⇔\(x\) x 1=\(\dfrac{8}{3}\)

⇔\(x\) =\(\dfrac{8}{3}\)

Vậy \(x\)=\(\dfrac{8}{3}\)

\(x\times3\times\dfrac{1}{3}=2\dfrac{2}{3}\\x\times3\times\dfrac{1}{3}=\dfrac{8}{3}\\ x\times3=\dfrac{8}{3}:\dfrac{1}{3}\\ x\times3=8\\ x=8:3\\ x=\dfrac{8}{3}\)

\(\Rightarrow x=\dfrac{8}{3}\)

\(\Rightarrow y\times\left(2+\dfrac{1}{5}\right)=\dfrac{8}{5}\\ \Rightarrow y\times\dfrac{11}{5}=\dfrac{8}{5}\\ \Rightarrow y=\dfrac{8}{5}:\dfrac{11}{5}=\dfrac{8}{5}\times\dfrac{5}{11}=\dfrac{8}{11}\)

a/\(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x=1+\dfrac{5}{7}+\dfrac{13}{14}\)
\(x=\dfrac{14}{14}+\dfrac{10}{14}+\dfrac{13}{14}\)
\(x=\dfrac{37}{14}\)
Vậy \(x=\dfrac{37}{14}\)
b/\(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{3}{5}+\dfrac{6}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{55}{15}-\dfrac{27}{15}\)
\(x=\dfrac{28}{15}\)
Vậy \(x=\dfrac{28}{15}\)
#kễnh

a) \(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)

\(x-\dfrac{23}{14}=1\)

\(x=1+\dfrac{23}{14}\)

\(x=\dfrac{37}{14}\)

b) \(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)

\(x+1+\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{11}{3}\)

\(x+\dfrac{9}{5}=\dfrac{11}{3}\)

\(x=\dfrac{11}{3}-\dfrac{9}{5}\)

\(x=\dfrac{28}{15}\)

`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)

giúp mik với đi ạ mik thực sự đang cần gấp

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

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

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

Đề bài yêu cầu gì?

a: Ta có: \(P=\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{\sqrt{x}-1}{\sqrt{x}-x}+\dfrac{\sqrt{x}+3}{x+5\sqrt{x}+6}\)

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

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

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

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

..

đc ko bạn

cảm ơn bạn nha