khong biet

\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)

\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)

\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)

\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)

=> -1,44444444444........... ≤ x ≤ 0,6111111111...........

Mà x ∈ Z

=> x ∈ { -1 ; 0 }

\(x\in\varnothing\) 

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

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

\(\Leftrightarrow\frac{3x-6-2-2x}{6}=-\frac{3x}{4}\)

\(\Leftrightarrow\frac{x-8}{6}=-\frac{3x}{4}\)

\(\Leftrightarrow4x-32=-18x\)

\(\Rightarrow x=\frac{16}{11}\)

\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)

\(=>\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)

\(=>\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)

\(=>\left(\frac{x+2016}{2015}+\frac{x+2016}{2014}\right)-\left(\frac{x+2016}{2013}+\frac{x+2016}{2012}\right)=0\)

\(=>\left(x+2016\right).\left[\left(\frac{1}{2015}+\frac{1}{2014}\right)-\left(\frac{1}{2013}+\frac{1}{2012}\right)\right]=0\)

\(=>\orbr{\begin{cases}x+2016=0\\\left(\frac{1}{2015}+\frac{1}{2014}\right)-\left(\frac{1}{2013}+\frac{1}{2012}\right)=0\end{cases}}\)

Do 1/2015 + 1/2014 < 1/2013 + 1/2012

=> (1/2015 + 1/2014) - (1/2013 + 1/2012) khác 0

=> x - 2016 = 0

=> x = 2016

Vậy x = 2016

Ủng hộ mk nha ^_-

\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)

\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2-\left(\frac{1}{4}\right)^2=0\)

\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}+\frac{1}{4}\right)\left(\frac{1}{x}-\frac{2}{3}-\frac{1}{4}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{x}-\frac{5}{12}\right)\left(\frac{1}{x}-\frac{11}{12}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}-\frac{5}{12}=0\\\frac{1}{x}-\frac{11}{12}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}=\frac{5}{12}\\\frac{1}{x}=\frac{11}{12}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{12}{11}\\x=\frac{12}{5}\end{cases}}\)

Vậy....

\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)

\(\Rightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)

\(\Rightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2=\left(\frac{1}{4}\right)^2\)

\(\Rightarrow\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\)

\(\Rightarrow\frac{1}{x}=\frac{11}{12}\)

\(\Rightarrow x=\frac{11}{12}\)

\(=>\left(\frac{1}{x}-\frac{2}{4}\right)^2=\frac{1}{16}\)

=> \(\orbr{\begin{cases}\frac{1}{x}-\frac{2}{4}=\frac{1}{4}\\\frac{1}{x}-\frac{2}{4}=-\frac{1}{4}\end{cases}}\)

=>\(\orbr{\begin{cases}\frac{1}{x}=\frac{3}{4}\\\frac{1}{x}=\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{4}{3}\\x=4\end{cases}}\)

học tốt

x-1/x+2=x-2/x+3 

theo tính chất DTSBN có

x-1/x+2=x-2/x+3=x-1-(x-2)/x+2-(x+3)=1/-1=-1

=>x-1=-(x+2)

=>x-1=-x-2

2x=-1

=>x=-1/2

nhân chéo lên (x-1).(x+3)=(x+2).(x-2)

<=> x²+2x-3=x²-4    <=> 2x=-1   <=>   x=-1/2

rõ ràng giúp

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{6}=\frac{z-3}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{2x-2}{4}=\frac{3y-6}{6}=\frac{z-3}{4}=\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+6-4}=\frac{2x-2+3y-6-z+3}{4+6-4}\)

\(=\frac{\left(2x+3y-z\right)+\left(-2+6+3\right)}{6}=\frac{50+\left(-5\right)}{6}=\frac{45}{6}=7,5\)

\(\frac{x-1}{2}=7,5\Rightarrow x-1=15\Rightarrow x=16\)

\(\frac{y-2}{3}=7,5\Rightarrow y-2=24,5\Rightarrow y=20,5\)

\(\frac{z-3}{4}=7,5\Rightarrow z-3=30\Rightarrow z=33\)

tìm x thì phải bt

\(\frac{2x+1}{x-3}\)bằng mấy chứ

ĐẶT A= BIỂU THỨC TRÊN:

A=2X-6+7/X-3

A=2X-6/X-3+7/X-3

A=2.(X-3)/X-3+7/X-3

A=2+7/X-3

SUY RA X-3 THUỘC Ư7

SỦY RA X-3=1 SUY RA X=4 SUY RA A=9

X-3=7 SUY RA X=10 SUY RA A=3

X-3=-(1) SUY RA X=2 SUY RA A=-(5)

X-3=-(7)  SUY RA X=-(4) SUY RA A=1

X-3