`1/2x^2(-2x^2y^2z).(-1/3)x^2y^3`

`=1/2 .(-2).(-1/3)x^{2+2+2}.y^{2+3}.z`

`=1/3x^6y^5z`

Ta có: \(\dfrac{1}{2}x^2\cdot\left(-2x^2y^2z\right)\cdot\left(-\dfrac{1}{3}\right)x^2y^3\)

\(=\left(\dfrac{1}{2}\cdot2\cdot\dfrac{1}{3}\right)\cdot\left(x^2\cdot x^2\cdot x^2\right)\cdot\left(y^2\cdot y^3\right)\cdot z\)

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

\(\dfrac{-3}{2}\times\dfrac{-15}{64}+\left(0,8-2\dfrac{4}{5}\right):3\dfrac{2}{3}\\ =\dfrac{45}{128}+\left(-2\right)\times\dfrac{3}{11}\\ =\dfrac{45}{128}-\dfrac{6}{11}\\ =-\dfrac{273}{1408}\)

Vô nghiệm

\(\dfrac{5}{7}\times x+\dfrac{2}{5}=\dfrac{2}{3}\\ \dfrac{5}{7}\times x=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{10}{15}-\dfrac{6}{15}=\dfrac{4}{15}\\ x=\dfrac{4}{15}:\dfrac{5}{7}=\dfrac{4\times7}{15\times5}=\dfrac{28}{75}\)

Vậy `x=28/75`

`HaNa☘D`

\(\dfrac{5}{7}\times x+\dfrac{2}{5}=\dfrac{2}{3}\)

\(\dfrac{5}{7}\times x=\dfrac{2}{3}-\dfrac{2}{5}\)

\(\dfrac{5}{7}\times x=\dfrac{10}{15}-\dfrac{6}{15}\)

\(\dfrac{5}{7}\times x=\dfrac{4}{15}\)

\(x=\dfrac{4}{15}:\dfrac{5}{7}\)

\(x=\dfrac{4}{15}\times\dfrac{7}{5}\)

\(x=\dfrac{28}{75}\)

#Toru

=>x*5/7=2/3-2/5=4/15

=>x=4/15:5/7=4/15*7/5=28/75

\(x^3-9x^2=x^3-9x^2+2x-3\)

⇔\(x^3-x^3-9x^2+9x^2-2x=-3\)

⇔\(-2x=-3\)

⇔\(x=\dfrac{3}{2}\)

`x^3 -9x^2 =x^3 -9x^2 +2x-3`

`=> x^3 -9x^2-x^3 +9x^2 -2x+3=0`

`=> (x^3-x^3) +(-9x^2+9x^2)-2x+3=0`

`=> -2x-3=0`

`=>-2x=-3`

`=>x=3/2`

\(=>x\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{10}{3}+1=\dfrac{13}{3}\)

\(=>x=\dfrac{13}{3}:\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{13}{3}:\dfrac{7}{6}=\dfrac{26}{7}\)

5/6x - 1 = 10/3

5/6x = 10/3 + 1 = 13/3

X = 13/3 : 5/6 = 26/5

\(\dfrac{1}{2}\times x+\dfrac{2}{3}\times x-1=3\dfrac{1}{3}\)

\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{10}{3}+1\)

\(\Rightarrow\dfrac{7}{6}x=\dfrac{13}{3}\)

\(\Rightarrow x=\dfrac{13}{3}:\dfrac{7}{6}\)

\(\Rightarrow x=\dfrac{26}{7}\)

\(\frac{2}{3}+\frac{8}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)

\(\frac{94}{105}< \frac{x}{105}< \frac{92}{105}\)

\(\Rightarrow94< x< 92\)

mà x là số tựu nhiên => \(x\in\varnothing\)