\(a,\Leftrightarrow5\left(x-2\right)-15x\le9+10\left(x+1\right)\)

\(\Leftrightarrow5x-10-15x\le9+10x+10\)

\(\Leftrightarrow-20x\le29\)

\(\Leftrightarrow x\ge-1,45\)

Vậy ...........

\(b,\Rightarrow\left(x+2\right)-3\left(x-3\right)=5\left(x-2\right)\)

\(\Leftrightarrow x+2-3x+9-5x+10=0\)

\(\Leftrightarrow-7x+21=0\)

\(\Leftrightarrow x=3\)

Vậy ..............

 \(\frac{x-2}{6}-\frac{x}{2}\le\frac{3}{10}+\frac{x+1}{3}\Leftrightarrow\frac{5\left(x-2\right)}{30}-\frac{15x}{30}\le\frac{9}{30}+\frac{10\left(x+1\right)}{30}\)

\(\Leftrightarrow5x-10-15x-9-10x-10\le0\) 

 \(\Leftrightarrow-20x-29\le0\Leftrightarrow\left(-20x\right)\cdot\frac{-1}{20}\ge29\cdot-\frac{1}{20}\)

 \(\Leftrightarrow x\ge-\frac{29}{20}\)

ĐKXĐ : \(\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)

\(\frac{x+2}{x^2-5x+6}-\frac{3}{x-2}=\frac{5}{x-3}\) 

 \(\Rightarrow\frac{x+2}{x-2x-3x+6}-\frac{3}{x-2}=\frac{5}{x-3}\)

 \(\Rightarrow\frac{x+2}{\left(x-2\right)\left(x-3\right)}-\frac{3}{x-2}=\frac{5}{x-3}\)

 \(\Rightarrow\frac{x+2}{\left(x-2\right)\left(x-3\right)}-\frac{3\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}=\frac{5\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

 \(\Rightarrow x+2-3x+9-5x+10=0\)

\(\Leftrightarrow-7x+21=0\Leftrightarrow x=3\) (nhân)

 tập nghiệm của phương trình là S= 3

Theo đề bài ta có: \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}-\frac{x-4}{5}-\frac{x-5}{6}>0\)

=> \(\frac{x-1}{2}+1+\frac{x-2}{3}+1+\frac{x-3}{4}+1-\left(\frac{x-4}{5}+1\right)-\left(\frac{x-5}{6}+1\right)>1\)

<=> \(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}>1\)

<=>\(\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)>1\)

<=> \(\left(x+1\right)\cdot\frac{43}{60}>1\)

<=>\(x+1>\frac{60}{43}\)

<=> x>\(\frac{17}{43}\)

Vậy x>17/43

Có phải đề bài là ......... + \(\frac{7}{x^2+5}\)ko bạn???

Ta có: ĐKXĐ : x thuộc R.

\(\frac{4x^2+16}{x^2+6}=\frac{3}{x^2+1}+\frac{5}{x^2+3}+\frac{7}{x^2+5}\)

<=> \(\frac{4x^2+16}{x^2+6}-3=\left(\frac{3}{x^2+1}-1\right)+\left(\frac{5}{x^2+3}-1\right)+\left(\frac{7}{x^2+5}-1\right)\)

<=> \(\frac{x^2-2}{x^2+6}=\frac{2-x^2}{x^2+1}+\frac{2-x^2}{x^2+3}+\frac{2-x^2}{x^2+5}\)

<=> \(\frac{x^2-2}{x^2+6}-\frac{2-x^2}{x^2+1}-\frac{2-x^2}{x^2+3}-\frac{2-x^2}{x^2+5}=0\)

<=> ( x² - 2 ) \(\left(\frac{1}{x^2+6}+\frac{1}{x^2+1}+\frac{1}{x^2+3}+\frac{1}{x^2+5}\right)\)= 0           ( vì nhân tử chung là x² - 2 nên 3 hạng tử sau đổi dấu )

<=> x² - 2 = 0.      ( vì biểu thức trong ngoặc > 0 với mọi x thuộc R )

<=> \(x=\sqrt{2}\)hoặc \(x=-\sqrt{2}\)

Vậy ..........

\(\frac{x}{3} + \frac{{2x + 1}}{6} = \frac{{4\left( {x - 2} \right)}}{5}\)

\(\frac{{10x}}{{3.10}} + \frac{{\left( {2x + 1} \right).5}}{{6.5}} = \frac{{6.4\left( {x - 2} \right)}}{{5.6}}\)

\(\frac{{10x}}{{30}} + \frac{{10x + 5}}{{30}} = \frac{{24x - 48}}{{30}}\)

\(10x + 10x + 5 = 24x - 48\)

\(10x + 10x - 24x =  - 5 - 48\)

\( - 4x =  - 53\)

\(x = \left( { - 53} \right):\left( { - 4} \right)\)

\(x = \frac{{53}}{4}\)

Vậy phương trình có nghiệm là \(x = \frac{{53}}{4}\).  

1)

a) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}< =>\frac{2\left(x+5\right)}{2\left(3x-6\right)}-\frac{3x-6}{2\left(3x-6\right)}=\frac{3\left(2x-3\right)}{3\left(2x-4\right)}.\)

(đk:x khác \(\frac{1}{2}\))

\(\frac{2x+10}{6x-12}-\frac{3x-6}{6x-12}=\frac{6x-9}{6x-12}< =>2x+10-3x+6=6x-9< =>x=\frac{25}{7}\)

Vậy x=\(\frac{25}{7}\)

b) /7-2x/=x-3 \(x\ge\frac{7}{2}\)

(đk \(x\ge3,\frac{7}{2}< =>x\ge\frac{7}{2}\))

\(\Rightarrow\orbr{\begin{cases}7-2x=x-3\\7-2x=-\left(x-3\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\left(< \frac{7}{2}\Rightarrow l\right)\\x=4\left(tm\right)\end{cases}}}\)

Vậy x=4

2)

\(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}>\frac{x-4}{5}+\frac{x-5}{6}\)

\(\Leftrightarrow\frac{30\left(x-1\right)}{60}+\frac{20\left(x-2\right)}{60}+\frac{15\left(x-3\right)}{60}-\frac{12\left(x-4\right)}{60}-\frac{10\left(x-5\right)}{60}>0\)

\(\Leftrightarrow30x-30+20x-40+15x-45-12x+48-10x+50>0\Leftrightarrow43x-17>0\Leftrightarrow x>\frac{17}{43}\)

\(\frac{x+2+1}{x+2}-\frac{x+3+1}{x+3}=\frac{x+4+1}{x+4}-\frac{x+5+1}{x+5}\)

=> \(1+\frac{1}{x+2}-1-\frac{1}{x+3}=1+\frac{1}{x+4}-1-\frac{1}{x+5}\)

=> \(\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{1}{\left(x+4\right)\left(x+5\right)}\)

Đến đây bạn tự giải tiếp nk

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

la`thu'hai nga`y 19 nhe

a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)

\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)

\(\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}=\frac{x-1}{5}+\frac{x-1}{6}\)

\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0

\(x-1=0\)

\(x=1\)