\(\left(x+\frac{1}{2}\right)\cdot\left(\frac{2}{3}-2x\right)=0\)

TH1 : \(\Rightarrow x+\frac{1}{2}=0\)

\(x=0-\frac{1}{2}\)

\(x=\frac{-1}{2}\)

TH2 : \(\Rightarrow\frac{2}{3}-2x=0\)

\(2x=0+\frac{2}{3}=\frac{2}{3}\)

\(x=\frac{2}{3}\div2=\frac{1}{3}\)

\(\Rightarrow x=\frac{-1}{2};\frac{1}{3}\)

\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

\(\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)

\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)

\(x=\frac{3}{5}^2-\frac{1}{5}^2\)

\(x=\frac{2}{5}\)

Các bạn ơi giúp mk với các bạn ơi mk sắp phải đi học rồi giúp mk với

các bạn có thương mk ko

\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)

 TH1:   \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)

TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)

\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)

\(\Rightarrow x=\frac{2}{5}\)

\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)

\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)

\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)

\(\Rightarrow3x=\frac{1}{9}\)

\(\Rightarrow x=\frac{1}{27}\)

\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)

\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)

a: =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5

=>x=3/5:2/3=3/5x3/2=9/10

b: \(\Leftrightarrow x\cdot2.8-50=34\)

=>2,8x=84

=>x=30

c: \(\Leftrightarrow\dfrac{1}{6}x=\dfrac{5}{12}\)

hay x=5/2

d: \(\Leftrightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}+\dfrac{7}{4}=\dfrac{41}{4}\)

=>2x-3/4=41/4 hoặc 2x-3/4=-41/4

=>2x=44/4=11 hoặc 2x=-19/2

=>x=11/2 hoặc x=-19/4

\(\frac{x-2}{27}\)+\(\frac{x-3}{26}\)+\(\frac{x-4}{25}\)+\(\frac{x-44}{5}\)=1

<=> \(\frac{x-2}{27}-1\)+\(\frac{x-3}{26}-1\)+\(\frac{x-4}{25}-1\)+\(\frac{x-44}{5}+3\)=1

<=> \(\frac{x-29}{27}\)+\(\frac{x-29}{26}\)+\(\frac{x-29}{25}\)+\(\frac{x-29}{5}\)=1

<=> ( x- 29 ) \(\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{5}\right)\)=1

phần sau tự làm tp nhé!

mk sắp phải đi học rồi các bạn giúp mình với có đc ko mk nhớ sẽ đền đáp công ơn của bạn 

a) (5 - x) +12 = -25

<-> 5 - x + 12 = -25

<-> 17 - x = - 25

<-> x = 42

b) 12 - 4(x - 2) = -4

<-> 12 - 4x + 8 = -4

<-> 20 - 4x = -4

<-> 4x = 24

<-> x = 6

(20-3.x):5=55-51

(20-3.x):5=4

20-3.x=4.5

20-3.x=20

3.x=20-20

3.x=0

x=0:3

x=0

không đc đừng ném mik nha

Câu b:

71 + (26 - 3\(x\)) : 5 = 75

(26 - 3\(x\)) : 5 = 75 - 71

(26 - 3\(x\)) : 5 = 4

(26 - 3\(x\)) = 4 x 5

26 - 3\(x\) = 20

3\(x\) = 26 - 20

3\(x\) = 6

\(x=6:3\)

\(x\) = 2

Vậy \(x=2\)

\(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)

\(\frac{2}{3}x\)            = \(\frac{1}{10}+\frac{1}{2}\)

\(\frac{2}{3}x\)            =          \(\frac{3}{5}\)

     \(x\)             =     \(\frac{3}{5}:\frac{2}{3}\)

     \(x\)             =         \(\frac{9}{10}\)

Câu b:

(2 4/5x - 50) : 2/3 = 51

2 4/5x - 50 = 51 x 2/3

14/5x - 50 = 34

14/5x = 34 + 50

14/5x = 84

x = 84 : 14/5

x = 30

Vậy x = 30