a) x² - 6x + 5 = 0

⇔ x² - x - 5x + 5 = 0

⇔ x(x - 1) - 5(x - 1) = 0

⇔ (x - 1)(x - 5) = 0

⇔ x - 1 = 0 ⇔ x = 1

Hoặc x - 5 = 0 ⇔ x = 5

Vậy nghiệm của pt là x = 1; x = 5

b) 2x² - x - 6 = 0

⇔ 2x² - 4x + 3x - 6 = 0

⇔ 2x(x - 2) + 3(x - 2) = 0

⇔ (x - 2)(2x + 3) = 0

⇔ x - 2 = 0 ⇔ x = 2

Hoặc 2x + 3 = 0 ⇔ x = \(\frac{-3}{2}\)

Vậy nghiệm của pt là x = 2; x = \(\frac{-3}{2}\)

a) ( 5x - 4)(4x + 6)=0

<=> \([^{5x-4=0}_{4x+6=0}< =>[^{x=\frac{4}{5}}_{x=\frac{-6}{4}}\)

Vậy S = \(\left\{\frac{4}{5};\frac{-6}{4}\right\}\)

b) ( 3,5x - 7 )( 2,1x - 6,3 ) = 0

<=> \([^{3,5x-7=0}_{2,1x-6,3=0}< =>[^{x=2}_{x=3}\)

Vậy S = \(\left\{2;3\right\}\)

c) ( 4x - 10 )( 24 + 5x ) = 0

<=> \([^{4x-10=0}_{24+5x=0}< =>[^{x=\frac{5}{2}}_{x=\frac{-24}{5}}\)

Vậy S = \(\left\{\frac{5}{2};\frac{-24}{5}\right\}\)

d) ( x - 3 )( 2x + 1 ) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=3\\x=\frac{-1}{2}\end{matrix}\right.\)

Vậy S = \(\left\{3;\frac{-1}{2}\right\}\)

e) ( 5x - 10 )( 8 - 2x ) = 0

<=> \(\left[{}\begin{matrix}5x-10=0\\8-2x=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

Vậy S = \(\left\{2;4\right\}\)

f) ( 9 - 3x )( 15 + 3x ) = 0

<=> \(\left[{}\begin{matrix}9-3x=0\\15+3x=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Vậy S = \(\left\{3;-5\right\}\)

Học tốt nhaaa !

Cảm ơn bn

a) Ta có: 5x(12x-7)-6(10x²+3) = 0

\(\Leftrightarrow\) 60x²-35x-60x²-18 = 0

\(\Leftrightarrow\) -35x = 18

\(\Leftrightarrow\) x = \(-\dfrac{18}{35}\)

\(o,x^2-9x+20=0\)

\(\Leftrightarrow x^2-4x-5x+20=0\)

\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)

\(n,3x^3-3x^2-6x=0\)

\(\Leftrightarrow3x\left(x^2-x-2\right)=0\)

\(\Leftrightarrow3x\left(x^2+x-2x-2\right)=0\)

\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)

\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\\x=2\end{cases}}\)

a/ \(3x-3=2x-7\)

\(\Leftrightarrow3x-2x=-7+3\)

\(\Leftrightarrow x=-4\)

b/ \(\left(2x+3\right)\left(3x-9\right)=0\)

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

Vậy ...

c/ \(\left(5x+2\right)\left(4x-6\right)=0\)

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

Vậy ..

\(A.\left(2,3x-6,5\right)\left(0,1x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2,3x-6,5=0\\0,1x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2,3x=6,5\\0,1x=-2\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{6,5}{2,3}\\x=-20\end{cases}}\)

\(2x^3-50x=0\)

<=>  \(2x\left(x^2-25\right)=0\)

<=>   \(2x\left(x-5\right)\left(x+5\right)=0\)

đến đây

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