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 \(x=-100\)

x/8=1/y<=>xy=1.8=8(nhân chéo)

=>x,y E Ư(8)

do đó

+)x=1=>y=8

+)x=8=>y=1

+)x=2=>y=4

+)x=4=>y=2

+)x=-1=>y=-8

+)x=-8=>y=-1

+)x=-2=>y=-4

+)x=-4=>y=-2

vậy ....

đề có bấy nhiu thui hả

tìm x,y biết\(\frac{x}{8}=\frac{1}{y}\)

\(\frac{9}{7}x\frac{14}{3}+5\)

=\(\frac{9}{7}x\frac{14}{3}=6+5\)

=\(6+5=11\)

k mình nha

lớp 1 ý

\(\frac{9}{7}\) \(\times\) \(\frac{14}{3}\) +     5

= \(\frac{9\times14}{7\times3}\) +      5

= 6     +      5 

= 11

Giải :

3x + 1 = 2x + 10

=> 3x - 2x = 10 - 1

=>     x     = 9

Bài này ta áp dụng quy tắc chuyển vế đổi dấu ở lớp 6 nha bạn !!

3x + 1 = 2x + 10 <=> 2x +x + 1 = 2x + 10

                               <=>       x + 1    =  10

                               <=>         x          =  10 - 1 = 9

Vậy x thỏa mãn là : x = 9

Gần chục năm rồi đấy

b) \(\frac{4}{9}x-\frac{1}{2}=\frac{-5}{9}\)

\(\Rightarrow\frac{4}{9}x=\frac{-5}{9}+\frac{1}{2}\)

\(\Rightarrow\frac{4}{9}x=\frac{-1}{18}\)

\(\Rightarrow x=\frac{-1}{18}:\frac{4}{9}\)

\(\Rightarrow x=\frac{-1}{8}\)

Ta có : \(\frac{2}{3}+\frac{-1}{5}=\frac{10+(-3)}{15}=\frac{7}{15}\)

Quy đồng : \(\frac{3}{5}=\frac{3\cdot3}{5\cdot3}=\frac{9}{15}\)

Mà \(\frac{9}{15}>\frac{7}{15}\)

\(\Rightarrow\frac{3}{5}>\frac{2}{3}+\frac{-1}{5}\)

P/S : Rõ ràng đây

Ta có:

\(\frac{2}{3}+\frac{-1}{5}=\frac{7}{15}\)

Vì \(\frac{3}{5}>\frac{7}{15}\)nên \(\frac{3}{5}>\frac{2}{3}+\frac{-1}{5}\)

Ta có:

\(\frac{2}{3}+\frac{-1}{5}=\frac{10}{15}+\frac{-3}{15}\)

\(=\frac{7}{15}\)

\(\frac{3}{5}=\frac{9}{15}\)

Mà \(\frac{7}{15}< \frac{9}{15}\)

\(\Rightarrow\frac{3}{5}>\left(\frac{2}{3}+\frac{-1}{5}\right)\)

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

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

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

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

\(\Rightarrow x=\pm1\)

\(\left(2^2\right)^{30}=4^{30}\)

\(4^{20}=4^{20}\)

Vì \(4^{20}< 4^{30}n\text{ê}n\)

\(2^{60}>4^{20}\)

2⁶⁰ và 4²⁰

2⁶⁰=(2²)³⁰=4³⁰

Vì 4 ³⁰> 4²⁰

Vậy 2⁶⁰>4²⁰

k mk nha