\(5x\left(x-2\right)-3\left(x-1\right)=20x^2-15x\left(2x+1\right)-24\)

\(\Rightarrow5x^2-10x-3x+3=20x^2-30x^2-15x-24\)

\(\Rightarrow5x^2-13x+3=-10x^2-15x-24\)

\(\Rightarrow5x^2+10x^2-13x+15x+3+24=0\)

\(\Rightarrow15x^2+2x+27=0\)

Ta có: 

\(\Delta=2^2-4\cdot15\cdot27==-1616< 0\)

Nên pt vô nghiệm

\(5x\left(x-2\right)-3\left(x-1\right)=20x^2-15x\left(2x+1\right)-24\\ \Leftrightarrow5x^2-10x-3x+3=20x^2-30x^2-15x-24\\ \Leftrightarrow5x^2-20x^2+30x^2-10x-3x+15x+3+24=0\\ \Leftrightarrow15x^2+2x+27=0\\ \Leftrightarrow15x^2-2.x.\sqrt{15}+\dfrac{2}{15}+\dfrac{403}{15}=0\\ \Leftrightarrow\left(\sqrt{15}x+\dfrac{\sqrt{30}}{15}\right)^2+\dfrac{403}{15}=0\left(Vô.lí\right)\\ Vậy:Không.có.x.thoả\)

x^2 -2x = 24

=> x^2 - 2x - 24=0

=>x^2 -8x+6x - 24 = 0

=> ( x^2- 8x)+( 6x-24) = 0

=> x(x-8) + 6(x-8) = 0

=> (x+6)(x-8)=0

=>\(\orbr{\begin{cases}x=-6\\x=8\end{cases}}\)

\(=\frac{\left(2.5\right)^4.3^4-2^4\left(3.5\right)^2}{2^8.5^2.3^3}=\frac{2^4.3^2.5^2\left(5^2.3^2-1\right)}{2^8.5^2.3^3}=\frac{255-1}{16.3}=\frac{14}{3}\)

xin lỗi mình làm lộn bài

x=6

1: =>\(5^{x-2}-9=2^4-\left(6^2-6^2\right)\)

=>\(5^{x-2}=16+9=25\)

=>x-2=2

=>x=4

2: \(\Leftrightarrow3^x+16=19^6:19^5-3=19-3=16\)

=>3^x=0

=>x=0

3: \(\Leftrightarrow2^x+2^x\cdot16=272\)

=>2^x*17=272

=>2^x=16

=>x=4

4: \(\Leftrightarrow2^{x-1}+3=24-\left(4^2-2^2+1\right)=24-\left(16-4+1\right)\)

=>\(2^{x-1}+3=24-16+4-1=8+4-1=12-1=11\)

=>2^x-1=8

=>x-1=3

=>x=4

\(a,x^2-2x=24\)

\(x^2-2x-24=0\)

\(x^2-2x+1-25=0\)

\(\left(x-1\right)^2=5^2=\left(-5\right)^2\)

\(x-1=5\)                     hoặc                           \(x-1=-5\)

\(\Rightarrow\hept{\begin{cases}x=6\\x=-4\end{cases}}\)

\(b,\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(2x+255=0\)

\(2x=-255\)

\(x=-\frac{255}{2}\)

a/ \(x^2-2x=24\)

<=> \(x^2-2x+1-1=24\)

<=> \(\left(x-1\right)^2=25\)

<=> \(\orbr{\begin{cases}x-1=25\\x-1=-25\end{cases}}\)<=> \(\orbr{\begin{cases}x=26\\x=-24\end{cases}}\)

b/ \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

<=> \(4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

<=> \(4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

<=> \(2x+255=0\)

<=> \(2x=-255\)

<=> \(x=-\frac{255}{2}\)

a) x² + 10x - 2x - 20 = 0

=> x(x + 10) - 2(x + 10) = 0

=> (x - 2)(x + 10) = 0

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+10=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-10\end{cases}}\)

b) \(x^2-5x-24=0\)

\(\Rightarrow x^2-5x+\frac{25}{4}-\frac{121}{4}=0\)

\(\Rightarrow\left(x-\frac{5}{2}\right)^2=\frac{121}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-\frac{5}{2}\right)^2=\left(-\frac{11}{2}\right)^2\\\left(x-\frac{5}{2}\right)^2=\left(\frac{11}{2}\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5}{2}=\left(-\frac{11}{2}\right)\\x-\frac{5}{2}=\frac{11}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{6}{2}=3\\x=\frac{16}{2}=8\end{cases}}\)

c) x² - 8x + 3x - 24 = 0

=> x(x - 8) + 3(x - 8) = 0

=> (x + 3)(x - 8) = 0

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-8=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)

`a,x^2+2x+1=9`

`<=>x^2+2.x.1+1^2=9`

`<=>(x+1)^2=3^2`

`<=>(x+1)^2=+-3`

\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

`b, x^2-4x-21=0`

`<=>x^2+3x-7x-21=0`

`<=>x(x+3) - 7(x+3)=0`

`<=>(x+3)(x-7)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

`c,x^2+10x-24=0`

`<=>x^2+12x-2x-24=0`

`<=>x(x+12)-2(x+12)=0`

`<=>(x+12)(x-2)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+12=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-12\\x=2\end{matrix}\right.\)

a: =>(x+1)^2=9

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

=>(x+4)(x-2)=0

=>x=2 hoặc x=-4

b: =>x^2-7x+3x-21=0

=>(x-7)(x+3)=0

=>x=7;x=-3

c: =>x^2+12x-2x-24=0

=>(x+12)(x-2)=0

=>x=2 hoặc x=-12

b/ x²-2x=24

=> x²-2x-24=0

=> (x-6)(x+4)=0

=>x=6 hoặc x =-4

a/ (x-3)² - 4 = 0

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

=> (x-5)(x-1)=0

=> x = 5 hoặc x=1

ý c đâu ạ