\(a,\left(x+2\right)^2+\left(x+3\right)^2-2\left(x-2\right)\left(x-3\right)=19\\ \Leftrightarrow x^2+4x+4+x^2+6x+9-2x^2+10x-12=19\\ \Leftrightarrow20x=20\\ \Leftrightarrow x=1\\ b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-5\right)=15\\ \Leftrightarrow x^3+8-x^3+5x=15\\ \Leftrightarrow5x=7\\ \Leftrightarrow x=\dfrac{7}{5}\\ c,\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\\ \Leftrightarrow x^3-3x^2+3x+1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=8\\ \Leftrightarrow x=\dfrac{8}{9}\)

a. (x + 2)² + (x + 3)² - 2(x - 2)(x - 3) = 19

<=> (x² + 4x + 4) + (x² + 6x + 9) - (2x + 4)(x - 3) = 19

<=> x² + 4x + 4 + x² + 6x + 9 - 2x² + 6x - 4x + 12 = 19

<=> x² + x² - 2x² + 4x + 6x + 6x - 4x + 9 + 4 + 12 - 19 = 0

<=> 12x + 6 = 0

<=> 6(2x + 1) = 0

<=> 2x + 1 = 0

<=> 2x = -1

<=> x = \(\dfrac{-1}{2}\)  

b. (x + 2)(x² - 2x + 4) - x(x² - 5) = 15

<=> x³ - 2x² + 4x + 2x² - 4x + 8 - x³ + 5x - 15 = 0

<=> x³ - x³ - 2x² + 2x² + 4x - 4x + 5x + 8 - 15 = 0

<=> 5x - 7 = 0

<=> 5x = 7

<=> x = \(\dfrac{7}{5}\)

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

gõ latex đi bn!

a. 2^x + 70 = 74

<=> 2^x = 4

<=> x = 2

b. 120 - \(\dfrac{4x}{2}\) = 80

<=> 120 - 2x = 80

<=> 120 - 80 = 2x

<=> 2x = 40

<=> x = 20

c. (3x + 5)² = 400

<=> \(|3x+5|=\sqrt{400}\)

<=> \(|3x+5|=20\)

<=> \(\left[{}\begin{matrix}3x+5=20\\3x+5=-20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-25}{3}\end{matrix}\right.\) 

a) (x + 3)^2 - 4x - 12 = 0

<=> (x + 3)^2 - 4(x + 3) = 0

<=> (x + 3)(x - 1) = 0

<=> x = -3 hoặc x = 1

b) x(x + 5)(x- 5) - (x - 3)(x^2 + 3x + 9) = 7

<=> x^3 - 25x - x^3 + 27 = 7

<=> -25x + 27 = 7

<=> x = 4/5

a/ \(\left(x+3\right)^2-4x-12=0\)

\(\left(x+3\right)^2-4\left(x+3\right)=0\)

\(\left(x+3\right)\left(x+3-4\right)=0\)

\(\left[{}\begin{matrix}x+3=0\Rightarrow x=-3\\x+3-4=0\Rightarrow x=1\end{matrix}\right.\)

---

b/ \(x\left(x+5\right)\left(x-5\right)-\left(x-3\right)\left(x^2+3x+9\right)=7\)

\(x\left(x^2-25\right)-\left(x^3-27\right)=7\)

\(x^3-25x-x^3+27=7\)

\(-25x=-20\)

\(x=\dfrac{20}{25}=\dfrac{4}{5}\)

a, <=>x² +6x+9-4x-12=0

<=> x² +2x -3=0

<=> x² +3x -x-3=0

<=> x.(x+3) - (x+3) =0

<=> (x-1)(x+3)=0

<=> x=1 hoặc x=-3

b, <=> x(x² -25) - (x-3)(x+3)² -7=0 

<=> x³ -25x + (9-x²) (x+3) -7=0

<=> x³ -25x+ 9x+27-x³ -3x² -7=0

<=> -3x² -16x +20=0

<=>(3x-10)(x-2) =0 (đoạn này tự phân tích nha ^ ^)

<=> x= 10/3 hoặc x=2

Chúc bạn học tốt nha!

\(a,\) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+6\)

\(< =>20x-5x^2+5x^2-12-x-6=0\)

\(< =>19x-18=0\)

\(< =>x=\dfrac{18}{19}\)

\(b,\left(2x-7\right)\left(5+4x\right)-8\left(x^2-4x+5\right)=-30\)

\(< =>10x+8x^2-35-28x-8x^2+24x-40+30=0\)

\(< =>6x-45=0< =>x=\dfrac{45}{6}=7,5\)

a) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+\Rightarrow6\\ \Leftrightarrow20x-5x^2+5x^2-12=x+6\\ \Leftrightarrow20x-12=x+6\\\Rightarrow20x-x=6+12\\ \Rightarrow19x=18\\ \Rightarrow x=\dfrac{18}{19}\)

b) \(\left(2x-7\right)\left(5+4x\right)-8\left(x^2-3x+5\right)=-30\\ \Rightarrow10x+8x^2-35-28x-8x^2+24x-40=-30\\ \Rightarrow6x-75=-30\\ \Rightarrow6x=45\\ \Rightarrow x=\dfrac{15}{2}\)

a)x=18/19

b)x=15/2

`a,x(x-1)-(x+2)^2=1`

`<=>x^2-x-x^2-4x-4=1`

`<=>-5x=5`

`<=>x=-1`

`b,(x+5)(x-3)-(x-2)^2=-1`

`<=>x^2+2x-15-x^2+4x-4+1=0`

`<=>6x-18=0`

`<=>x-3=0`

`<=>x=3`

`c,x(2x-4)-(x-2)(2x+3)=0`

`<=>2x(x-2)-(x-2)(2x+3)=0`

`<=>(x-2)(2x-2x-3)=0`

`<=>-3(x-2)=0`

`<=>x-2=0`

`<=>x=2`

`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`

`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`

`<=>4x+26=-12`

`<=>4x=-38`

`<=>x=-19/2`

a)x.(5-2x)-2x.(1-x)=15
   x [ 5 - 2x -2.(1-x) ] = 15
   x ( 5 - 2x -2 + 2x ) =15
   x . 3 =15
   x = 5
b)(3x+2)²+(1+3x).(1-3x)=2
   9x²+12x+4+1-9x²=2
   12x + 5 = 2
    12x = -3
        x = -1/4

a)\(\Leftrightarrow\)\(5x-2x^2-2x+2x^2=15\)

\(\Leftrightarrow\)\(3x=15\)

\(\Leftrightarrow\)\(x=5\)

b)\(\Leftrightarrow\)\(9x^2+12x+4+1-9x^2-2=0\)

\(\Leftrightarrow\)\(12x+3=0\)

\(\Leftrightarrow\)\(x=-0,25\)

a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)

\(\Leftrightarrow x\in\left\{-2;12\right\}\)

b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)

a) \(\sqrt{x-3}>2\left(đk:x\ge3\right)\)

\(\Leftrightarrow x-3>4\Leftrightarrow x>7\)

b) \(\sqrt{36x^2-12x+1}=5\)

\(\Leftrightarrow\sqrt{\left(6x-1\right)^2}=5\)

\(\Leftrightarrow\left|6x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}6x-1=5\\6x-1=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-\dfrac{2}{3}\end{matrix}\right.\)

c) \(\sqrt{9\left(5x^2-2x+16\right)}=3x+12\left(đk:x\ge-4\right)\)

\(\Leftrightarrow9\left(5x^2-2x+16\right)=9x^2+72x+144\)

\(\Leftrightarrow36x^2-90x=0\)

\(\Leftrightarrow18x\left(2x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)

\(a,\Leftrightarrow x^2\left(x+5\right)-9\left(x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\\x=-3\end{matrix}\right.\\ b,\Leftrightarrow\left[{}\begin{matrix}1-2x=3x-2\\2x-1=3x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=1\end{matrix}\right.\)

a) \(\Leftrightarrow x^2\left(x+5\right)-9x-45=0\)
    \(\Leftrightarrow x^2\left(x+5\right)-9x\left(x+5\right)=0\)
    \(\Leftrightarrow\left(x+5\right)\left(x^2-9\right)=0\)
    \(\Leftrightarrow\left(x+5\right)\left(x-3\right)\left(x+3\right)=0\)
    \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-3=0\\x+3=0\end{matrix}\right.\)
    \(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\\x=-3\end{matrix}\right.\)
Vậy...
b) \(\Leftrightarrow\left(1-2x\right)^2-\left(3x-2\right)^2=0\)
    \(\Leftrightarrow\left(1-2x-3x+2\right)\left(1-2x+3x-2\right)=0\)
    \(\Leftrightarrow\left(3-5x\right)\left(x-1\right)=0\)
    \(\Leftrightarrow\left[{}\begin{matrix}3-5x=0\\x-1=0\end{matrix}\right.\)
    \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=1\end{matrix}\right.\)
Vậy...