`a, 2/3 +3/4 = (8+9)/12=17/12.`

`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.

`=> x=2.`

`b, 5/6-1/4=(20-6)/24=7/12`.

`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.

`=> x=1`.

a) \(\dfrac{2}{3}+\dfrac{3}{4}=\dfrac{8+9}{12}=\dfrac{17}{12}\)

-> 1 1/3 + 4/5 = 4/3 + 4/5 =  20+12/15 = 32/15

vậy x có thể = 14/14 = 1 (x thuộc N)

A bn lướt xuống dưới mà xem cách làm 

nhưng của bn là cho 3 ra ngoài nha

Giải:

A=3/1.3+3/3.5+3/5.7+...+3/49.51

A=3/2.(2/1.3+2/3.5+2/5.7+...+2/49.51)

A=3/2.(1/1-1/3+1/3-1/5+1/5-1/7+...+1/49-1/51)

A=3/2.(1/1-1/51)

A=3/2.50/51

A=25/17

B=1/3+1/3²+1/3³+...+1/3⁸

3B=1+1/3+1/3²+...+1/3⁷

3B-B=(1+1/3+1/3²+...+1/3⁷)-(1/3+1/3²+1/3³+...+1/3⁸)

2B=1-1/3⁸

   B=1-1/3⁸ /2

Chúc bạn học tốt!

\(a,P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{1-x}\right)\left(dkxd:x\ge0,x\ne1\right)\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)

\(=\dfrac{\sqrt{x}.\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

\(=\dfrac{x-2}{\sqrt{x}}\)

\(b,x=4+2\sqrt{3}\Rightarrow P=\dfrac{\left(4+2\sqrt{3}\right)-2}{\sqrt{4+2\sqrt{3}}}\)

\(=\dfrac{2\sqrt{3}+4-2}{\sqrt{\sqrt{3}^2+2\sqrt{3}+1}}\)

\(=\dfrac{2\sqrt{3}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}}\)

\(=\dfrac{2\left(\sqrt{3}+1\right)}{\left|\sqrt{3}+1\right|}\)

\(=\dfrac{2\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=2\)

a: \(P=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{x-1}\)

\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x-2}{\sqrt{x}}\)

b: Khi x=4+2căn 3 thì \(P=\dfrac{2+2\sqrt{3}}{\sqrt{3}+1}=2\)

\(1\dfrac{4}{5}+2\dfrac{5}{7}+3\dfrac{4}{5}+4\dfrac{5}{7}\)

\(\text{=}\left(1\dfrac{4}{5}+3\dfrac{4}{5}\right)+\left(2\dfrac{5}{7}+4\dfrac{5}{7}\right)\)

\(\text{=}1+3+\left(\dfrac{4}{5}+\dfrac{4}{5}\right)+2+4+\left(\dfrac{5}{7}+\dfrac{5}{7}\right)\)

\(\text{=}10+\dfrac{8}{5}+\dfrac{10}{7}\text{=}131\dfrac{1}{35}\)

\(1\dfrac{1}{2}x1\dfrac{1}{3}x1\dfrac{1}{4}x1\dfrac{1}{5}x1\dfrac{1}{6}x1\dfrac{1}{7}x1\dfrac{1}{8}x1\dfrac{1}{9}\)

\(=\dfrac{3}{2}x\dfrac{4}{3}x\dfrac{5}{4}x\dfrac{6}{5}x\dfrac{7}{6}x\dfrac{8}{7}x\dfrac{9}{8}x\dfrac{10}{9}\)

\(=x^7.\dfrac{3.4.5.6.7.8.9.10}{2.3.4.5.6.7.8.9}\)

\(=x^7.\dfrac{10}{2}\)

\(=5x^7\)

\(=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times...\times\dfrac{9}{8}\times\dfrac{10}{9}=\dfrac{10}{2}=5\)

\(\left(-\dfrac{2}{5}\right)^2\cdot\left|\dfrac{1}{3}-\dfrac{3}{5}\right|-\dfrac{2}{5}\cdot\sqrt{\dfrac{1}{25}}+\dfrac{4}{3}\)

\(=\dfrac{4}{25}\cdot\dfrac{4}{15}-\dfrac{2}{5}\cdot\dfrac{1}{5}+\dfrac{4}{3}\)

\(=\dfrac{16}{375}-\dfrac{2}{25}+\dfrac{4}{3}\)

\(=\dfrac{16}{375}-\dfrac{30}{375}+\dfrac{500}{375}\)

\(=\dfrac{486}{375}=\dfrac{162}{125}\)

\(\dfrac{3}{x+2}\)=\(\dfrac{5}{2x+1}\)(x khác -1/2 và x khác -2)

=>6x+3=5x+10

<=>x=7 tm 

\(\dfrac{5}{x+2}-\dfrac{x-1}{x-2}=\dfrac{12}{x^2-4}+1\left(x\ne-2;x\ne2\right)\)

\(< =>\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

suy ra

`5x-10-(x^2 +2x-x-2)=12+x^2 -4`

`<=>5x-10-x^2 -2x+x+2-12-x^2 +4=0`

`<=>-x^2 -x^2 +5x-2x+x-10+2+4=0`

`<=>-x^2 +4x-4=0`

`<=>x^2 -4x+4=0`

`<=>(x-2)^2 =0`

`<=>x-2=0`

`<=>x=2(ktmđk)`

vậy phương trình vô nghiệm

ĐKXĐ: \(x\ne\pm2\)

\(\dfrac{5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow5\left(x-2\right)-\left(x-1\right)\left(x+2\right)=12+\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow5x-10-\left(x^2+x-2\right)=12+x^2-4\)

\(\Leftrightarrow-x^2+4x-8=x^2+8\)

\(\Leftrightarrow2x^2-4x+16=0\)

\(\Leftrightarrow2\left(x-1\right)^2+14=0\)

Do \(\left\{{}\begin{matrix}2\left(x-1\right)^2\ge0\\14>0\end{matrix}\right.\) ;\(\forall x\)

\(\Rightarrow2\left(x-1\right)^2+14>0\)

Vậy phương trình đã cho vô nghiệm

\(3\dfrac{1}{5}\times\dfrac{10}{11}+1\dfrac{2}{11}\\ =\dfrac{16}{5}\times\dfrac{10}{11}+\dfrac{13}{11}\\ =\dfrac{16\times10}{5\times11}+\dfrac{13}{11}\\ =\dfrac{160}{55}+\dfrac{13}{11}\\ =\dfrac{32}{11}+\dfrac{13}{11}\\ =\dfrac{32+13}{11}\\ =\dfrac{45}{11}\)

\(5\dfrac{1}{3}:1\dfrac{2}{3}-1\dfrac{1}{5}\\ =\dfrac{16}{3}:\dfrac{5}{3}-\dfrac{6}{5}\\ =\dfrac{16}{3}\times\dfrac{3}{5}-\dfrac{6}{5}\\ =\dfrac{16\times3}{3\times5}-\dfrac{6}{5}\\ =\dfrac{48}{15}-\dfrac{6}{5}\\ =\dfrac{16}{5}-\dfrac{6}{5}\\ =\dfrac{16-6}{5}\\ =\dfrac{10}{5}\\ =2\)

\(3\dfrac{1}{5}\times\dfrac{10}{11}+1\dfrac{2}{11}\\ =\dfrac{3\times5+1}{5}\times\dfrac{10}{11}+\dfrac{1\times11+2}{11}\\ =\dfrac{16}{5}\times\dfrac{10}{11}+\dfrac{13}{11}\\ =\dfrac{160}{55}+\dfrac{13}{11}\\ =\dfrac{32}{11}+\dfrac{13}{11}\\ =\dfrac{45}{11}\)

\(5\dfrac{1}{3}:1\dfrac{2}{3}-1\dfrac{1}{5}\\ =\dfrac{5\times3+1}{3}:\dfrac{1\times3+2}{3}-\dfrac{1\times5+1}{5}\\ =\dfrac{16}{3}:\dfrac{5}{3}-\dfrac{6}{5}\\ =\dfrac{16}{3}\times\dfrac{3}{5}-\dfrac{6}{5}\\ =\dfrac{16}{5}-\dfrac{6}{5}\\ =\dfrac{10}{5}=2\)