1.2x^2+x-6=2x^2+4x-3x+6=(2x^2+4x)-(3x+6)=2x(x+2)-3(x+2)=(x+2)(2x-3)

2.x^3-9x^2+14x

=x*(x^2-9x+14)

=x*(x^2-7x-2x+14)

=x*((x^2-7x)-(2x-14))

=x*(x(x-7)-2(x-7))

=x*((x--7)(x-2))

=x*(x-7)(x-2)

a,

\(\left(x-6\right)^2=9\\ \Rightarrow\left[{}\begin{matrix}x-6=-3\\x-6=3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=9\end{matrix}\right.\)

b,

\(\left|x\right|=3\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)

c,

\(\left|x+5\right|=15\\ \Rightarrow\left[{}\begin{matrix}x+5=-15\\x+5=15\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-20\\x=10\end{matrix}\right.\)

d, 

\(2^{x-1}=16\\ \Rightarrow2^{x-1}=2^4\\ \Rightarrow x-1=4\\ \Rightarrow x=5\)

e, 

\(5^{x+1}=125\\ \Rightarrow5^{x+1}=5^3\\ \Rightarrow x+1=3\\ \Rightarrow x=2\)

a: Ta có: \(\left(x-6\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=3\\x-6=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=3\end{matrix}\right.\)

b: Ta có: \(\left|x\right|=3\)

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

c: Ta có: \(\left|x+5\right|=15\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-20\end{matrix}\right.\)

Đặt x² + 10x + 24 = y

pt đã cho trở thành ( y + 4x ).y - 165x² = 0

<=> y² + 4xy - 165x² = 0

<=> y² - 11xy + 15xy - 165x² = 0

<=> y( y - 11x ) + 15x( y - 11x ) = 0

<=> ( y - 11x )( y + 15x ) = 0

=> ( x² + 10x + 24 - 11x )( x² + 10x + 24 + 15x ) = 0

<=> ( x² - x + 24 )( x² + 25x + 24 ) = 0

<=> ( x² - x + 24 )( x² + 24x + x + 24 ) = 0

<=> ( x² - x + 24 )[ x( x + 24 ) + ( x + 24 ) ] = 0

<=> ( x² - x + 24 )( x + 24 )( x + 1 ) = 0

Vì x² - x + 24 > 0 ∀ x

nên pt <=> ( x + 24 )( x + 1 ) = 0 <=> x = -24 hoặc x = -1

Vậy ...

Đặt t = \(x^2+14x+24\)

\(\Rightarrow\)\(t\left(t-4x\right)-165x^{^2}=0\)

\(\Leftrightarrow t^2-4xt-165x^2=0\)

\(\Leftrightarrow t^2+11xt-15xt-165x^2=0\)

\(\Leftrightarrow t\left(t+11x\right)-15x\left(t+11x\right)=0\)

\(\Leftrightarrow\left(t+11x\right)\left(t-15x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t+11x=0\\t-15x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}t=-11x\\t=15x\end{cases}}}\)

với t= -11x

\(\Rightarrow x^2+14x+24=-11x\)

\(\Leftrightarrow x^2+25x+24=0\)

\(\Leftrightarrow x^2+x+24x+24=0\)

\(\Leftrightarrow x\left(x+1\right)+24\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+24\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+24=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-24\end{cases}}}\)

với t=15x

\(\Rightarrow x^2+14x+24=15x\)

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

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{95}{4}=0\)(Vô Lí)

vậy....

\(a.\dfrac{1}{2}:3+x=\dfrac{14}{5}\)

\(\dfrac{1}{6}+x=\dfrac{14}{5}\)

\(=>x=\dfrac{79}{30}\)

\(b.\dfrac{8}{5}:x:\dfrac{7}{4}=\dfrac{11}{6}\)

\(\left(\dfrac{8}{5}\cdot\dfrac{4}{7}\right):x=\dfrac{11}{6}\)

\(\dfrac{32}{35}:x=\dfrac{11}{6}\)

\(x=\dfrac{192}{385}\)

\(c.\dfrac{24}{10}+x:\dfrac{3}{4}=\dfrac{11}{3}\)

\(x:\dfrac{3}{4}=\dfrac{11}{3}-\dfrac{24}{10}\)

\(x:\dfrac{3}{4}=\dfrac{38}{30}\)

\(=>x=\dfrac{19}{20}\)

\(a,\dfrac{1}{2}:3+x=\dfrac{14}{5}\\ \Leftrightarrow x+\dfrac{1}{6}=\dfrac{14}{5}\\ \Leftrightarrow x=\dfrac{79}{30}\\ b,\dfrac{8}{5}:x:\dfrac{7}{4}=\dfrac{11}{6}\\ \Leftrightarrow x=\dfrac{192}{385}\\ c,\dfrac{24}{10}+x:\dfrac{3}{4}=\dfrac{11}{3}\\ \Leftrightarrow\dfrac{4}{3}x=\dfrac{19}{15}\\ \Leftrightarrow x=\dfrac{19}{20}\)

a   1/6+x=14/5

     x=14/5-1/6

     x= 79/30

b    8/5:x=11/6*7/4

     8/5:x=77/24

     x=77/24:8/5

     x=385/192

c  x:3/4=11/3-24/10

  x:3/4=19/5

  x=19/5*3/4

  x=57/20

Giúp

(x^2 +24+14x) (x^2+24+10x) =165x^2

Đặt t = x^2 + 24+12x

(t-2x)(t+2x) = 165x^2

t^2 - 4x^2 =165x^2

t^2 = 169x^2

t = 13x hay t = -13x

Nếu t = 13x thì 

x^2 +12x + 24= 13x

x^2 - x + 24 = 0 (Vô nghiệm vì vế trái > 0)

Nếu t = -13x thì:

x^2 +12x+24 = -13x

x^2 +25x +24=0

(x+1)(x+24) = 0

x + 1 =0 hay x+24 = 0

x = -1 hay x= -24

Vậy... 

Học tốt!

a: =>(x-2)^3*[(x-2)^8-1]=0

=>(x-2)(x-3)(x-1)=0

=>\(x\in\left\{2;3;1\right\}\)

b: (x-5)^24=(x-5)^9

=>\(\left(x-5\right)^9\cdot\left[\left(x-5\right)^{15}-1\right]=0\)

=>x-5=0 hoặc x-5=1

=>x=6 hoặc x=5

c: =>(x-5)^4*[(x-5)^21-1]=0

=>x-5=0 hoặc x-5=1

=>x=5 hoặc x=6

a) \(\left(x-2\right)^{11}=\left(x-2\right)^3\)

\(\Rightarrow\left(x-2\right)^{11}-\left(x-2\right)^3=0\)

\(\Rightarrow\left(x-2\right)^3\left[\left(x-2\right)^8-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^3=0\\\left(x-2\right)^8-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x-2\right)^8=1\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

b) \(\left(x-5\right)^{24}=\left(x-5\right)^9\)

\(\Rightarrow\left(x-5\right)^{24}-\left(x-5\right)^9=0\)

\(\Rightarrow\left(x-5\right)^9\left[\left(x-5\right)^{15}-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^9=0\\\left(x-5\right)^{15}-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{15}=1\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)

c) \(\left(x-5\right)^{25}=\left(x-5\right)^4\)

\(\Rightarrow\left(x-5\right)^{25}-\left(x-5\right)^4\)

\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^{21}-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\\\left(x-5\right)^{21}-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{21}=1\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)

\(\frac{4}{5}\cdot\frac{3}{x}\cdot\frac{2}{11}=\frac{24}{165}\)

        \(\frac{3}{x}\)     \(=\frac{24}{165}:\frac{2}{11}:\frac{4}{5}\)

        \(\frac{3}{x}\)     \(=1\)

         \(x\)     \(=3\)

\(\frac{4}{3}:\frac{x}{5}=\frac{5}{13}\cdot\frac{4}{9}\)

\(\frac{4}{3}:\frac{x}{5}=\frac{20}{117}\)

      \(\frac{x}{5}=\frac{4}{3}:\frac{20}{117}\)

     \(\frac{x}{5}=\frac{39}{5}\)

     \(x=39\)