a: \(\left(x+5\right)^2>=0\forall x\)

\(\left(2y-8\right)^2>=0\forall y\)

Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)

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

=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)

Tách ?

`a, 70 -5.(x-3) =45`

`=> 5.(x-3)= 70-45`

`=> 5.(x-3)=25`

`=>x-3=25:5`

`=>x-3=5`

`=>x= 5+3`

`=>x=8`

______

`b,10 + 2.x = 4^5:4^3`

`=> 10 + 2.x = 4^(5-3)`

`=> 10 + 2.x =4^2=16`

`=> 2.x=16-10`

`=>2.x=6`

`=>x=6:2`

`=>x=3`

_____

`c,60-3.x-2=51`

`=>  60-3.x= 51+2`

`=> 60-3.x=53`

`=>3.x=60-53`

`=> 3.x= 7`

`=>x= 7/3`

____

`d, 4.x-20=2^5:2^3`

`=> 4.x-20=2^(5-3)`

`=> 4.x-20=2^2`

`=> 4.x= 4+20`

`=>4.x=24`

`=>x=24:4`

`=>x=6`

____

`2^x . 4=16`

`=> 2^x=16:4`

`=>2^x= 4`

`=>2^x=2^2`

`=>x=2`

____

`f, 3^x . 3=243`

`=>3^x=243:3`

`=> 3^x=81`

`=> 3^x=3^3`

`=>x=3`

_____

`g, 64. 4^x =16^8`

`=> 4^3 . 4^x=(4^2)^8`

`=> 4^3 . 4^x = 4^(16)`

`=> 4^x= 4^(16-3)`

`=>4^x=4^(13)`

`=>x=13`

_____

`2^x . 16^2 =1024`

`=> 2^x= 1024 : 16^2`

`=>2^x=4`

`=>2^x=2^2`

`=>x=2`

a: =>5(x-3)=25

=>x-3=5

=>x=8

b: =>2x=16-10=6

=>x=3

c: =>58-3x=51

=>3x=7

=>x=7/3

d: =>4x-20=4

=>4x=24

=>x=6

e: =>2^x=4

=>2^x=2^2

=>x=2

f: =>3^x=81

=>3^x=3^4

=>x=4

g: =>4^x*4^3=4^16

=>x+3=16

=>x=13

h: =>2^x=1024/256=4=2^2

=>x=2

\(\left(x+\frac{2}{3}\right)\left(\frac{5}{4}-2x\right)>0\)

th1 : 

\(\hept{\begin{cases}x+\frac{2}{3}>0\\\frac{5}{4}-2x>0\end{cases}\Rightarrow\hept{\begin{cases}x>-\frac{2}{3}\\-2x>-\frac{5}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x>-\frac{2}{3}\\x>\frac{5}{8}\end{cases}\Rightarrow}x>\frac{5}{8}}\)

th2 : 

\(\hept{\begin{cases}x+\frac{2}{3}< 0\\\frac{5}{4}-2x< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -\frac{2}{3}\\-2x< -\frac{5}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x< -\frac{2}{3}\\x< \frac{5}{8}\end{cases}\Rightarrow}x< -\frac{2}{3}}\)

vậy_

\(\Leftrightarrow\left(2x+3-2x+5\right)^2=x^2+6x+64\)

=>x^2+6x=0

=>x(x+6)=0

=>x=0 hoặc x=-6

\(6^2-\left(x+3\right)=45\)

\(36-\left(x+3\right)=45\)

\(x+3=35-45\)

\(x+3=-10\)

\(x=-13\)

lớp 6 mà bạn

\(6^2-\left(x+3\right)=45\)

\(\Rightarrow36-\left(x+3\right)=45\)

\(\Rightarrow x+3=36-45\)

\(\Rightarrow x+3=-9\)

\(\Rightarrow x=-9-3=-12\)

\(125-5\left(3x-1\right)=5^5.5^3\)

\(\Rightarrow5^3-5\left(3x-1\right)=5^5.5^3\)

\(\Rightarrow5^3-15x+5=5^5.5^3\)

\(\Rightarrow5\left(5^2-3x+1\right)=5^5.5^3\)

\(\Rightarrow5^2-3x+1=5^5.5^3:5\)

\(\Rightarrow25-3x+1=5^7\)

Từ đây làm nốt nhé

\(4^x+1+4^0=65\)

\(\Rightarrow4^{x+1}+1=65\)

\(\Rightarrow4^{x+1}=64\)

\(\Rightarrow4^{x+1}=4^3\)

\(\Rightarrow x+1=3\)

\(\Rightarrow x=2\)

\(70-5\left(x-3\right)=5.3^2\)

\(\Rightarrow70-5\left(x-3\right)=5.9=45\)

\(\Rightarrow5\left(x-3\right)=70-45=25\)

\(\Rightarrow x-3=25:5=5\)

\(\Rightarrow x=8\)

Chúc em học tốt hơn nhé!!

a) \(\dfrac{x}{-6}=\dfrac{-15}{45}\)

    \(\dfrac{-x}{6}=\dfrac{-15}{45}\)

    \(\dfrac{x}{6}=\dfrac{15}{45}\)

    \(x=\dfrac{\left(15\cdot6\right)}{45}\)

    \(x=2\)

b) \(\dfrac{x}{5}=\dfrac{16}{25}\)

    \(x=\dfrac{\left(16\cdot5\right)}{25}\)

    \(x=\dfrac{80}{25}\)

   \(x=\dfrac{16}{5}\)

c) \(\dfrac{5}{x-3}=\dfrac{20}{-12}\)

  \(x-3=\dfrac{\left(5\cdot-12\right)}{20}\)

 \(x-3=-3\)

 \(x=\left(-3\right)+3\)

\(x=0\)

d) \(\dfrac{2}{5}\cdot x=\dfrac{6}{35}\)

         \(x=\dfrac{6}{35}\div\dfrac{2}{5}\)

        \(x=\dfrac{3}{7}\)