\(4\left(x-3\right)-8x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(4-8x\right)=0\\ \Leftrightarrow2\left(1-2x\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ 5x\left(x-7\right)-10\left(7-x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(5x+10\right)=0\\ \Leftrightarrow5\left(x+2\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\\ 2x-8=3x\left(x-4\right)\\ \Leftrightarrow2\left(x-4\right)-3x\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(2-3x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\\ 3x\left(x-5\right)=10-2x\\ \Leftrightarrow3x\left(x-5\right)+2\left(x-5\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=5\end{matrix}\right.\\ 6x\left(x-3\right)-3\left(3-x\right)=0\\ \Leftrightarrow\left(6x+3\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

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

\(\left(4-8x\right)\left(x-3\right)=0\)

\(\left[{}\begin{matrix}4-8x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\3\end{matrix}\right.\)

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

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

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

\(x^8+x^7+1\)

\(=x^8+x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-xx+1\)

\(=\left(x^8-x^6+x^5-x^3+x^2\right)\)

\(+\left(x^7-x^5+x^4-x^2+x\right)\)

\(+\left(x^6-x^4+x^3-x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

\(x^5+x+1\)

\(=x^5-x^2+x^2+x+1\)

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

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

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

\(x^4+2x^2-24\)

\(=x^4+2x^2+1-25\)

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

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

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

a) ( x - 3)⁴ + ( x - 5)⁴ = 82

Đặt : x - 4 = a , ta có :

( a + 1)⁴ + ( a - 1)⁴ = 82

⇔ a⁴ + 4a³ + 6a² + 4a + 1 + a⁴ - 4a³ + 6a² - 4a + 1 = 82

⇔ 2a⁴ + 12a² - 80 = 0

⇔ 2( a⁴ + 6a² - 40) = 0

⇔ a⁴ - 4a² + 10a² - 40 = 0

⇔ a²( a² - 4) + 10( a² - 4) = 0

⇔ ( a² - 4)( a² + 10) = 0

Do : a² + 10 > 0

⇒ a² - 4 = 0

⇔ a = + - 2

+) Với : a = 2 , ta có :

x - 4 = 2

⇔ x = 6

+) Với : a = -2 , ta có :

x - 4 = -2

⇔ x = 2

KL.....

b) ( n - 6)( n - 5)( n - 4)( n - 3) = 5.6.7.8

⇔ ( n - 6)( n - 3)( n - 5)( n - 4) = 1680

⇔ ( n² - 9n + 18)( n² - 9n + 20) = 1680

Đặt : n² - 9n + 19 = t , ta có :

( t - 1)( t + 1) = 1680

⇔ t² - 1 = 1680

⇔ t² - 41² = 0

⇔ ( t - 41)( t + 41) = 0

⇔ t = 41 hoặc t = - 41

+) Với : t = 41 , ta có :

n² - 9n + 19 = 41

⇔ n² - 9n - 22 = 0

⇔ n² + 2n - 11n - 22 = 0

⇔ n( n + 2) - 11( n + 2) = 0

⇔ ( n + 2)( n - 11) = 0

⇔ n = - 2 hoặc n = 11

+) Với : t = -41 ( giải tương tự )

@Giáo Viên Hoc24.vn

@Giáo Viên Hoc24h

@Giáo Viên

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@Akai Haruma

5: \(\Leftrightarrow9\left(x^2-5x-4\right)=36\left(x+1\right)+8\left(x^2-10x\right)\)

\(\Leftrightarrow9x^2-45x-36-36x-36-8x^2+80x=0\)

\(\Leftrightarrow x^2-x-72=0\)

=>(x-9)(x+8)=0

=>x=9 hoặc x=-8

6: \(\Leftrightarrow x^2-9=9x-x^2-9+x\)

\(\Leftrightarrow2x^2-10x=0\)

=>2x(x-5)=0

=>x=0 hoặc x=5

5, <=> 9x^2 - 45x - 36 = 36x + 36 + 8x^2 - 80x 

<=> x^2 - x - 72 = 0 <=> x = 9 ; x = -8 

6, <=> x^2 - 9 = 9x - x^2 - 9 + x = 10x - x^2 - 9 

<=> 2x^2 - 10x = 0 <=> x = 0 ; x = 5 

7, <=> (x-1)^2 = (3x+3)^2 

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

<=> (-2x-4)(4x+2) = 0 <=> x = -2;x=-1/2

8, = (x^2-10x-15)(x^2-10x+25)

D

a: \(x^4-2x^3+x^2-2x\)

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

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

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

b: \(x^4+x^3-8x-8\)

\(=\left(x^4+x^3\right)-\left(8x+8\right)\)

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

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

\(=\left(x+1\right)\left(x-2\right)\left(x^2+2x+4\right)\)

a) \(\sqrt[]{x^2-4x+4}=x+3\)

\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)

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

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

\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)

\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)

\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)

Giải pt (1)

\(\Delta=9+32=41>0\)

Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)

Giải pt (2)

\(\Delta=9+48=57>0\)

Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)

Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)