$(2x+\dfrac 3 5)^2-\dfrac{24}{25}=1\\\Leftrightarrow (2x+\dfrac{3}{5})^2=\dfrac{49}{25}\\\Leftrightarrow \left[\begin{array}{1}2x+\dfrac{3}{5}=\dfrac{7}{5}\\2x+\dfrac{3}{5}=-\dfrac{7}{5}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{1}2x=\dfrac{4}{5}\\2x=-2\end{array}\right.\\\Leftrightarrow \left[\begin{array}{1}x=\dfrac{2}{5}\\x=-1\end{array}\right.$

Vậy $x=\dfrac{2}{5},x=-1$

GIải

\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{24}{25}=1\)

\(\left(2x+\dfrac{3}{5}\right)^2\)           \(=1+\dfrac{24}{25}\)

\(\left(2x+\dfrac{3}{5}\right)^2\)           \(=\dfrac{49}{25}\)

\(4x+\dfrac{9}{25}\)                 \(=\dfrac{49}{25}\)

\(4x\)                           \(=\dfrac{49}{25}-\dfrac{9}{25}\)

\(4x\)                           \(=\dfrac{8}{5}\)

 \(x\)                            \(=4:\dfrac{8}{5}\)

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

\(\left(\frac{2013}{2011}+\frac{2014}{2012}+\frac{2015}{2013}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)=0\)

thanks 

2x=3y=5z <=>\(\frac{x}{3}=\frac{y}{5}=\frac{z}{2}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\frac{x}{3}=\frac{y}{5}=\frac{z}{2}=\frac{x+y-z}{3+5-2}=\frac{95}{6}\)

Từ đó bạn có thế => x,y,z=

2x  = 3y = 5z 

\(\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)

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

\(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{15}=5\\\frac{y}{10}=5\\\frac{z}{6}=5\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=15.5=75\\y=5.10=50\\z=6.5=30\end{cases}}\)

Vậy x = 75 ; y = 50 và z = 30

@@ Học tốt@@
## Chiyuki Fujito

CẢM ƠN Ạ ^-^

\(x\cdot y=3\Rightarrow x=\dfrac{3}{y}\\ \Rightarrow\dfrac{3}{y}+y=5\\ \Rightarrow y^2-5y+1=0\\ \Leftrightarrow\left[{}\begin{matrix}y=\dfrac{5+\sqrt{21}}{2}\Rightarrow x=\dfrac{15-3\sqrt{21}}{2}\\y=\dfrac{5-\sqrt{21}}{2}\Rightarrow x=\dfrac{15+3\sqrt{21}}{2}\end{matrix}\right.\)

\(B=\left(2x-3y\right)\left(3y-2x\right)=-\left(2x-3y\right)^2\\ \Rightarrow\left[{}\begin{matrix}B\simeq-172,176\\B\simeq-790,823\end{matrix}\right.\)

\(C=x^5+y^5\\ \Rightarrow\left[{}\begin{matrix}C\simeq2525,096\\C\simeq613574,904\end{matrix}\right.\)

Em xem lại đề xem, bài này số xấu

Bằng 0 chứ nhỉ em ?

(x-5) . (2x-4)= 0

\(\left[{}\begin{matrix}x-5=0\\2x-4=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=5\\2x=4\end{matrix}\right.< =>\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

\(A=x^2-12x+7=x^2-12x+36-29\)

\(=\left(x-6\right)^2-29\ge-29\)

Vậy \(A_{min}=-29\Leftrightarrow x=6\)

\(C=x-x^2-4=-\left(x^2-x+4\right)\)

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

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

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

Vậy \(C_{min}=\frac{-3}{4}\Leftrightarrow x=\frac{1}{2}\)

a) 5x.(x+3/4) = 0

=> x = 0

x+3/4 = 0 => x = -3/4

b) \(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}.\)

\(\Rightarrow\frac{x+7}{2010}+\frac{x+6}{2011}-\frac{x+5}{2012}-\frac{x+4}{2013}=0\)

\(\frac{x+7}{2010}+1+\frac{x+6}{2011}+1-\frac{x+5}{2012}-1-\frac{x+4}{2013}-1=0\)

\(\left(\frac{x+7}{2010}+1\right)+\left(\frac{x+6}{2011}+1\right)-\left(\frac{x+5}{2012}+1\right)-\left(\frac{x+4}{2013}+1\right)=0\)

\(\frac{x+2017}{2010}+\frac{x+2017}{2011}-\frac{x+2017}{2012}-\frac{x+2017}{2013}=0\)

\(\left(x+2017\right).\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)

=> x + 2017 = 0

x = -2017

a) để 2x - 3 > 0

=> 2x > 3

x > 3/2

b) 13-5x < 0

=> 5x < 13

x < 13/5

c) \(\frac{x+3}{2x-1}>0\)

=> x + 3 > 0

x > -3

d) \(\frac{x+7}{x+3}=\frac{x+3+4}{x+3}=1+\frac{4}{x+3}\)

Để x+7/x+3 < 1

=> 1 + 4/x+3 < 1

=> 4/x+3 < 0

=> không tìm được x thỏa mãn điều kiện