a) Ta có: 5x(12x-7)-6(10x²+3) = 0

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

\(\Leftrightarrow\) -35x = 18

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

a) \(2x+6=0\Leftrightarrow2x=-6\Leftrightarrow x=-3 \)
b)\(4x+20=0\Leftrightarrow4x=-20\Leftrightarrow x=-5\)
c)\(2\left(x+1\right)=5x+7\Leftrightarrow2x+1=5x+7\Leftrightarrow2x-5x=7-1\Leftrightarrow-3x=6\Leftrightarrow x=-2\)
d)\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
e)\(3x-1=2x-5\Leftrightarrow3x-2x=-5+1\Leftrightarrow x=-4\)
f)\(15-7x=9-3x\Leftrightarrow-7x+3x=9-15\Leftrightarrow-4x=-6\Leftrightarrow x=\frac{3}{2} \)
g)\(x-3=18\Leftrightarrow x=18+3\Leftrightarrow x=21\)

h)\(2x+1=15-5x\Leftrightarrow2x+5x=15-1\Leftrightarrow7x=14\Leftrightarrow x=2\)
i)\(3x-2=2x+5\Leftrightarrow3x-2x=5+2\Leftrightarrow x=7\)
k)\(-4x+8=0\Leftrightarrow-4x=-8\Leftrightarrow x=2\)
l)\(2x+3=0\Leftrightarrow2x=-3\Leftrightarrow x=-\frac{3}{2}\)
m)\(4x+5=3x\Leftrightarrow4x-3x=-5\Leftrightarrow x=-5\)

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 !

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 = \(\frac{-3}{2}\)

c) -4x² + 3x + 7 = 0

⇔ -4x² - 4x + 7x + 7 = 0

⇔ - 4x(x +1) + 7(x + 1) = 0

⇔ (x + 1)(-4x + 7) = 0

⇔ x + 1 = 0 ⇔ x = -1

Hoặc -4x + 7 = 0 ⇔ x = \(\frac{7}{4}\)

Vậy nghiệm của pt là x = -1 và x = \(\frac{7}{4}\)

CÂU NÀO CÁC BẠN LÀM ĐC THÌ GIÚP MK NHA !!!! 

a, x=-505

b, x=35/8 hoac -37/8

nhung cau con lai thi tong tu

a. \(\left|4x+2020\right|=0\)

\(\Rightarrow4x+2020=0\)

\(\Rightarrow4x=-2020\)

\(\Rightarrow x=-505\)

b. \(\left|2x+\frac{1}{4}\right|+\left|-5\right|=\left|-14\right|\)

\(\Rightarrow\left|2x+\frac{1}{4}\right|+5=14\)

\(\Rightarrow\left|2x+\frac{1}{4}\right|=9\)

\(\Rightarrow\orbr{\begin{cases}2x+\frac{1}{4}=9\\2x+\frac{1}{4}=-9\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{35}{4}\\2x=-\frac{37}{4}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{35}{8}\\x=-\frac{37}{8}\end{cases}}\)

c. \(\left|2020-5x\right|-\left|3\right|=-\left|-8\right|\)

\(\Rightarrow\left|2020-5x\right|-3=-8\)

\(\Rightarrow\left|2020-5x\right|=-5\left(vl\right)\)

=> x vô nghiệm

d. \(\left|x^2+4x\right|=0\)

\(\Rightarrow x^2+4x=0\)

\(\Rightarrow x\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)

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

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

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

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

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

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

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

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

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

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

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

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

\(m,x^2-11x+28=0\)

\(\Leftrightarrow x^2-4x-7x+28=0\)

\(\Leftrightarrow x\left(x-4\right)-7\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-7=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=7\end{cases}}\)

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

\(\Leftrightarrow16x^2+24x+9=4x^2-8x+4\)

\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)

\(\Leftrightarrow12x^2+32x+5=0\)

\(\Leftrightarrow\left(x+\frac{1}{6}\right)\left(x+\frac{5}{2}\right)=0\)

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

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 ..

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

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

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