a: \(\left|x-3\right|\ge0\forall x\)

\(\left|2y-6\right|\ge0\forall y\)

Do đó: \(\left|x-3\right|+\left|2y-6\right|\ge0\forall x,y\)

=>\(\left|x-3\right|+\left|2y-6\right|+10\ge10\forall x,y\)

\(\left(y-3\right)^2\ge0\forall y\)

=>\(\left(y-3\right)^2+3\ge3\forall y\)

=>\(\frac{30}{\left(y-3\right)^2+3}\le\frac{30}{3}=10\forall y\)

Ta có: \(\left|x-3\right|+\left|2y-6\right|+10=\frac{30}{\left(y-3\right)^2+3}\)

mà \(\left|x-3\right|+\left|2y-6\right|+10\ge10\forall x,y\)

và \(\frac{30}{\left(y-3\right)^2+3}\le\frac{30}{3}=10\forall y\)

nên dấu '=' xảy ra khi x-3=0 và y-3=0

=>x=3 và y=3

b: \(\left(2x+6\right)^{2020}\ge0\forall x\)

=>\(\left(2x+6\right)^{2020}+51\ge51\forall x\)

Ta có: \(\left|x+3\right|\ge0\forall x\)

=>\(3\left|x+3\right|\ge0\forall x\)

=>\(3\left|x+3\right|+2\ge2\forall x\)

=>\(\frac{102}{3\left|x+3\right|+2}\le\frac{102}{2}=51\forall x\)

Ta có: \(\left(2x+6\right)^{2020}+51=\frac{102}{3\left|x+3\right|+2}\)

mà \(\left(2x+6\right)^{2020}+51\ge51\forall x\)

và \(\frac{102}{3\left|x+3\right|+2}\le\frac{102}{2}=51\forall x\)

nên dấu '=' xảy ra khi x+3=0

=>x=-3

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

`(1 1/3 - 25% - 5/12) + 2x=1,6 : 3/5`

`(4/3 - 1/4 - 5/12) +2x=8/5 : 3/5`

`2/3 + 2x = 8/3`

`2x=2`

`x=1`

(\(\dfrac{4}{3}-\dfrac{1}{4}-\dfrac{5}{12}\))+2x=\(\dfrac{8}{5}:\dfrac{3}{5}\)

(\(\dfrac{13}{12}\)\(-\dfrac{5}{12}\))+2x=\(\dfrac{8}{3}\)

\(\dfrac{2}{3}\)+2x=\(\dfrac{8}{3}\)

2x=\(\dfrac{8}{3}\):\(\dfrac{2}{3}\)

2x=4

⇒x=2

\(\Leftrightarrow2x-\dfrac{1}{3}=\left(\dfrac{12}{30}-\dfrac{4}{15}\right):\dfrac{3}{5}=\dfrac{2}{9}\)

=>2x=5/9

hay x=5/18

x= 5/18

a) đk: x khác 1; \(\dfrac{3}{2}\)

 \(P=\left[\dfrac{2x}{\left(2x-3\right)\left(x-1\right)}-\dfrac{5}{2x-3}\right]:\left(\dfrac{3-3x+2}{1-x}\right)\)

= \(\dfrac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}:\dfrac{5-3x}{1-x}\)

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

b) Có \(\left|3x-2\right|+1=5\)

<=> \(\left|3x-2\right|=4\)

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

TH1: Thay x = 2 vào P, ta có:

P = \(\dfrac{-1}{2.2-3}=-1\)

TH2: Thay x = \(\dfrac{-2}{3}\)vào P, ta có:

P = \(\dfrac{-1}{2.\dfrac{-2}{3}-3}=\dfrac{3}{13}\)

c) Để P > 0

<=> \(\dfrac{-1}{2x-3}>0\)

<=> 2x - 3 <0

<=> x < \(\dfrac{3}{2}\) ( x khác 1)

d) P = \(\dfrac{1}{6-x^2}\)

<=> \(\dfrac{-1}{2x-3}=\dfrac{1}{6-x^2}\)

<=> \(\dfrac{-1}{2x-3}=\dfrac{-1}{x^2-6}\)

<=> 2x - 3 = x² - 6

<=> x² - 2x - 3 = 0

<=> (x-3)(x+1) = 0

<=> \(\left[{}\begin{matrix}x=-1\left(Tm\right)\\x=3\left(Tm\right)\end{matrix}\right.\)

Học thầy chẳng học được 

\(\Leftrightarrow2\left(2x-3\right)-9=5-3x-2\)

=>4x-6-9=-3x+3

=>7x=18

hay x=18/7

\(\Leftrightarrow\dfrac{2x-3}{3}-\dfrac{3}{2}-\dfrac{5-3x}{6}+\dfrac{1}{3}=0\)

\(\Leftrightarrow\dfrac{2\left(2x-3\right)-9-5+3x+2}{6}=0\)

\(\Leftrightarrow4x-6-9-5+3x+2=0\)

\(\Leftrightarrow7x-18=0\)

\(\Leftrightarrow7x=18\)

\(\Leftrightarrow x=\dfrac{18}{7}\)

a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

Ta có: \(A=\frac{2}{1\cdot5}+\frac{3}{5\cdot11}+\frac{4}{11\cdot19}+\frac{5}{19\cdot29}+\frac{6}{29\cdot41}\)

\(=\frac12\left(\frac{4}{1\cdot5}+\frac{6}{5\cdot11}+\frac{8}{11\cdot19}+\frac{10}{19\cdot29}+\frac{12}{29\cdot41}\right)\)

\(=\frac12\left(1-\frac15+\frac15-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{41}\right)\)

\(=\frac12\left(1-\frac{1}{41}\right)=\frac12\cdot\frac{40}{41}=\frac{20}{41}\)

Ta có: \(B=\frac{40}{31\cdot39}+\frac{35}{39\cdot46}+\frac{30}{46\cdot52}+\frac{25}{52\cdot57}+\frac{20}{57\cdot61}\)

\(=5\left(\frac{8}{31\cdot39}+\frac{7}{39\cdot46}+\frac{6}{46\cdot52}+\frac{5}{52\cdot57}+\frac{4}{57\cdot61}\right)\)

\(=5\left(\frac{1}{31}-\frac{1}{39}+\frac{1}{39}-\frac{1}{46}+\frac{1}{46}-\frac{1}{52}+\frac{1}{52}-\frac{1}{57}+\frac{1}{57}-\frac{1}{61}\right)\)

\(=5\left(\frac{1}{31}-\frac{1}{61}\right)=5\cdot\frac{30}{61\cdot31}=\frac{150}{1891}\)

\(\frac{A}{B}=\frac{20}{41}:\frac{150}{1891}=\frac{20}{41}\cdot\frac{1891}{150}=\frac{2}{15}\cdot\frac{1891}{41}=\frac{3782}{615}\)