Question

Tìm x, biết:

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

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

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

Answer 1

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

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

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

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

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

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

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

\(\)

Answer 2

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

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

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

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

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

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

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

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

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

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

Answer 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

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