\(a,\left(\dfrac{31}{35}-\dfrac{4}{7}\right)\times\dfrac{8}{7}:2\\ =\left(\dfrac{31}{35}-\dfrac{4\times5}{35}\right)\times\dfrac{8}{7}:2\\ =\dfrac{11}{35}\times\dfrac{8}{7}:2\\ =\dfrac{88}{245}:2\\ =\dfrac{44}{245}\\ b,\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\\ =\left(\dfrac{2-1}{2}\right)\times\left(\dfrac{3-1}{3}\right)\times\left(\dfrac{4-1}{4}\right)\times\left(\dfrac{5-1}{5}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)

a, ( \(\dfrac{31}{35}\) - \(\dfrac{4}{7}\)) \(\times\) \(\dfrac{8}{7}\): 2

= \(\left(\dfrac{31}{35}-\dfrac{20}{35}\right)\) \(\times\) \(\dfrac{8}{7}\) : 2

= \(\dfrac{11}{35}\) \(\times\) \(\dfrac{8}{7}\) \(\times\) \(\dfrac{1}{2}\)

= \(\dfrac{44}{35}\) \(\times\) \(\dfrac{4}{7}\)

= \(\dfrac{44}{245}\)

b, ( 1 - \(\dfrac{1}{2}\)) \(\times\) ( 1 - \(\dfrac{1}{3}\)) \(\times\) ( 1 - \(\dfrac{1}{4}\)) \(\times\) ( 1 - \(\dfrac{1}{5}\))

= \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{4}\) \(\times\) \(\dfrac{4}{5}\)

= \(\dfrac{1}{5}\) \(\times\) \(\dfrac{2\times3\times4}{2\times3\times4}\)

= \(\dfrac{1}{5}\)

`(x+2)/3=(x-4)/5`

`<=> (x+2)*5=3(x-4)`

`<=> 5x+10=3x-12`

`<=> 5x-3x=-12-10`

`<=> 2x=-22`

`<=> x=-11`

cảm ơn bn nhiều nhé

\(\Rightarrow x+\dfrac{1}{6}=\dfrac{3}{4}\\ \Rightarrow x=\dfrac{7}{12}\)

X+

\(x+\dfrac{1}{6}=\dfrac{3}{4}\)

nên x=7/12

\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)

x = nhân ạ

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\)

\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\times\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{4}{4}-\dfrac{1}{4}\right)\times\left(\dfrac{5}{5}-\dfrac{1}{5}\right)\)

\(=\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\)

\(=\dfrac{1\times2\times3\times4}{2\times3\times4\times5}\)

\(=\dfrac{1}{5}\)

a,sửa đề : đk x khác -2;  2 

 \(x^2+x-2+5x-10=12+x^2-4\)

\(\Leftrightarrow6x-20=0\Leftrightarrow x=\dfrac{10}{3}\left(tm\right)\)

b, \(3x-12+5+5x=105\Leftrightarrow8x=112\Leftrightarrow x=14\)

c, \(3x^2+14x-49=-\left(x^2+2x-15\right)\)

\(\Leftrightarrow4x^2+16x-34=0\Leftrightarrow x=\dfrac{-4\pm5\sqrt{2}}{2}\)

a. ko hỉu đề lắm :v

b.\(\dfrac{x-4}{5}+\dfrac{1+x}{3}=7\)

\(\Leftrightarrow\dfrac{3\left(x-4\right)+5\left(1+x\right)}{15}=\dfrac{105}{15}\)

\(\Leftrightarrow3\left(x-4\right)+5\left(1+x\right)=105\)

\(\Leftrightarrow3x-12+5+5x-105=0\)

\(\Leftrightarrow8x-112=0\)

\(\Leftrightarrow8x=112\)

\(\Leftrightarrow x=14\)

c.\(\left(3x-7\right)\left(x+7\right)=\left(5+x\right)\left(3-x\right)\)

\(\Leftrightarrow3x^2+21x-7x-49=15-5x+3x-x^2\)

\(\Leftrightarrow4x^2+16x-64=0\)

Nghiệm xấu lắm bạn

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)\(\left|x+\dfrac{2}{3}\right|=\dfrac{5}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=\dfrac{-5}{6}\\x+\dfrac{2}{3}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)

b) \(\left(x-\dfrac{1}{3}\right)^2=\dfrac{4}{9}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{2}{3}\\x-\dfrac{1}{3}=\dfrac{-2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{3}\end{matrix}\right.\)

a) Ta có: \(\left|x+\dfrac{2}{3}\right|=\dfrac{5}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=-\dfrac{5}{6}\\x+\dfrac{2}{3}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)

b) Ta có: \(\left(x-\dfrac{1}{3}\right)^2=\dfrac{4}{9}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{2}{3}\\x-\dfrac{1}{3}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{3}\end{matrix}\right.\)

a) \(\dfrac{2}{3}+\dfrac{3}{4}< x< 1\dfrac{1}{3}+\dfrac{4}{5}\)

\(\dfrac{2\times4}{3\times4}+\dfrac{3\times3}{4\times3}< x< \dfrac{\left(1\times3+1\right)\times5}{3\times5}+\dfrac{4\times3}{5\times3}\)

\(\dfrac{8}{12}+\dfrac{9}{12}< x< \dfrac{20}{15}+\dfrac{12}{15}\\ \dfrac{17}{12}< x< \dfrac{32}{15}\)

Ước tính: \(\dfrac{17}{12}=1,4\) và \(\dfrac{32}{15}=2,1\). Vậy số tự nhiên x = 2 sẽ thõa mãn 1,4 < x < 2,1

b)

 \(\dfrac{5}{6}-\dfrac{1}{4}< x< 2\dfrac{1}{3}-\dfrac{2}{5}\\ \dfrac{5\times4}{6\times4}-\dfrac{1\times6}{4\times6}< x< \dfrac{\left(2\times3+1\right)\times5}{3\times5}-\dfrac{2\times3}{5\times3}\\ \dfrac{20}{24}-\dfrac{6}{24}< x< \dfrac{35}{15}-\dfrac{6}{15}\\ \dfrac{14}{24}< x< \dfrac{29}{15}\)

Ước tính \(\dfrac{14}{24}=0,5\) và \(\dfrac{29}{15}=1,9\)

Vậy với x là số tự nhiên x = 1 sẽ thõa mãn 0,5 < x < 1,9

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

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

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