)

  \(-x=-23+14=-9\)

​  \(x=9\)

f) \(x+2+3\left(1-x\right)-5\left(2-x\right)=6\left(1-x\right)+\left(3-x\right)\)

  \(x+2+3-3x-10+5x=6-6x+3-x\)

  \(x-3x+5x+6x+x+2+3-10=6+3\)

  \(10x-7=9\)

  \(10x=9+7=16\)

  \(x=\frac{16}{10}=\frac{8}{5}\)

 720 : ( x . 2 + x . 3 ) = 3.2
720 : ( x . 2 + x.3 ) = 6
( x .2 + x.3 )           = 720 : 6 
x.2+x.3 = 120
x . ( 2 + 3 ) = 120
x . 5 = 120
     x     = 120 : 5 
    x      = 24