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\(a,\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)

\(\Rightarrow\left[\frac{x+1}{65}+1\right]+\left[\frac{x+2}{64}+1\right]=\left[\frac{x+3}{63}+1\right]+\left[\frac{x+4}{62}+1\right]\)

\(\Rightarrow\frac{x+1+65}{65}+\frac{x+2+64}{64}=\frac{x+3+63}{63}+\frac{x+4+62}{62}\)

\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}\)

\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}=0\)

\(\Rightarrow\left[x+66\right]\left[\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\right]=0\)

Mà \(\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\ne0\)

\(\Rightarrow x+66=0\)

\(\Rightarrow x=0-66=-66\)

Auto làm nốt câu b

a,  Cộng cả 2 vế với 2 

Ta có \(\frac{x+1}{64}+\frac{x+2}{63}+2=\frac{x+3}{62}+\frac{x+4}{61}+2\)

\(\left(\frac{x+1}{64}+\frac{64}{64}\right)+\left(\frac{x+2}{63}+\frac{63}{63}\right)=\left(\frac{x+3}{62}+\frac{62}{62}\right)+\left(\frac{x+4}{61}+\frac{61}{61}\right)\)

=>  \(\frac{x+65}{64}+\frac{x+65}{63}=\frac{x+65}{62}+\frac{x+65}{61}\)\(\)

=> \(\frac{x+65}{64}+\frac{x+65}{63}-\frac{x+65}{62}-\frac{x+65}{61}=0\)

=> \(\left(x+65\right)\left(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\right)=0\)

Do \(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\ne0\)=> \(x+65=0\)

=> \(x=-65\)

b ,  Lm tương tự như Câu a

Chúc bn hok tốt

a) \(\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)

\(\Leftrightarrow\frac{x+1}{65}+\frac{x+2}{64}+2=\frac{x+3}{63}+\frac{x+4}{62}+2\)

\(\Leftrightarrow\left(\frac{x+1}{65}+1\right)+\left(\frac{x+2}{64}+1\right)=\left(\frac{x+3}{63}+1\right)+\left(\frac{x+4}{62}+1\right)\)

\(\Leftrightarrow\frac{x+1+65}{65}+\frac{x+2+64}{64}=\frac{x+3+63}{63}+\frac{x+4+62}{62}\)

\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}\)

\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{64}-\frac{x+66}{63}-\frac{x+66}{62}=0\)

\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{64}-\frac{1}{63}-\frac{1}{62}\right)=0\)

\(\Leftrightarrow x+66=0\)

\(\Leftrightarrow x=-66\)

CÂu b) làm tương tự:

- Trừ 3 cho hai vế ( câu a) mk cộng 2 cho hai vế)

- Tách -3 = -1-1-1 rồi kết hợp với mỗi hạng tử

CỐ LÊN NHÉ

NẾU bạn KHÔNG HIỂU thì câu b) mik sẽ làm kĩ càng và rõ ràng hơn cho bạn hiểu

\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

<=>  \(\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)

<=>  \(\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)

<=>  \(\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

Nhận thấy:   \(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\)

=>  \(x-105=0\)

<=>  \(x=105\)

\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

\(\Leftrightarrow\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}-\frac{x-100}{5}-\frac{x-101}{4}-\frac{x-102}{3}=0\)

\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)-\left(\frac{x-100}{5}-1\right)-\left(\frac{x-101}{4}-1\right)-\left(\frac{x-102}{3}-1\right)=0\)

\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}-\frac{x-105}{5}-\frac{x-105}{4}-\frac{x-105}{3}=0\)

\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

\(\Leftrightarrow x-105=0\left(Vì\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)\)

\(\Leftrightarrow x=105\)

ai làm ơn trả lời hộ mình câu này với

a) \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)=\left(\frac{x-100}{5}-1\right)+\left(\frac{x-101}{4}-1\right)+\left(\frac{x-102}{3}-1\right)\)
\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)
\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x=105\)
b) \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Leftrightarrow\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Leftrightarrow x=50\)

a. \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

\(\Rightarrow\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)

\(\Rightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}-\frac{x-105}{5}-\frac{x-105}{4}-\frac{x-105}{3}=0\)

\(\Rightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

\(\Rightarrow x-105=0\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)\)

\(\Rightarrow x=105\)

b. \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)

\(\Rightarrow\frac{29-x}{21}+1+\frac{27-x}{23}+1+\frac{25-x}{25}+1+\frac{23-x}{27}+1+\frac{21-x}{29}+1=0\)

\(\Rightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)

\(\Rightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)

\(\Rightarrow50-x=0\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\ne0\right)\)

\(\Rightarrow x=50\)

a) \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

\(\Leftrightarrow\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)

\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)

\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

Dễ dàng thấy nhân tử thứ hai luôn bé thua 0 nên \(x-105=0\)\(\Leftrightarrow x=105\)

b) Kĩ thuật làm tương tự câu a cộng mỗi phân số VT với 1 thì VP=0 và ta có nhân tử chung 50-x

1+1=3 :)))

Theo đề \(\Rightarrow\left(\frac{301-x}{103}+1\right)+\left(\frac{302-x}{102}+1\right)=\left(\frac{303-x}{101}+1\right)+\left(\frac{304-x}{100}+1\right)\)

\(\Leftrightarrow\left(\frac{301-x}{103}+1\right)+\left(\frac{302-x}{102}+1\right)-\left(\frac{303-x}{101}+1\right)-\left(\frac{304-x}{100}+1\right)=0\)

Sau khi đã quy đồng các phân số với các số 1, ta có :

\(\frac{301-x+103}{103}+\frac{302-x+102}{102}-\frac{303-x+101}{101}-\frac{304-x+100}{100}=0\)

\(\Rightarrow\frac{404-x}{103}+\frac{404-x}{102}-\frac{404-x}{101}-\frac{404-x}{100}=0\)

\(\Leftrightarrow\left(404-x\right)\times\frac{1}{103}+\left(404-x\right)\times\frac{1}{102}-\left(404-x\right)\times\frac{1}{101}-\left(404-x\right)\times\frac{1}{100}=0\)

\(\Leftrightarrow\left(404-x\right)\times\left(\frac{1}{103}+\frac{1}{102}-\frac{1}{101}-\frac{1}{100}\right)=0\)

Vì \(\frac{1}{103}< \frac{1}{102}< \frac{1}{101}< \frac{1}{100}\Rightarrow\frac{1}{103}+\frac{1}{102}-\frac{1}{101}-\frac{1}{100}\ne0\)

Để \(\left(404-x\right)\times\left(\frac{1}{103}+\frac{1}{102}-\frac{1}{101}-\frac{1}{100}\right)=0\)thì \(404-x=0\)

\(404-x=0\)

\(\Rightarrow x=404\)

Vậy x=404

Phương trình \(\Leftrightarrow\left(\frac{301-x}{103}+1\right)+\left(\frac{302-x}{102}+1\right)=\left(\frac{303-x}{101}+1\right)+\left(\frac{304-x}{100}+1\right)\)

\(\Leftrightarrow\frac{404-x}{103}+\frac{404-x}{102}=\frac{404-x}{101}+\frac{404-x}{100}\)

\(\Leftrightarrow\left(404-x\right)\left(\frac{1}{103}+\frac{1}{102}-\frac{1}{101}-\frac{1}{100}\right)=0\)

\(\Leftrightarrow404-x=0\)vì \(\left(\frac{1}{103}+\frac{1}{102}-\frac{1}{101}-\frac{1}{100}\right)\ne0\)

\(\Leftrightarrow x=404\)

Vậy phương trình có nghiệm x=404

a: \(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot\left(x-1\right)+\dfrac{1}{10}x-x=-\dfrac{9}{10}\)

\(\Leftrightarrow\dfrac{9}{10}x-\dfrac{9}{10}-\dfrac{9}{10}x=-\dfrac{9}{10}\)

=>-9/10=-9/10(luôn đúng)

b: \(\Leftrightarrow\dfrac{195x+195+130x+195+117x+195+100x+195}{195}=\dfrac{22\cdot39+4\cdot65+6\cdot39+40\cdot5}{195}\)

=>347x+780=1552

=>347x=772

hay x=772/347

đề hơi lạ xem lại

Ta có :5/x = 1/8 - y/4 = (1-2y)/8 
<=> x = 5.8/(1-2y) ; thấy 1-2y là số lẻ nên ƯCLN(8,1-2y) = 1 
do đó x/8 = 5/(1-2y) 
Để x, y nguyên khi 1-2y phải là ước của 5 
*Xét 1-2y = -1 => y = 1 => x = -40 
*Xét 1-2y = 1 => y = 0 => x = 40 
*Xét 1-2y = -5 => y = 3 => x = -8 
*Xét 1-2y = 5 => y = -2 => x = 8 
Vậy có 4 cặp (x,y) nguyên (-40,1) ; (40, 0) ; (-8, -5) ; (8, 5)