\(\left|x-3,5\right|=7,5\)

\(\Leftrightarrow x-3,5=7,5\Rightarrow x=11\)

\(\Leftrightarrow x-3,5=-7,5\Rightarrow x=-4\)

\(\left|x+\dfrac{4}{5}\right|-\dfrac{1}{2}=0\)

\(\Leftrightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow x+\dfrac{4}{5}=\dfrac{1}{2}\Rightarrow x=\dfrac{-3}{10}\)

\(\Leftrightarrow x+\dfrac{4}{5}=-\dfrac{1}{2}\Rightarrow x=\dfrac{-13}{10}\)

\(3,6-\left|x-0,4\right|=0\)

\(\Leftrightarrow\left|x-0,4\right|=3,6\)

\(\Leftrightarrow x-0,4=3,6\Rightarrow x=4\)

\(\Leftrightarrow x-0,4=-3,6\Rightarrow x=-3,2\)

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.....................

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)

\(3x^2(x-5)+12(5-x)=0\)

\(\Leftrightarrow 3x^2(x-5)-12(x-5)=0\)

\(\Leftrightarrow (x-5)(3x^2-12)=0\)

\(\Leftrightarrow 3(x-5)(x^2-4)=0\)

\(\Leftrightarrow 3(x-5)(x-2)(x+2)=0\)

\(\Rightarrow \left[\begin{matrix} x-5=0\\ x-2=0\\ x+2=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=5\\ x=2\\ x=-2\end{matrix}\right.\)

b)

\((x-2)^2-(3x-1)^2=0\)

\(\Leftrightarrow [(x-2)-(3x-1)][(x-2)+(3x-1)]=0\)

\(\Leftrightarrow (-2x-1)(4x-3)=0\Rightarrow \left[\begin{matrix} -2x-1=0\\ 4x-3=0\end{matrix}\right. \)

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

c)

\(\frac{(x+1)^2}{3}-\frac{(x-2)^2}{2}=\frac{2x+1}{2}-\frac{(x-3)^2}{6}\)

\(\Leftrightarrow \frac{2(x+1)^2}{6}-\frac{3(x-2)^2}{6}=\frac{3(2x+1)}{6}-\frac{(x-3)^2}{6}\)

\(\Leftrightarrow 2(x+1)^2-3(x-2)^2=3(2x+1)-(x-3)^2\)

\(\Leftrightarrow 2(x^2+2x+1)-3(x^2-4x+4)=6x+3-(x^2-6x+9)\)

\(\Leftrightarrow -x^2+16x-10=-x^2+12x-6\)

\(\Leftrightarrow 4x=4\Rightarrow x=1\)

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

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

+, Xét \(\left\{{}\begin{matrix}x>0\\x-\dfrac{4}{7}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\x>\dfrac{4}{7}\end{matrix}\right.\Rightarrow x>\dfrac{4}{7}\)(1)

+, Xét \(\left\{{}\begin{matrix}x< 0\\x-\dfrac{4}{7}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 0\\x< \dfrac{4}{7}\end{matrix}\right.\Rightarrow x< 0\)(2)

Vậy \(x>\dfrac{4}{7}\) và \(x< 0\)

b, \(\left(x-\dfrac{2}{5}x^2\right)< 0\)

\(\Rightarrow x.\left(1-\dfrac{2}{5}x\right)< 0\)

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

\(\Rightarrow x>2,5\)

Vậy \(x>2,5\)

c, \(\left(x-\dfrac{2}{5}\right).\left(x+\dfrac{3}{7}\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{2}{5}>0\\x+\dfrac{3}{7}>0\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x-\dfrac{2}{5}< 0\\x+\dfrac{3}{7}< 0\end{matrix}\right.\)

+, Xét \(\left\{{}\begin{matrix}x-\dfrac{2}{5}>0\\x+\dfrac{3}{7}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>\dfrac{2}{5}\\x>\dfrac{-3}{7}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{2}{5}\)

+, Xét \(\left\{{}\begin{matrix}x-\dfrac{2}{5}< 0\\x+\dfrac{3}{7}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< \dfrac{2}{5}\\x< \dfrac{-3}{7}\end{matrix}\right.\) \(\Rightarrow x< \dfrac{-3}{7}\)

Vậy \(x>\dfrac{2}{5};x< \dfrac{-3}{7}\)

Chúc bạn học tốt!!!

a: =>11(x-3)=6(x-5)

=>11x-33=6x-30

=>5x=3

=>x=3/5

b: =>(4/3-1/4x-5/12)-2x=8/5*5/3=8/3

=>-9/4x+11/12=8/3

=>-9/4x=32/12-11/12=21/12=7/4

=>x=-7/9

c: =>1/2x-1/3-2/3x-1=x

=>-1/6x-4/3=x

=>-7/6x=4/3

=>x=-4/3:7/6=-4/3*6/7=-24/21=-8/7

d: =>1-2x-3x+1=7/2

=>-5x=3/2

=>x=-3/10

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

a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3

=>x=-1/3+3/4=-4/12+9/12=5/12

b: =>x(1/2-5/6)=7/2

=>-1/3x=7/2

hay x=-21/2

c: (4-x)(3x+5)=0

=>4-x=0 hoặc 3x+5=0

=>x=4 hoặc x=-5/3

d: x/16=50/32

=>x/16=25/16

hay x=25

e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4

=>2x=-7/4+3=5/4

hay x=5/8