\(\dfrac{2+x}{5}-0,5x=\dfrac{1-2x}{4}+0,25\)

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

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

\(\Leftrightarrow\dfrac{2+x}{5}-\dfrac{x}{2}=\dfrac{2-2x}{4}\)

\(\Leftrightarrow\dfrac{2+x}{5}=\dfrac{1-x}{2}+\dfrac{x}{2}\)

\(\Leftrightarrow\dfrac{2+x}{5}=\dfrac{1-x+x}{2}\)

\(\Leftrightarrow\dfrac{2+x}{5}=\dfrac{1}{2}\)

\(\Leftrightarrow2\left(2+x\right)=5\\ \Leftrightarrow2x+4-5=0\\ \Leftrightarrow2x-1=0\\ \Leftrightarrow x=\dfrac{1}{2}\)

\(PT.\Rightarrow\) \(\dfrac{8+4x-10x-5+10x-5}{20}=0.\Rightarrow4x=2.\Leftrightarrow x=\dfrac{1}{2}.\)

\(\dfrac{2+x}{5}-0,5x=\dfrac{1-2x}{4}+0,25\\ \Leftrightarrow\dfrac{4.\left(2+x\right)}{20}-\dfrac{20.0,5x}{20}=\dfrac{5.\left(1-2x\right)}{20}+\dfrac{0,25.20}{20}\\ \Leftrightarrow8+4x-10x=5-10x+5\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ \Rightarrow S=\left\{\dfrac{1}{2}\right\}\)

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Lời giải:

PT $\Leftrightarrow 0,4+0,2x-0,5x=0,25-0,5x+0,25$

$\Leftrightarrow 0,4-0,3x=0,5-0,5x$

$\Leftrightarrow 0,2x=0,1\Rightarrow x=0,5$

Bạn quy đồng là ra thôi

\(5+\frac{x+4}{5}< x-\frac{x-2}{2}+\frac{x+3}{3}\)

\(\Rightarrow5.30+6\left(x+4\right)< 30x-15\left(x-2\right)+10\left(x+3\right)\)

\(\Rightarrow150+6x+24< 30x-15x+30+10x+30\)

\(\Rightarrow19x>114\Rightarrow x>6\)

                                                     Vậy x > 6

\(\hept{\begin{cases}\text{|}0,5x\text{|}=0,5x\\\sqrt{\left(0,5x\right)^2}=0,5x\\\left(0,5x\right)^2=\left(0,5x\right)^2\end{cases}}\)

2, tương tự

\(\hept{\begin{cases}\text{|}-\frac{2}{3}x\text{|}=\frac{2}{3}x\\\sqrt{\left(-\frac{2}{3}x\right)^2}=\frac{2}{3}x\\\left(-\frac{2}{3}x\right)^2=\left(\frac{2}{3}x\right)^2\end{cases}}\)

4, tương tự 

a) Ta có: \(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow7-2x-4=-x-4\)

\(\Leftrightarrow-2x+3+x+4=0\)

\(\Leftrightarrow-x+7=0\)

\(\Leftrightarrow-x=-7\)

hay x=7

Vậy: S={7}

b) Ta có: \(\dfrac{2+x}{5}-0.5x=\dfrac{1-2x}{4}+0.25\)

\(\Leftrightarrow\dfrac{4\left(2+x\right)}{20}-\dfrac{0.5x\cdot20}{20}=\dfrac{5\left(1-2x\right)}{20}+\dfrac{20\cdot0.25}{20}\)

\(\Leftrightarrow4\left(2+x\right)-10x=5\left(1-2x\right)+5\)

\(\Leftrightarrow8+4x-10x=5-10x+5\)

\(\Leftrightarrow-6x+8=-10x+10\)

\(\Leftrightarrow-6x+8+10x-10=0\)

\(\Leftrightarrow4x-2=0\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

d) Ta có: \(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-59}{1}+\dfrac{x-58}{2}+\dfrac{x-57}{3}\)

\(\Leftrightarrow\dfrac{x-1}{59}-1+\dfrac{x-2}{58}-1+\dfrac{x-3}{57}-1=\dfrac{x-59}{1}-1+\dfrac{x-58}{2}-1+\dfrac{x-57}{3}-1\)

\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}=\dfrac{x-60}{1}+\dfrac{x-60}{2}+\dfrac{x-60}{3}\)

\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}\right)-\left(x-60\right)\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)

mà \(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\ne0\)

nên x-60=0

hay x=60

Vậy: S={60}

Đây là những bài cơ bản mà bạn!

\(\frac{5x-2}{3}=\frac{5-3x}{2}\)

\(< =>\frac{\left(5x-2\right).2}{6}=\frac{\left(5-3x\right).3}{6}\)

\(< =>\left(5x-2\right).2=\left(5-3x\right).3\)

\(< =>10x-4=15-9x\)

\(< =>10x+9x=15+4\)

\(< =>19x=19< =>x=1\)

\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)

\(< =>\frac{\left(10x+3\right).3}{36}=\frac{36}{36}+\frac{\left(6+8x\right).4}{36}\)

\(< =>\left(10x+3\right).3=36+\left(6+8x\right).4\)

\(< =>30x+9=36+24+32x\)

\(< =>32x-30x=9-36-24\)

\(< =>2x=9-60=-51< =>x=-\frac{51}{2}\)