Bài 3: 

a: =>x=13-24=-11

b: =>3x=12

hay x=4

\(-52+\frac{2}{3}x=-46\)

\(\frac{2}{3}x=-46+52\)

\(\frac{2}{3}x=6\)

\(x=6:\frac{2}{3}\)

\(x=9\)

x=9 nha bn

đúng nha

Happy new year!!

\(-52+\frac{2}{3}x=-46\)

\(\Rightarrow\frac{2}{3}x=-46-\left(-52\right)\)

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

\(\Rightarrow x=9\)

-52 + \(\frac{2}{3}\)x X = -46

\(\frac{2}{3}\)X = \(-46+52\)

\(\frac{2}{3}x=6\)

\(x=6:\frac{2}{3}=9\)

2 / 3 x X = -46 -- 52

2/3 xX = 6

X = 6 : 2/3

X = 9

2/3.x = -46 - ( - 52)

2/3.x = 6

x=  6 : 2/3

x = 9

2/3.x = -46 - ( - 52)

2/3.x = 6

x=  6 : 2/3

x = 9

-52+ 2/3 x= -46

=>\(\frac{2x}{3}=6\)

=>2x=3*6

=>2x=18

=>x=9

Phương trình đầu bài tương đương với 
\(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)\(\Leftrightarrow\frac{x+43+57}{57}+\frac{x+46+54}{54}=\frac{x+49+51}{51}+\frac{x+52+48}{48}\)\(\Leftrightarrow\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)

\(\Leftrightarrow\orbr{\begin{cases}x+100=0\\\frac{1}{57}+\frac{1}{54}=\frac{1}{51}+\frac{1}{48}\left(sai\right)\end{cases}\Leftrightarrow x+100=0\Leftrightarrow x=-100}\)

Vậy phương trình có nghiệm duy nhất là x=-100

<=> \(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)

<=> \(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)

<=> \(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)

vi \(\frac{1}{57}< \frac{1}{51};\frac{1}{54}< \frac{1}{48}\Rightarrow\frac{1}{57}-\frac{1}{51}+\frac{1}{54}-\frac{1}{48}< 0\)

=> x+100=0 => x= -100

vay pt co nghiem \(x=-100\)

Ta thấy:\(\frac{x+43}{57}\)\(+\)\(\frac{x+46}{54}\)\(+\)\(2\)\(=\)\(\frac{x+49}{51}\)\(+\)\(\frac{x+52}{48}\)\(+\)\(2\)

\(\Rightarrow\)\(\frac{x+43}{57}\)\(+\)\(\frac{57}{57}\)\(+\)\(\frac{x+46}{54}\)\(+\)\(\frac{54}{54}\)\(=\)\(\frac{x+49}{51}\)\(+\)\(\frac{51}{51}\)\(+\)\(\frac{x+48}{52}\)\(+\)\(\frac{52}{52}\)

\(\Leftrightarrow\)\(\frac{x+100}{57}\)\(+\)\(\frac{x+100}{54}\)\(=\)\(\frac{x+100}{51}\)\(+\)\(\frac{x+100}{52}\)

\(\Leftrightarrow\)\((\)\(x+100)\)\((\frac{1}{57}\)\(+\)\(\frac{1}{54}\)\()\)\(=\)\((x+100)\)\((\frac{1}{52}\)\(+\)\(\frac{1}{51})\)

\(\Leftrightarrow\)\((x+100)\)\((\frac{1}{57}\)\(+\)\(\frac{1}{54}\)\(-\)\(\frac{1}{52}\)\(-\)\(\frac{1}{51}\)\()\)\(=\)\(0\)\((1)\)

Ta thấy: \(\frac{1}{57}\)< \(\frac{1}{52}\)

          \(\frac{1}{54}\)<\(\frac{1}{51}\)

\(\Rightarrow\)\(\frac{1}{57}\)\(+\)\(\frac{1}{54}\)< \(\frac{1}{52}\)\(+\)\(\frac{1}{51}\)

\(\Rightarrow\)\(\frac{1}{57}\)\(+\)\(\frac{1}{54}\)\(-\)\(\frac{1}{52}\)\(-\)\(\frac{1}{51}\)< 0 \((2)\)

Từ \((1)\)và \(\left(2\right)\)\(\Rightarrow\)\(x+100\)\(=0\)

\(\Leftrightarrow x=-100\)

Vậy phương trình có tập nghiệm \(x=-100\)

\(D=\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+\frac{1}{336}+\frac{1}{546}\)

\(D=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+\frac{1}{21.26}\)

\(D=\frac{1}{5}\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}\right)\)

\(D=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{21}-\frac{1}{26}\right)\)

\(D=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{26}\right)\)

\(D=\frac{1}{5}.\frac{25}{26}=\frac{5}{26}\)

\(x+\frac{5}{3}=\frac{1}{81}\)

\(x=\frac{1}{81}-\frac{5}{3}\)

\(x=\frac{-134}{81}\)

\(x+\frac{5}{3}=\frac{1}{81}\)

\(\Leftrightarrow x=\frac{1}{81}-\frac{5}{3}\)

\(\Leftrightarrow x=-\frac{134}{81}\)

\(-52+\frac{2}{3}x=-46\)

\(\Leftrightarrow\frac{2}{3}x=-46+52\)

\(\Leftrightarrow\frac{2}{3}x=6\)

<=> 2x = 6.3

<=> 2x = 18

=> x = 9

\(\frac{x}{45}=\frac{5}{6}+\frac{-29}{30}\)

(có thể hơi khó hiểu từ bước 2, bạn lấy mẫu số chung là 30 nhé)

\(\Leftrightarrow\frac{x}{45}=\frac{25}{30}-\frac{29}{30}\)

\(\Leftrightarrow\frac{x}{45}=\frac{25-29}{30}\)

\(\Leftrightarrow\frac{x}{45}=\frac{-4}{30}\)

\(\Leftrightarrow\frac{x}{45}=\frac{-2}{15}\)

\(\Leftrightarrow x=45\left(-\frac{2}{15}\right)\)

=> x = -6

a)\(\frac{x}{17}=\frac{60}{204}=\frac{5}{17}\Rightarrow x=5\)

b)\(\frac{6+x}{33}=\frac{7}{11}\Rightarrow11\left(6+x\right)=7.33\Rightarrow11.6+11x=231\Rightarrow66+11x=231\)

\(\Rightarrow11x=231-66\Rightarrow11x=165\Rightarrow x=\frac{165}{11}=15\)

c)\(\frac{12+x}{43-x}=\frac{2}{3}\Rightarrow2\left(43-x\right)=3\left(12+x\right)\Rightarrow2.43-2x=3.12+3x\)

\(86-2x=36+3x\Rightarrow86-36=3x+2x\Rightarrow50=5x\Rightarrow x=\frac{50}{5}=10\)

 \(\frac{15}{26}+\frac{x}{16}=\frac{46}{52}\)

\(\frac{x}{16}=\frac{46}{52}-\frac{15}{26}\)

\(\frac{x}{16}=\frac{4}{13}\)

\(< =>\frac{x}{208}=\frac{64}{208}=\frac{4}{13}\)

                    \(x=4\)

\(\frac{15}{26}+\frac{x}{16}=\frac{46}{52}\)

\(\Rightarrow\frac{15}{26}+\frac{x}{16}=\frac{23}{26}\)

\(\Rightarrow\frac{x}{16}=\frac{23}{26}-\frac{15}{26}\)

\(\Rightarrow\frac{x}{16}=\frac{4}{13}\)

\(\Rightarrow13x=4.16\)

\(\Rightarrow13x=64\)

\(\Rightarrow x=64:13\)

\(\Rightarrow x=\frac{64}{13}\)

Vậy x = \(\frac{64}{13}\)

_Chúc bạn học tốt_

Thanh Hiền lm đúng r đấy^^

<=>\(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)

<=>\(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)

<=>\(\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)

<=>\(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)

Vì \(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\ne0\)

=>x+100=0

<=>x=-100

k nha bạn

\(\Leftrightarrow\frac{37x+1648}{1026}=\frac{11x+556}{272}\Rightarrow\left(37x+1648\right)272=1026\left(11x+556\right)\)

<=>(37x+1648)272=272(37x+1648)

=>272(37x+1648)=1026(11x+556)

=>10064x+448256=11286x+570456

<=>-1222x=122200

=>x=122200:-1222

=>x=-100 ( dễ hiểu chưa hả )

\(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)

\(\Leftrightarrow\)\(\frac{x+43}{57}+1+\frac{x+46}{54}=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)

\(\Leftrightarrow\)\(\frac{x+43}{57}+\frac{57}{57}+\frac{x+46}{54}+\frac{54}{54}=\frac{x+49}{51}+\frac{51}{51}+\frac{x+52}{48}+\frac{48}{48}\)

\(\Leftrightarrow\)\(\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)

Vì \(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\ne0\)nên \(x+100=0\Leftrightarrow x=-100\)

Vậy \(x=-100\)