Câu 3                           

 a, x= 53/5                                                x= 0

=> cái vế  trong ngoặc là 0.

rồi bạn làm tiếp nha

\(\dfrac{-12}{25}.\left(\dfrac{3}{4}-x+\dfrac{6}{-11}-\dfrac{5}{6}\right)=0\)

\(\dfrac{3}{4}-x+\dfrac{-6}{11}-\dfrac{5}{6}=0\)

\(\dfrac{3}{4}-x+\dfrac{-91}{66}=0\)

\(\dfrac{3}{4}-x=0-\left(\dfrac{-91}{66}\right)\)

\(\dfrac{3}{4}-x=\dfrac{91}{66}\)

\(x=\dfrac{-83}{132}\)

a: Ta có: \(-\dfrac{12}{25}\cdot\left(\dfrac{3}{4}-x+\dfrac{6}{-11}-\dfrac{5}{6}\right)=0\)

\(\Leftrightarrow x+\dfrac{83}{132}=0\)

hay \(x=-\dfrac{83}{132}\)

a. \(=\left(\dfrac{12}{15}+\dfrac{3}{15}\right)+\left(\dfrac{4}{3}+\dfrac{5}{3}\right)+\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\\ =1+2+1=4\)

b. \(x\times\dfrac{3}{4}=\dfrac{1\times3\times6\times5}{6\times5\times5\times3}\\ x\times\dfrac{3}{4}=\dfrac{1}{5}\\ x=\dfrac{1}{5}:\dfrac{3}{4}\\ x=\dfrac{4}{15}\)

a.

(12/15+3/15)+(4/3+5/3)+(1/4+3/4)

=1+3+1=5

b.X x 3/4=1/5

           X=1/5:3/4

           X=4/15

3/4x + 1/2x  -15 = 35

(3/4 + 1/2) x X = 15 + 35 = 50

5/4 x X= 50

X = 50 : 5/4

X = 40 

Câu b:

\(\frac{27,5\times8,4}{x}\times7=21\)

\(\frac{27,5\times8,4}{x}=3\)

\(\frac{231}{x}=3\)

x = 231 : 3

x = 77 

a) 2/3x+4=-12

2/3x=-16

x=-24

b) 3/4+1/4:x=-3

3/4+1/4.1/x=-3

3/4+1/4x=-3

1/4x=-15/4

=>-60x=4

x=-1/15

Lời giải:
PT $\Leftrightarrow (x^3-2x^2)+(x^2-4)=0$

$\Leftrightarrow x^2(x-2)+(x-2)(x+2)=0$

$\Leftrightarrow (x-2)(x^2+x+2)=0$

$\Rightarrow x-2=0$ hoặc $x^2+x+2=0$
Nếu $x-2=0\Leftrightarrow x=2$ (tm)

Nếu $x^2+x+2=0$
$\Leftrightarrow (x+\frac{1}{2})^2=-\frac{7}{4}<0$ (vô lý)

Vậy pt có nghiệm duy nhất $x=2$