số số hạng có trong biểu thức trên là:

(58 - 40 ) : 1 + 1 = 19

=>( 40 - 41 ) + ( 42 - 43 ) + ... + ( 56 - 57 ) + 58

=> (-1) x  [( 19 - 1 ) : 2 ] + 58

=> (-1) x 9 + 58

=> (-9) + 58
=> 49
HT~~~

Đú đởn ăn chơi học hành sa sút đến nỗi có bài toán đơn giản như zậy mà cũng phải hỏi !

40-41+42-43+44-45+46-47+48-49+50-51+52-53+54-55+56-57+58=

=40+(42-41)+(44-43)+(46-45)+(48-47)+(50-49)+(52-51)+(54-53)+(56-55)+(58-57)=49

chơi ff ko xin id

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

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

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

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

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

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

Vì\(\frac{1}{57}< \frac{1}{51},\frac{1}{54}< \frac{1}{48}\Rightarrow\left(\frac{1}{57}+\frac{1}{54}\right)< \left(\frac{1}{51}+\frac{1}{48}\right)\)

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

Phương trình (*) xảy ra khi: \(100-x=0\Leftrightarrow x=100\)

Vậy phương trình có nghiệm duy nhất là x = 100

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 \(x=-100\)

a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+2\right)=\left(\frac{x+3}{7}+1\right)+\left(\frac{x+4}{6}+1\right)\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Mà \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

\(\Rightarrow x+10=0\)

\(\Rightarrow x=-10\)

Vậy x = -10

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

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

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

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

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

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

\(\Rightarrow x+100=0\)

\(\Rightarrow x=-100\)

Vậy x = -100

a.\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

=>\(\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

<=> \(\frac{x+1+9}{9}+\frac{x+2+8}{8}=\frac{x+3+7}{7}+\frac{x+4+6}{6}\)

<=>\(\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

<=> \(\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

<=> \(\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

<=> x+10=0

<=> x=-10

Vậy tập nghiệm của phương trình trên là S=\(\left\{-10\right\}\)

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

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

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

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

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

<=>x+100=0

<=>x= -100

Vậy tập nghiệm của phương trình trên là S=\(\left\{-100\right\}\)

Câu a là đúng

300

300

= 300