a, \(\left|2x-3\right|-\dfrac{1}{3}=0\Leftrightarrow\left|2x-3\right|=\dfrac{1}{3}\)

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

b, tương tự 

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

TH1 : \(2x-1=x+\dfrac{1}{3}\Leftrightarrow x=\dfrac{4}{3}\)

TH2 : \(2x-1=-x-\dfrac{1}{3}\Leftrightarrow3x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{2}{9}\)

d, \(3x-\left|x+15\right|=\dfrac{5}{4}\Leftrightarrow\left|x+15\right|=3x-\dfrac{5}{4}\)ĐK : x >= 5/12

TH1 : \(x+15=3x-\dfrac{5}{4}\Leftrightarrow-2x=-\dfrac{65}{4}\Leftrightarrow x=\dfrac{65}{8}\)( tm )

TH2 : \(x+15=\dfrac{5}{3}-3x\Leftrightarrow4x=-\dfrac{40}{3}\Leftrightarrow x=-\dfrac{10}{3}\)

TH2 x = -10/3 ( ktm ) nhé

a, \(\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{5}\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{5}>0\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{5}< 0\end{matrix}\right.\)

+,Xét \(\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{5}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{3}\\x>-\dfrac{2}{5}\end{matrix}\right.\)

\(\Rightarrow x>\dfrac{1}{3}\)

+, Xét \(\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{5}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{1}{3}\\x< -\dfrac{2}{5}\end{matrix}\right.\)

\(\Rightarrow x< -\dfrac{2}{5}\)

Vậy...........

b, \(\left(x+\dfrac{3}{5}\right)\left(x+1\right)< 0\)

Vì \(x+\dfrac{3}{5}< x+1\) với mọi \(x\in R\)

\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{5}< 0\\x+1>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< -\dfrac{3}{5}\\x>-1\end{matrix}\right.\)

Vậy...........

c, \(\dfrac{3}{7}x-\dfrac{2}{5}x=\dfrac{-17}{35}\)

\(\Rightarrow\dfrac{1}{35}x=\dfrac{-17}{35}\)

\(\Rightarrow x=-17\)

d, \(\left(\dfrac{3}{4}x-\dfrac{9}{10}\right)\left(\dfrac{1}{3}+\dfrac{-3}{5}x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{10}=0\\\dfrac{1}{3}+\dfrac{-3}{5}x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{10}\\-\dfrac{3}{5}x=-\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=\dfrac{5}{9}\end{matrix}\right.\)

Vậy.........

Chúc bạn học tốt!!!

a/ \(\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{5}\right)>0\)

TH1:\(\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{5}>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{3}\\x>-\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow x>\dfrac{1}{3}\)

TH2:\(\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{5}< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{1}{3}\\x< -\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow x< -\dfrac{2}{5}\)

Vậy \(x>\dfrac{1}{3}\) hoặc \(x< -\dfrac{2}{5}\) thì tm

b/ \(\left(x+\dfrac{3}{5}\right)\left(x+1\right)< 0\)

TH1:\(\left\{{}\begin{matrix}x+\dfrac{3}{5}< 0\\x+1>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< -\dfrac{3}{5}\\x>-1\end{matrix}\right.\) \(\Rightarrow-1< x< -\dfrac{3}{5}\)

TH2:\(\left\{{}\begin{matrix}x+\dfrac{3}{5}>0\\x+1< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>-\dfrac{3}{5}\\x< -1\end{matrix}\right.\)(vô lý)

Vậy....................

c/ \(\dfrac{3}{7}x-\dfrac{2}{5}x=-\dfrac{17}{35}\)

\(\Rightarrow\left(\dfrac{3}{7}-\dfrac{2}{5}\right)x=-\dfrac{17}{35}\)

\(\Rightarrow\dfrac{1}{35}x=-\dfrac{17}{35}\)

\(\Rightarrow x=-\dfrac{17}{35}:\dfrac{1}{35}=-17\)

Vậy.............

d/ \(\left(\dfrac{3}{4}x-\dfrac{9}{10}\right)\left(\dfrac{1}{3}+\dfrac{-3}{5}x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{10}=0\\\dfrac{1}{3}-\dfrac{3}{5}x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{10}\\\dfrac{3}{5}x=\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=\dfrac{5}{9}\end{matrix}\right.\)

Vậy.....................

Bài 1:

a, \(2y.\left(y-\dfrac{1}{7}\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}2y=0\\y-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=0\\y=\dfrac{1}{7}\end{matrix}\right.\)

Vậy \(y\in\left\{0;\dfrac{1}{7}\right\}\)

b, \(\dfrac{-2}{5}+\dfrac{2}{3}y+\dfrac{1}{6}y=\dfrac{-4}{15}\)

\(\Rightarrow\dfrac{5}{6}y=\dfrac{-4}{15}+\dfrac{2}{5}\)

\(\Rightarrow\dfrac{5}{6}y=\dfrac{2}{15}\)

\(\Rightarrow y=\dfrac{4}{25}\)

Vậy \(y=\dfrac{4}{25}\)

Chúc bạn học tốt!!!

Bài 1:

a, \(2y\left(y-\dfrac{1}{7}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2y=0\\y-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{1}{7}\end{matrix}\right.\)

Vậy...

b, \(\dfrac{-2}{5}+\dfrac{2}{3}y+\dfrac{1}{6}y=\dfrac{-4}{15}\)

\(\Rightarrow\dfrac{5}{6}y=\dfrac{2}{15}\)

\(\Rightarrow y=\dfrac{4}{25}\)

Vậy...

Bài 2:

a, \(x\left(x-\dfrac{4}{7}\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}x>0\\x-\dfrac{4}{7}>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x< 0\\x-\dfrac{4}{7}< 0\end{matrix}\right.\)

\(\Rightarrow x>\dfrac{4}{7}\left(x\ne0\right)\) hoặc \(x< \dfrac{4}{7}\left(x\ne0\right)\)

Vậy...

Các phần còn lại tương tự nhé

a: \(\left|x\right|=3+\dfrac{1}{5}=\dfrac{16}{5}\)

mà x<0

nên x=-16/5

b: \(\left|x\right|=-2.1\)

nên \(x\in\varnothing\)

c: \(\left|x-3.5\right|=5\)

=>x-3,5=5 hoặc x-3,5=-5

=>x=8,5 hoặc x=-1,5

d: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

=>|x+3/4|=1/2

=>x+3/4=1/2 hoặc x+3/4=-1/2

=>x=-1/4 hoặc x=-5/4

Câu 1: 

b: \(\Leftrightarrow\left|x-1\right|=-3x+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{1}{3}\\\left(-3x+1-x+1\right)\left(-3x+1+x-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{1}{3}\\\left(-4x+2\right)\cdot\left(-2x\right)=0\end{matrix}\right.\Leftrightarrow x=0\)

c: \(\Leftrightarrow\left[{}\begin{matrix}2x-1=2x+3\\2x+3=1-2x\end{matrix}\right.\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\)

d: \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

e: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\\left[x\left(x^2-\dfrac{5}{4}\right)-x\right]\left[x\left(x^2-\dfrac{5}{4}\right)+x\right]=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x\left(x^2-\dfrac{9}{4}\right)\cdot x\cdot\left(x^2-\dfrac{1}{4}\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;\dfrac{1}{2};\dfrac{3}{2}\right\}\)

a.

| x | = 5,6

=>\(\left[{}\begin{matrix}x=5,6\\x=-5,6\end{matrix}\right.\)

Vậy \(x\in\left\{-5,6;5,6\right\}\)

b, \(\left|x-3,5\right|=5\)

=>\(\left[{}\begin{matrix}x-3,5=5\\x-3,5=-5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8,5\\x=-1,5\end{matrix}\right.\)

Vậy \(x\in\left\{-1,5;8,5\right\}\)

c,\(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

=> \(\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)

=>\(\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{1}{4};\dfrac{5}{4}\right\}\)

d,\(\left|4x\right|-\left(\left|-13,5\right|\right)=\left|\dfrac{1}{4}\right|\)

=> \(\left|4x\right|-13,5=\dfrac{1}{4}\)

=> \(\left|4x\right|=13,75\)

=>\(\left[{}\begin{matrix}4x=13,75\\4x=-13,75\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=3,4375\\x=-3,4375\end{matrix}\right.\)

Vậy \(x\in\left\{-3,4375;3,4375\right\}\)

e, ( x - 1 ) ³ = 27

=> x - 1 = 3

=> x = 4

Vậy x = 4

f, ( 2x - 3)² = 36

=> \(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=4,5\\x=-1,5\end{matrix}\right.\)

Vậy x\(\in\left\{-1,5;4,5\right\}\)

g, \(5^{x+2}=625\)

=> \(5^{x+2}=5^4\)

=> x + 2 = 4

=> x = 2

Vậy x = 2

h, ( 2x - 1)³ = -8

=> 2x - 1 = -2

=> x = \(\dfrac{-1}{2}\)

Vậy x = \(\dfrac{-1}{2}\)

i, \(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}...\dfrac{30}{62}.\dfrac{31}{64}=2^x\)

=> \(\dfrac{1.2.3.4.5...30.31}{4.6.8.10.12...62.64}=2^x\)

=>\(\dfrac{1.2.3.4.5...30.31}{\left(2.3.4.5...30.31.32\right)\left(2.2.2.2...2.2_{ }\right)}=2^x\)(có 31 số 2)

=> \(\dfrac{1}{32.2^{31}}=2^x\)

=> \(\dfrac{1}{2^{36}}=2^x\)

=> x = -36

Vậy x = -36

1.

a) \(-\dfrac{4}{9}+\left(-\dfrac{5}{6}\right)-\dfrac{17}{4}=-\dfrac{16}{36}-\dfrac{30}{36}-\dfrac{153}{36}\)

\(=-\dfrac{199}{36}\)

b) \(5\dfrac{1}{2}+\left(-3\right)=5\dfrac{1}{2}-3=\dfrac{11}{2}-\dfrac{6}{2}=\dfrac{5}{2}\)

c) \(4\dfrac{9}{11}+\left(-2\dfrac{1}{11}\right)=\dfrac{53}{11}-\dfrac{23}{11}=\dfrac{30}{11}\)

2.

a) \(4,3-\left(1,2\right)=3,1\)

b) \(0-\left(-0,4\right)=0+0,4=0,4\)

c) \(-\dfrac{2}{3}-\dfrac{1}{3}=-\dfrac{3}{3}=-1\)

d) \(-\dfrac{1}{2}-\dfrac{-1}{6}=-\dfrac{1}{2}+\dfrac{1}{6}=-\dfrac{3}{6}+\dfrac{1}{6}=-\dfrac{2}{6}=-\dfrac{1}{3}\)

3a) A=\(\dfrac{5}{x+xy+xyz}+\dfrac{5}{y+yz+1}+\dfrac{5xyz}{z+xz+xyz}\)

=\(\dfrac{5}{x\left(1+y+yz\right)}+\dfrac{5}{y+yz+1}+\dfrac{5xy}{1+x+xy}\)

=\(\dfrac{5}{x\left(1+y+zy\right)}+\dfrac{5x}{x\left(1+zy+y\right)}+\dfrac{5xy}{x\left(1+y+zy\right)}\)

=\(\dfrac{5+5x+5xy}{x\left(1+yz+y\right)}\)

=\(\dfrac{5x\left(yz+1+y\right)}{x\left(1+yz+y\right)}=5\)

Thank you!!!!!

a) |x-3,5| = 7,5

TH1:  => x - 3,5 = 7,5

=> x = 7,5 + 3,5 = 11

TH2 : x - 3,5 = -7,5

=> x = -7,5 + 3.5 = -4

b) 3,6 - | x - 0,4| = 0 

=> | x - 0,4| = 3,6 - 0 = 3,6

Th1: x - 0,4 = 3,6

=> x = 0,4 + 3,6 = 4

th2: x - 0,4 = -3,6

=> x = 0,4 + (-3,6) = -3,2

c) |x - 3,5| + |4,5 - x | = 0

= a + a = 0 ( loại bỏ vì nếu vậy thì phép tính trên sẽ ko hợp lí)

= -a + a = 0 

Ta có: x - 3,5 = -a

         4,5 - x = a 

=> 3,5 + -a = 4,5 - a = 4,5 + (-a)

Vậy , không có số x nào thỏa mãn đk trên (theo mk là thế!)

Tíc nhá!