\(\Rightarrow\left(3x^2-3x-2\right)⋮\left(x+1\right)\\ \Rightarrow\left(3x^2+3x-6x-6+4\right)⋮\left(x+1\right)\\ \Rightarrow\left[3x\left(x+1\right)-6\left(x+1\right)+4\right]⋮\left(x+1\right)\\ \Rightarrow x+1\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \Rightarrow x\in\left\{-5;-3;-2;0;1;3\right\}\)

12

2x+2x-6=42

=>4x=48

=>x=12 nhé

2x + 2(x-3) = 42

2x+2x-6=42

4x-6=42

4x=42+6

4x=48

x=48:4

x=12

= x³ -3x -x-2 -8x³+4x

= -7x³ -2

\(x\times9,6+x:2,5=45\\ x\times9,6+x\times0,4=45\\ x\times\left(9,6+0,4\right)=45\\ x\times10=45\\ x=45:10\\ x=4,5\)

tìm x nhá 

\(9,6x+x:2,5=45\\ \Leftrightarrow\left(9,6+0,4\right)x=45\\ \Leftrightarrow x=4,5\)

Vậy x = 4,5.

a, 2/3 - 2/9 + 7/9 = 6/9 - 2/9 + 7/9 = 10/9
b, 7/6 + 3/5 : 6 = 7/6 + 3/5 x 1/6 = 7/6 + 1/10 = 70/60 + 6/10 = 76/10 = 38/5
c, x : 6/25 = 18

x = 18 x 6/25

x = 36/25
d, ( 2/5 + 4/7) : x = 17/5

29/35 : x = 17/5

x = 29/35 : 17/5

x = 29/119

\(a.\dfrac{2}{3}-\dfrac{2}{9}+\dfrac{7}{9}=\dfrac{6}{9}-\dfrac{2}{9}+\dfrac{7}{9}=\dfrac{4}{9}+\dfrac{7}{9}=\dfrac{11}{9}\\ b.\dfrac{7}{6}+\dfrac{3}{5}:6=\dfrac{7}{6}+\dfrac{3}{5}\times\dfrac{1}{6}=\dfrac{7}{6}+\dfrac{1}{10}=\dfrac{70}{60}+\dfrac{6}{60}=\dfrac{76}{60}=\dfrac{19}{15}\\ c.x:\dfrac{6}{25}=18\\ x=18\times\dfrac{6}{25}\\ x=\dfrac{108}{25}\\ d.\left(\dfrac{2}{5}+\dfrac{4}{7}\right):x=\dfrac{17}{5}\\ \dfrac{34}{35}:x=\dfrac{7}{5}\\ x=\dfrac{34}{35}:\dfrac{7}{5}\\ x=\dfrac{34}{49}\)

Bài làm:

a) Ta có: \(\left(-\frac{3}{8}x^2z\right).\left(\frac{2}{3}xy^2z^2\right).\left(\frac{4}{5}x^3y\right)\)

\(=-\frac{1}{5}x^6y^3z^3\)

b) Tại x=-1 ; y=-2 ; z=3 thì giá trị đơn thức là:

\(-\frac{1}{5}.\left(-1\right)^6.\left(-2\right)^3.3^3=\frac{216}{5}\)

a) Ta có : \(\left(\frac{-3}{8}x^2z\right)\cdot\frac{2}{3}xy^2z^2\cdot\frac{4}{5}x^3y=\left(-\frac{3}{8}\cdot\frac{2}{3}\cdot\frac{4}{5}\right)\cdot x^2xx^3\cdot y^2y\cdot zz^2=-\frac{1}{5}x^6y^3z^3\)

b) Với x = -1 ; y = -2 , z = 3

Thế vào ba đơn thức trên và đơn thức tích ta được :

\(\frac{-3}{8}x^2z=\frac{-3}{8}\left(-1\right)^2\cdot3=\frac{-3}{8}\cdot1\cdot3=\frac{-9}{8}\)

\(\frac{2}{3}xy^2z^2=\frac{2}{3}\cdot\left(-1\right)\cdot\left(-2\right)^2\cdot3^2=\frac{2}{3}\left(-1\right)\cdot4\cdot9=-24\)

\(\frac{4}{5}x^3y=\frac{4}{5}\left(-1\right)^3\cdot\left(-2\right)=\frac{4}{5}\left(-1\right)\left(-2\right)=\frac{8}{5}\)

\(-\frac{1}{5}x^6y^3z^3=-\frac{1}{5}\left(-1\right)^6\left(-2\right)^3\cdot3^3=-\frac{1}{5}\cdot1\cdot\left(-8\right)\cdot27=\frac{216}{5}\)

NHANH LÊN MIK ĐANG CẦN GẤP NHA CÁC BAN

\(\frac{1}{6}\)(2x-3) = \(\frac{-1}{2}\)(x-\(\frac{1}{4}\))-\(\frac{2}{3}\)
\(\frac{1}{3}\)x - \(\frac{1}{2}\)=\(\frac{-1x}{2}\)-\(\frac{-1}{8}\)-\(\frac{2}{3}\)
\(\frac{1x}{3}\)-\(\frac{1}{2}\)=\(\frac{-1x}{2}\) - \(\frac{19}{24}\)
\(\frac{1x}{3}\) - \(\frac{1}{2}\) - \(\frac{-1x}{2}\) - \(\frac{19}{24}\) =0
\(\frac{5x}{6}\) - \(\frac{7}{24}\)=0
\(\frac{5x}{6}\) = \(\frac{7}{24}\)
x = \(\frac{7}{20}\)

\(\left(x-36\right):\left(2.3^2\right)=2^2.3\)

\(\Rightarrow\left(x-36\right):18=12\)

\(\Rightarrow\left(x-36\right)=12.18\)

\(\Rightarrow x-36=216\)

\(\Rightarrow x=216+36\)

\(\Rightarrow x=252\)

(x-36):(2.3²)=2.2³

⇒ (x-36):(2.9)=2.8

⇒ (x-36):18=16

⇒ x-36=18.16=288

⇒ x=288+36=324

\(\left(x-36\right):\left(2\cdot3^2\right)=2^2\cdot3\\ x-36=2^2\cdot3\cdot2\cdot3^2\\ x-36=3^3\cdot2^3\\ x-36=216\\ x=216+36\\ x=252\)