a) 3 x 1/5 + 2/5 = 3/5 + 2/5 = 1

b) 3 : 1/5 - 2/5 = 15 - 2/5 = 73/5

bài 2 :
a) X x 1/2 = 1 - 1/5

    X x 1/2 = 4/5 

           X   = 4/5 : 1/2

           X   = 8/5 

vậy x =...

b) x : 1/3 = 1 + 1/5

    x : 1/3 = 6/5

          x   = 6/5 x 1/3

          x   = 2/5

vậy x =...

\(a,50\%x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x+x=\dfrac{4}{5}+0,2\)

\(\Leftrightarrow\dfrac{3}{2}x=\dfrac{4}{5}+\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{3}{2}x=1\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

\(b,\left(x-\dfrac{3}{4}\right):\dfrac{1}{2}+\dfrac{3}{2}=\dfrac{25}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{25}{2}-\dfrac{3}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{22}{2}\)

\(\Leftrightarrow x-\dfrac{3}{4}=11:2\)

\(\Leftrightarrow x=\dfrac{11}{2}+\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{25}{4}\)

a x=340

b x=6800

a) (x + 52400) : 5 = (52400 + 340) : 5

x = 340

b) ( x – 5480) × 6 = (6800 – 5480) × 6

x = 6800.

6800 nhé !

b) Ta có: \(x-43=\left(57-x\right)-50\)

\(\Leftrightarrow x-43=57-x-50\)

\(\Leftrightarrow x+x=7+43\)

\(\Leftrightarrow2x=50\)

hay x=25

Vậy: x=25

c) Ta có: (x-2)(8-x)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\8-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=8\end{matrix}\right.\)

Vậy: \(x\in\left\{2;8\right\}\)

a) \(\dfrac{9}{7}\div x=\dfrac{3}{5}\)

           \(x=\dfrac{9}{7}\div\dfrac{3}{5}\)

          \(x=\dfrac{15}{7}\)

b) \(x+\dfrac{1}{3}=\dfrac{8}{6}-\dfrac{1}{2}\)

    \(x+\dfrac{1}{3}=\dfrac{5}{6}\)

    \(x=\dfrac{5}{6}-\dfrac{1}{3}\)

   \(x=\dfrac{1}{2}\)

a) x = 3/5 x 9/7

x = 27/35

b) x+ 1/3 = 5/6

x = 5/6 - 1/3

x = 1/2

a)\(=>x=\dfrac{9}{7}:\dfrac{3}{5}=\dfrac{15}{7}\)

b)\(x=\left(\dfrac{8}{6}-\dfrac{1}{2}\right)-\dfrac{1}{3}=\dfrac{1}{2}\)

a)3(x-2)+2(x-3)=5

=>3x-6+2x-6=5

=>5x=17

=>x=17/5

b)(2x-8)^2=16

TH1:2x-8=4=>x=6

TH2:2x-8=-4=>x=2

\(a,\Leftrightarrow3x-6+2x-6=5\\ \Leftrightarrow5x-12=5\\ \Leftrightarrow x=\dfrac{17}{5}\\ b,\left(2x-8\right)^2-16=0\\ \Leftrightarrow\left(2x-12\right)\left(2x-4\right)=0\\ \Leftrightarrow4\left(x-6\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-4x+1-4x^2+11x+3=3\\ \Leftrightarrow7x=-1\\ \Leftrightarrow x=-\dfrac{1}{7}\)

a) Ta có: \(\left|4-5x\right|=24\)

\(\Leftrightarrow\left[{}\begin{matrix}-5x+4=24\\-5x+4=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-5x=20\\-5x=-28\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{-4;\dfrac{28}{5}\right\}\)

b) Ta có: \(\left(8+x\right)\left(6-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)

Vậy: \(x\in\left\{-8;6\right\}\)

a) \(\left|4-5x\right|=24\)

\(\Leftrightarrow\left[{}\begin{matrix}4-5x=24\\4-5x=-24\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)

b) \(\left(8+x\right)\left(6-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}8+x=0\\6-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)

a.|4-5.x|=24

TH1: 4-5.x=24

        ⇔5x=-20

        ⇔x=-4

TH1: 4-5.x=-24

        ⇔5x=28

        ⇔x=28/5

b.(8+x).(6-x)=0

\(\left\{{}\begin{matrix}8+x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)

a) x = 5/8 : 3/4

x = 5/6

b) x = 5/6 - 1/12

x = 3/4

a)x=\(\dfrac{5}{8}:\dfrac{3}{4}=\dfrac{5}{6}\)

b) x=\(\dfrac{5}{6}-\dfrac{1}{12}=\dfrac{3}{4}\)

a \(x=\dfrac{5}{8}:\dfrac{3}{4}\)

\(x=\dfrac{5}{8}\times\dfrac{4}{3}\)

\(x=\dfrac{5}{6}\)

b \(x=\dfrac{5}{6}-\dfrac{1}{12}\)

\(x=\dfrac{10}{12}-\dfrac{1}{12}\)

\(x=\dfrac{9}{12}\)

\(x=\dfrac{3}{4}\)

a: \(x\cdot3\cdot5=2,7\)

=>\(x\cdot15=2,7\)

=>\(x=\dfrac{2,7}{15}=0,18\)

b: \(1,2:x\cdot4=20\)

=>\(1,2:x=20:4=5\)

=>\(x=1,2:5=0,24\)

a: \(\Leftrightarrow x^2+10x+25-x^2+4x=55\)

=>14x=30

hay x=15/7

b: \(\Leftrightarrow\left(x-7\right)\left(x-3\right)=0\)

hay \(x\in\left\{7;3\right\}\)