a, Ta có : \(-\dfrac{3}{2}-2x+\dfrac{3}{4}=-2\)

\(\Rightarrow-2x=-2+\dfrac{3}{2}-\dfrac{3}{4}=-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{5}{8}\)

Vậy ...

b, Ta có : \(\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)\left(-\dfrac{3}{2}-\dfrac{10}{3}\right)=\dfrac{2}{5}\)

\(\Rightarrow-\dfrac{29}{6}\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)=\dfrac{2}{5}\)

\(\Rightarrow-\dfrac{2}{3}x-\dfrac{3}{5}=-\dfrac{12}{145}\)

\(\Rightarrow-\dfrac{2}{3}x=-\dfrac{12}{145}+\dfrac{3}{5}=\dfrac{15}{29}\)

\(\Rightarrow x=-\dfrac{45}{58}\)

Vậy...

- Tham khảo thanh này để đánh nha bạn ;-;

a) Ta có: \(\dfrac{-3}{2}-\dfrac{2x+3}{4}=-2\)

\(\Leftrightarrow\dfrac{-6-2x-3}{4}=-2\)

\(\Leftrightarrow-2x-9=-8\)
\(\Leftrightarrow-2x=1\)

hay \(x=-\dfrac{1}{2}\)

b) Ta có: \(\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)\left(\dfrac{3}{2}-\dfrac{10}{3}\right)=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{-2}{3}x-\dfrac{3}{5}=\dfrac{2}{5}:\dfrac{-11}{6}=-\dfrac{12}{55}\)

\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{21}{55}\)

hay \(x=-\dfrac{63}{110}\)

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

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

x - 18 - 42 = 23 - 43 - 70 - x

\(\rightarrow\) 2x = -30

\(\rightarrow\) x = -15

\(\left(-\left(x+15\right)\right)^2\) - 19 = \(3^2\) .5

\(\rightarrow\)\(x^2+30x+225\) - 19 = 45

\(\rightarrow x^2+30x+161=0\)

\(\rightarrow x^2+23x+7x+161=0\)

\(\rightarrow x\left(x+23\right)+7\left(x+23\right)=0\)

\(\rightarrow\left(x+7\right)\left(x+23\right)=0\)

\(\rightarrow\) x = -7 hoặc x = -23

ok:

(2x+2⁴).5³=4.5⁵

(2x+2⁴)=4.5⁵:5³

(2x+2⁴)=4.5²

(2x+2⁴)=100

  2x=100-16

  2x=84

    x=84:2

    x=42

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3>=-5\\2x-3< =5\end{matrix}\right.\Leftrightarrow-1< =x< =4\)

3/5 - 1/3 x (2,48 + 0,52) x y : 60 : 5 = 1/5

       1/3 x (2,48+0,52) x y : 60 : 5 = 2/5

    1/3 x 3 x y : 60 : 5                     = 2/5

                      y: 60 : 5                    =2/5

                      y: 60                          =2/5 x 5

                      y : 60                         =2 

                      y                                =  2 x 60

                      y                                  =120

Ta có: \(\dfrac{3}{5}-\dfrac{1}{3}\left(2.48+0.52\right)\cdot y:60:5=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{3}{5}-\dfrac{1}{3}\cdot3\cdot y\cdot\dfrac{1}{60}\cdot\dfrac{1}{5}=\dfrac{1}{5}\)

\(\Leftrightarrow y\cdot\dfrac{1}{300}=\dfrac{2}{5}\)

hay y=120

Ta có \(\left(x+y\right)^2=x^2+2xy+y^2=49\Leftrightarrow xy=\dfrac{49-25}{2}=12\)

\(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=25^2-2\cdot12^2=337\)

Ta có \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=7^3-3\cdot12\cdot7=91\)

\(\left(x^2+y^2\right)\left(x^3+y^3\right)=91\cdot25=2275\\ \Leftrightarrow x^5+y^5+2x^2y^2\left(x+y\right)=2275\\ \Leftrightarrow x^5+y^5=2275-2\cdot144\cdot7=259\)