\(\dfrac{1}{x^2+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)

\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)

\(\Leftrightarrow\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)+\left(\dfrac{1}{x+2}-\dfrac{1}{x+4}\right)+\left(\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)+\left(\dfrac{1}{x+6}-\dfrac{1}{x+8}\right)=\dfrac{8}{105}\)

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)

\(\Leftrightarrow\dfrac{8}{x\left(x+8\right)}=\dfrac{8}{105}\)

\(\Leftrightarrow x\left(x+8\right)=105\)

\(\Leftrightarrow x^2+8x-105=0\)

\(\Leftrightarrow x^2-7x+15x-105=0\)

\(\Leftrightarrow x\left(x-7\right)+15\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)

Thử lại ta có nghiệm của phương trình trên là \(x=7\text{v}à\text{x}=15\)

\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

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

⇔\(\dfrac{2\left(x+1\right)}{8}-\dfrac{5+2x}{8}=\dfrac{4\left(3-4x\right)}{8}\) 

⇔ 2x + 2 - 5 - 2x = 12 -16x

⇔ 16x = 15 

⇔ x = 15/16

b) \(\dfrac{4-3x}{5}-\dfrac{4-x}{10}=\dfrac{x+2}{2}\)

⇔\(\dfrac{2\left(4-3x\right)}{10}-\dfrac{4-x}{10}=\dfrac{5\left(x+2\right)}{10}\)

⇔ 8 - 6x - 4 + x = 5x + 10

⇔ 10x = -6

⇔ x = -6/10

Câu 1:

x + 1/4 - 5 + 2x/8 = 3 - 4x/2

<=> 2x + 2/8 - 5 + 2x/8 = 12 - 16x/8

<=> 2x + 2 - 5 - 2x = 12 - 16x

<=> -3 = 12 - 16x <=> 15 = 16x <=> x = 15/16

Câu 2:

4 - 3x/5 - 4 - x/10 = x + 2/2

<=> 8 - 6x/10 - 4 - x/10 = 5x + 10/10

<=> 8 - 6x - 4 + x = 5x + 10

<=> 4 - 5x = 5x + 10

<=> 4 = 10x + 10 <=> 10x = -6 <=> x = -3/5

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

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 - \(\dfrac{1}{5}\))

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

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

= \(\dfrac{1}{5}\)

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

\(-\dfrac{1}{3}< \dfrac{x}{24}< \dfrac{3}{4}\)

ta tìm quy đòng mẫu

\(-\dfrac{1}{3}=-\dfrac{1x8}{3x8}=-\dfrac{8}{24}\)

\(\dfrac{3}{4}=\dfrac{3x6}{4x6}=\dfrac{18}{24}\)

=> \(-\dfrac{8}{24}< \dfrac{x}{24}< \dfrac{18}{24}\)

=> x={.. bn tự ghi kết quả nha :))

\(\Leftrightarrow\dfrac{x}{-2}=\dfrac{1}{-2}=\dfrac{-18}{y}=\dfrac{z}{-24}\)

=>x=1; y=36; z=12

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10