\(x=\dfrac{7}{25}+\dfrac{-1}{5}=\dfrac{7}{25}-\dfrac{1}{5}=\dfrac{2}{25}.\\ x=\dfrac{5}{11}+\dfrac{4}{-9}=\dfrac{5}{11}-\dfrac{4}{9}=\dfrac{1}{99}.\\ \dfrac{5}{9}-\dfrac{x}{-1}=\dfrac{-1}{3}\Leftrightarrow\dfrac{5}{9}+x=-\dfrac{1}{3}.\Leftrightarrow x=-\dfrac{8}{9}.\)

\(x=\dfrac{7}{25}+-\dfrac{1}{5}=>\dfrac{7}{25}+-\dfrac{5}{25}=>x=\dfrac{2}{25}\)

\(x=\dfrac{5}{11}+\dfrac{4}{-9}=>\dfrac{-45}{-99}+\dfrac{44}{-99}=>x=\dfrac{-1}{-99}=\dfrac{1}{99}\)

\(\dfrac{5}{9}-\dfrac{x}{-1}=-\dfrac{1}{3}=>-\dfrac{1}{3}-\dfrac{5}{9}=>\dfrac{x}{-1}=-\dfrac{8}{9}=>x=-\dfrac{8}{9}\)

\(x=\dfrac{7}{25}+\dfrac{-1}{5}\\ \Rightarrow x=\dfrac{7}{25}+\dfrac{-5}{25}\\ \Rightarrow x=\dfrac{2}{25}\\ x=\dfrac{5}{11}+\dfrac{4}{-9}\\ \Rightarrow x=\dfrac{-45}{-99}+\dfrac{44}{-99}\\ \Rightarrow x=\dfrac{45}{99}+\dfrac{44}{99}\\ \Rightarrow x=\dfrac{95}{99}\\ \dfrac{5}{9}-\dfrac{x}{-1}=\dfrac{-1}{3}\\ \Rightarrow\dfrac{x}{-1}=\dfrac{5}{9}-\dfrac{-1}{3}\\ \Rightarrow\dfrac{x}{-1}=\dfrac{5}{9}-\dfrac{-3}{9}\\ \Rightarrow\dfrac{x}{-1}=\dfrac{8}{9}\Rightarrow x\cdot9=-1\cdot8=-8\\ \Rightarrow x\cdot9=-8\\ \Rightarrow x=\dfrac{-8}{9}\)

Ta có: 

\(x^4=y^4\)

\(\Rightarrow x^4-y^4=0\)

\(\Rightarrow\left(x^2\right)^2-\left(y^2\right)^2=0\)

\(\Rightarrow\left(x^2-y^2\right)\left(x^2+y^2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2-y^2=0\\x^2+y^2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-y=0\\x+y=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

_______________

Ta có: 

\(x^5=y^5\)

\(\Rightarrow x^5-y^5=0\)

\(\Rightarrow x-y=0\)

\(\Rightarrow x=y\)

\(\dfrac{x^2}{20}=\dfrac{4}{5}\)

\(\Leftrightarrow x^2=16\)

hay \(x\in\left\{4;-4\right\}\)

Bài 1 :

\(3x+5=2\left(x-\frac{1}{4}\right)\)

\(\Leftrightarrow3x+5=2x-\frac{1}{2}\)

\(\Leftrightarrow5+\frac{1}{2}=2x-3x\)

\(\Leftrightarrow\frac{11}{2}=-x\)

\(\Leftrightarrow\frac{-11}{2}=x\)

Vậy \(x=\frac{-11}{2}\)

Bài 2:

a, \(\left|x+\frac{19}{5}\right|+\left|y+\frac{2018}{2019}\right|+\left|z-3\right|=0\)

Vì \(\hept{\begin{cases}\left|x+\frac{19}{5}\right|\ge0\\\left|y+\frac{2018}{2019}\right|\ge0\\\left|z-3\right|\ge0\end{cases}}\)

       Mà \(\left|x+\frac{19}{5}\right|+\left|y+\frac{2018}{2019}\right|+\left|z-3\right|=0\)

\(\Rightarrow+,\left|x+\frac{19}{5}\right|=0\)

\(\Leftrightarrow x+\frac{19}{5}=0\)

\(\Leftrightarrow x=\frac{-19}{5}\)

\(\Rightarrow+,\left|y+\frac{2018}{2019}\right|=0\)

\(\Leftrightarrow y+\frac{2018}{2019}=0\)

\(\Leftrightarrow y=\frac{-2018}{2019}\)

\(\Rightarrow+,\left|z-3\right|=0\)

\(\Leftrightarrow z-3=0\)

\(\Leftrightarrow z=3\)

Vậy \(\hept{\begin{cases}x=\frac{-19}{5}\\y=\frac{-2018}{2019}\\z=3\end{cases}}\)

b, Ta có : \(\left|x-\frac{1}{2}\right|+\left|2y+4\right|+\left|z-5\right|\ge0\)

Vì : \(\hept{\begin{cases}\left|x-\frac{1}{2}\right|\ge0\\\left|2y+4\right|\ge0\\\left|z-5\right|\ge0\end{cases}}\)

Mà : \(\left|x-\frac{1}{2}\right|+\left|2y+4\right|+\left|z-5\right|\ge0\)

\(\Rightarrow+,\left|x-\frac{1}{2}\right|\ge0\)

\(\Rightarrow x\inℚ\)

\(\Rightarrow+,\left|2y+4\right|\ge0\)

\(\Rightarrow y\inℚ\)

\(\Rightarrow+,\left|z-5\right|\ge0\)

\(\Rightarrow z\inℚ\)

Vậy chỉ cần \(\hept{\begin{cases}x\inℚ\\y\inℚ\\z\inℚ\end{cases}}\)thì thỏa mãn.

234*(-26)+134*26

\(a)6/4=9/4\) =)??

thiếu dữ liệu

bn ê .; là j vậy

1b)\(\frac{7}{19}x\frac{8}{11}+\frac{3}{11}:\frac{19}{7}-\frac{2}{-19}=\frac{7}{19}x\frac{8}{11}+\frac{3}{11}x\frac{7}{19}+\frac{2}{19}=\left(\frac{8}{11}+\frac{3}{11}\right)\frac{7}{19}+\frac{2}{19}=\frac{7}{19}+\frac{2}{19}=\frac{9}{19}\)

c)\(4\left(\frac{4}{9}+\frac{7}{11}-\frac{4}{9}\right)=4\frac{7}{11}\)

từ rồi làm tiếp

2)

a)\(\frac{5}{6}-x=-\frac{5}{12}-\frac{1}{-4}=-\frac{5}{12}+\frac{1}{4}=-\frac{5}{12}+\frac{3}{12}=-\frac{1}{6}.\)

\(x=\frac{5}{6}--\frac{1}{6}=\frac{5}{6}+\frac{1}{6}=1\)

b)\(\frac{x-2}{-5}=\frac{6}{7}\)

\(\frac{\left(x-2\right)\left(-7\right)}{35}=\frac{30}{35}\)

\(\Rightarrow\left(x-2\right)\left(-7\right)=30\)

c)\(\frac{1}{2}x-\frac{3}{4}x=-\frac{5}{6}\)

\(\left(\frac{1}{2}-\frac{3}{4}\right)x=-\frac{5}{6}\)

\(-\frac{1}{4}x=-\frac{5}{6}\)

\(x=\frac{20}{6}=\frac{10}{3}\)

vậy \(x=\frac{10}{3}\)

a) 150 : [17 - (2x - 3)] = 350

=> 17 - (2x - 3) = \(\frac{150}{350}=\frac{3}{7}\)

=> 2x - 3 = \(17-\frac{3}{7}=\frac{116}{7}\)

=> 2x = \(\frac{116}{7}+3\)

=> 2x = \(\frac{137}{7}\)

=> x = \(\frac{137}{7}:2=\frac{137}{14}\)

b) 4 . [105 - (x - 9)] - 486 = 0

=> 4. [105 - (x - 9)] = 486

=> 105 - (x - 9) = 243/2

=> x - 9 = \(105-\frac{243}{2}=-\frac{33}{2}\)

=> x = \(-\frac{33}{2}+9=-\frac{15}{2}\)

c) 1000 - 2{600 - 4.[198 - 7.(21 - 7)]}.x = 200

=> 1000 - 2{600 - 4.[198 - 7.14]}.x = 200

=> 1000 - 2{600 - 4.[198 - 98].x = 200

=> 1000 - 2{600 - 4.100}.x = 200

=> 1000 - 2{600 -400}.x = 200

=> 1000 - 2.200.x = 200

=> 2.200.x = 800

=> 400.x = 800

=> x = 2

Câu 2 :

a) 350 : [19 - (7x - 3)]  = 350

=> 19 - (7x - 3) = 1

=> 7x - 3 = 18

=> 7x = 21

=> x = 3

b) 7.[109 - (x - 9)] - 420 = 0

=> 7.[109 - (x - 9)] = 420

=> 109 - (x - 9) = 60

=> x - 9 = 49

=> x = 49 + 9 = 58

c) 254 - 2{200 - 4.[90 - 7.(19 - 9)]}.x = 54

=> 254 - 2{200 - 4.[90 - 7.10]}.x = 54

=> 254 - 2{200 - 4[90 - 70]} . x= 54

=> 254 - 2{200 - 4.20} . x = 54

=> 254 - 2{200 - 80}.x = 54

=> 254 - 2.120 . x=  54

=> 2.120.x = 200

=> 240.x = 200

=> \(x=\frac{5}{6}\)