a)\(\frac{1}{4}+\frac{1}{3}:2x=-5\)

   \(\frac{1}{3}:2x=-5-\frac{1}{4}\)

   \(\frac{1}{3}:2x=-\frac{21}{3}\)

   \(2x=\frac{1}{3}:\left(\frac{-21}{3}\right)\)

   \(2x=-\frac{1}{21}\)

   \(x=\frac{-1}{42}\)

b)\(\left(3x-\frac{1}{4}\right).\left(x+\frac{1}{2}\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}3x=\frac{1}{4}\\x=-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{12}\\x=-\frac{1}{2}\end{array}\right.\)

c)\(\left(2x-5\right).\left(\frac{3}{2}x+9\right).\left(0,3x-12\right)=0\)

   \(\Rightarrow\left[\begin{array}{nghiempt}2x-5=0\\\frac{3}{2}x+9=0\\0,3x-12=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x=5\\\frac{3}{2}x=-9\\0,3x=12\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-6\\x=40\end{array}\right.\)

a) 1/4 + 1/3 : 2x = -5

=> 1/3 : 2x = -5 - 1/4

=> 1/3 : 2x = -21/4

=> 2x = 1/3 : (-21/4) = -4/63

=> x = -4/63 : 2 = -2/63

\(\frac{1}{4}+\frac{1}{3}:2x=-5\)

\(\frac{1}{3}:2x=-5-\frac{1}{4}\)

\(\frac{1}{3}:2x=\frac{-21}{4}\)

\(2x=\frac{1}{3}:\frac{-21}{4}\)

\(2x=\frac{-4}{63}\)

\(x=\frac{-4}{63}:2\)

\(x=\frac{-2}{63}\)

\(\)

\(\frac{1}{4}+\frac{1}{3}:2x=-5\)

\(\Rightarrow\frac{1}{3}:2x=-\frac{21}{4}\)

\(\Rightarrow2x=\frac{-4}{63}\)

\(\Rightarrow x=\frac{-2}{63}\)

\(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}}\)

\(\left(2x-5\right)\left(\frac{3}{2}x+9\right)\left(0,3x-12\right)=0\)

Th1 : \(2x-5=0\Rightarrow x=\frac{5}{2}\)

Th2 : \(\frac{3}{2}x+9=0\Rightarrow x=-6\)

Th3 : \(0,3x-12=0\Rightarrow x=\frac{12}{0,3}\)

a,1/4 + 1/3:2x = -5

1/3 : 2x = -5-1/4

1/3 : 2x = -21/4

2x = 1/3 : -21 / 14

2x = -2/9

x= -2/9 : 2

x= -1/9

B, (3x-1/4) . (x+1/2) = 0

T/h1: <=> 3x-1/4 = 0 

<=> 3x = 1/4

=> x= 1/12

T/h2: <=> x+1/2 =0

=> x= -1/2

Vậy x= 1/2 và x = -1/2

C, Th1: 

2x-5 = 0

=> 2x= 5

=> x = 5/2

Th2: 

<=> 3/2 x+9 =0

=> 3/2 x = -9

=> x= -6

TH3:

<=> 0.3x-12 = 0

       0.3x=12

           x= 12:0,3

=> x= 40

Vậy x= 5/2, x= -6 và x=40

Chọn đúng nhen bạn

\(\left(3x-\frac{2}{4}\right)\cdot\left(x+\frac{1}{2}\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}3x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-\frac{1}{2}\end{cases}}\)

còn lại tương tự bài trên

Mọi người giúp mình nhanh nhé ! Mình đang cần gấp

e, 3x(2-x) =15(x-2)

\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)

\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

Vậy..

f, (x+5)(x+4)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)

Vậy..

g, x(x+4)=0

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

,h, (2x -4)(x-2)=0

\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)

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

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

i, (x+1/5)(2x-3)=0

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

k, x²-4x=0

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

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

m, 4x²-1=0

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

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

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

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

n, x²-6x+9=0

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

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

<=> x=3

l, (3x-5)²-(x+4)²=0

\(\Leftrightarrow\left(3x-5-x-4\right)\left(3x-5+x+4\right)=0\)

\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)

Vậy ..

o, 7x(x+2)-5(x+2)=0

\(\Leftrightarrow\left(x+2\right)\left(7x-5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)

Vậy....

p, 3x(2x-5)-4x+10=0

\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)

\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(3x-2\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)

Vậy...

q, (2-2x)-x²+1=0

\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)

\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)

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

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

Vậy ....

r, x(1-3x)=5(1-3x)

\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)

\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)

s, 2x-3/4+x+1/6=3

\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)

r, x(1-3x)=5(1-3x)

➜x(1-3x)-5(1-3x)=0

➜(x-5)(1-3x)=0

➜\(\left[{}\begin{matrix}x-5=0\\1-3x=0\end{matrix}\right.\text{➜}\left[{}\begin{matrix}x=5\\x=\frac{1}{3}\end{matrix}\right.\)

Mk lười lắm mai nha!!!~~~~~~~~~~~~

Làm dần:

a, 3x+6=0

➜3x=-6

➜x=2

b, 2x-10=0

➜2x=10

➜x=5

c, 3x-7=11

➜3x=11+7

➜3x=18

➜x=6

d, 3x-9=0

➜3x=9

➜x=3

a)(x+2).(x+3)-(x-2).(x+5)=10

  ( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10

 x^2 +3x+2x+6-x^2 -5x+2x+10-10=0

 2x+6=0

2x=-6

x=-3