1.

 \(x^2-5x+6=0\\ \Rightarrow x^2-2x-3x+6=0\\ \Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\\ \Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

2.

\(\left(x+4\right)^2-\left(3x-1\right)^2=0\\ \Rightarrow\left(x+4-3x+1\right)\left(x+4+3x-1\right)=0\\ \Rightarrow\left(-2x+5\right)\left(4x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2x+5=0\\4x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

3.

\(x^2-2x+24=0\\ \Rightarrow\left(x^2-2x+1\right)+23=0\\ \Rightarrow\left(x-1\right)^2+23=0\)

Vì (x-1)²≥0

23>0

\(\Rightarrow\left(x-1\right)^2+23>0\)

Vậy x vô nghiệm

4.

\(9x^2-4=0\\ \Rightarrow\left(3x-4\right)\left(3x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-4=0\\3x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

5.

\(x^2+2x-8=0\\ \Rightarrow\left(x^2+2x+1\right)-9=0\\ \Rightarrow\left(x+1\right)^2-3^2=0\\ \Rightarrow\left(x-2\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

a. (x - 3)² - 4 = 0

<=> (x - 3)² - 2² = 0

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

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

<=> \(\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\)

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

b. x² - 2x = 24

<=> x² - 2x - 24 = 0

<=> x² - 6x + 4x - 24 = 0

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

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

<=> \(\left[{}\begin{matrix}x+4=0\\x-6=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-4\\x=6\end{matrix}\right.\)

a) \(\left(x-3\right)^2-4=0\)

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

TH1:\(x-3=2\text{⇒}x=5\)

TH2:\(x-3=-2\text{⇒}x=1\)

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

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

\(\Rightarrow\left(x-3-2\right).\left(x-3+2\right)=0\)

\(\Rightarrow\left(x-5\right).\left(x-1\right)=0\)

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

Vậy \(x\in\left\{1;5\right\}\)

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

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

\(\Rightarrow x^2+4x-6x-24=0\)

\(\Rightarrow x.\left(x+4\right)-6.\left(x+4\right)\)

\(\Rightarrow\left(x-6\right).\left(x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Vậy \(x\in\left\{-4;6\right\}\)

Hoctot

a) x = 8 3 .                            b) x = − 9 20 .  

a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)

\(\Leftrightarrow2x-1=0\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x^3-\left(x+1\right)\left(x^2-x+1\right)=x\)

\(\Leftrightarrow x^3-x^3-1=x\)

hay x=-1

c: Ta có: \(56x^4+7x=0\)

\(\Leftrightarrow7x\left(8x^3+1\right)=0\)

\(\Leftrightarrow x\left(2x+1\right)=0\)

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

d: Ta có: \(x^2-5x-24=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

a) x = 1

b) x = -2

c) x = -10

a) Ta có: \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2+4\left(x^2-5x\right)+6\left(x^2-5x\right)+24=0\)

\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x+4\right)+6\left(x^2-5x+4\right)=0\)

\(\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0\)

\(\Leftrightarrow\left(x^2-2x-3x+6\right)\left(x^2-x-4x+4\right)=0\)

\(\Leftrightarrow\left[x\left(x-2\right)-3\left(x-2\right)\right]\left[x\left(x-1\right)-4\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\\x=4\end{matrix}\right.\)

Vậy: S={1;2;3;4}

b) Ta có: \(\left(2x+1\right)^2-2x-1=2\)

\(\Leftrightarrow\left(2x+1\right)^2-\left(2x+1\right)-2=0\)

\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)+\left(2x+1\right)-2=0\)

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

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

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

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

Vậy: \(S=\left\{-1;\dfrac{1}{2}\right\}\)

c) Ta có: \(x\left(x-1\right)\left(x^2-x+1\right)-6=0\)

\(\Leftrightarrow x\left(x^3-x^2+x-x^2+x-1\right)-6=0\)

\(\Leftrightarrow x\left(x^3-2x^2+2x-1\right)-6=0\)

\(\Leftrightarrow x^4-2x^3+2x^2-x-6=0\)

\(\Leftrightarrow x^4-2x^3+2x^2-4x+3x-6=0\)

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

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

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

\(\Leftrightarrow\left(x-2\right)\left[x\left(x^2-1\right)+3\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left[x\left(x-1\right)\left(x+1\right)+3\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2-x+3\right)=0\)

mà \(x^2-x+3>0\forall x\)

nên (x-2)(x+1)=0

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

Vậy: S={2;-1}

d) Ta có: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)

\(\Leftrightarrow\left(x^2+1\right)^2+2x\left(x^2+1\right)+x\left(x^2+1\right)+2x^2=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x^2+1+2x\right)+x\left(x^2+1+2x\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2+x+1\right)=0\)

mà \(x^2+x+1>0\forall x\)

nên x+1=0

hay x=-1

Vậy: S={-1}

a) (1) ⇔ 45x – 6 – 4x2 + 4 ≥ 2x – 4x2 + 6x – 18

⇔ 37x ≥ –16 ⇔ x ≥ -16/37

Tập nghiệm: S = {x|x ≥ -16/37}

b) (2) ⇔ x2 – 6x + 9 + 2x – 2 ≤ x2 + 3 ⇔ –4x ≤ –4 ⇔ x ≥ 1

Tập nghiệm: S = {x | x ≥ 1}.

\(\left(5+\sqrt{24}\right)^{x^2-2x-2}=49-10\sqrt{24}\)

=>\(\left(5+\sqrt{24}\right)^{x^2-2x-2}=\left(5-\sqrt{24}\right)^2\)

=>\(\left(5+\sqrt{24}\right)^{x^2-2x-2}=\left(5+\sqrt{24}\right)^{-2}\)

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

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

=>x(x-2)=0

=>x=0 hoặc x=2

=>x1-x2=0-2=-2

ở oooo

e: \(\Leftrightarrow3x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)