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\(pt\Leftrightarrow\frac{29}{21}-\frac{x}{21}+\frac{27}{23}-\frac{x}{23}+\frac{25}{25}-\frac{x}{25}+\frac{23}{27}-\frac{x}{27}+\frac{21}{29}-\frac{x}{29}=-5\Leftrightarrow-x\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=-5-\frac{29}{21}-\frac{27}{23}-\frac{25}{25}-\frac{23}{27}-\frac{21}{29}\Leftrightarrow-x=\frac{-5-\frac{29}{21}-\frac{27}{23}-\frac{25}{25}-\frac{23}{27}-\frac{21}{29}}{\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}}=-50\Leftrightarrow x=50\\ \Rightarrow S=\left\{50\right\}\)
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) \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Leftrightarrow\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Leftrightarrow x=50\)
\(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5.\)
\(\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)\)\(=0\)
\(\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\left(50-x\right).\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
=> 50 - x = 0 \(\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\ne0\right)\)
=> 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
pạn -1 vào mỗi phân số là xong. Rùi ra x\(\frac{x-2015}{1986}\)+\(\frac{x-2015}{1988}\)+ \(\frac{x-2015}{1990}\)+...+\(\frac{x-2015}{x1996}\)-\(\frac{x-2015}{29}\)-\(\frac{x-2015}{27}\)-...\(\frac{x-2015}{19}\)=0
<=>(x-2015)(\(\frac{1}{1986}\)+\(\frac{1}{1988}\)+... -\(\frac{1}{19}\))=0...(mà \(\frac{1}{1986}\)+...- \(\frac{1}{19}\) khác 0)
=>x-2015=0
<=> x=2015
1/Bạn cộng tất cả các phân số ở 2 vế với 1, tất cả các phân số sẽ có chung tử, cậu nhóm tử đó lại thành PT tích..với mẫu =0 tìm đc x
2/Trừ 1 vào từng phân thức đc
\(\frac{x-b-c}{a}-1+\frac{x-a-c}{b}-1+\frac{x-a-b}{c}-1=0\)
\(\Leftrightarrow\frac{x-\left(a+b+c\right)}{a}+\frac{x-\left(a+b+c\right)}{b}+\frac{x-\left(a+b+c\right)}{c}=0\)
\(\Leftrightarrow\left(x-\left(a+b+c\right)\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)
\(\Rightarrow x=a+b+c\)
Các câu na ná chắc nên mk làm mẫu 2 bài thui nha !
a, pt <=> x-23/24 + x-23/25 - x-23/26 - x-23/27 = 0
<=> (x-23).(1/24+1/25-1/26-1/27) = 0
<=> x-23=0 ( vì 1/24+1/25-1/26-1/27 > 0 )
<=> x=23
b, pt <=> (201-x/99 + 1)+(203-x/97 + 1)+(205-x/95 + 1) = 0
<=> 300-x/99 + 300-x/97 + 300-x/95 = 0
<=> (300-x).(1/99+1/97+1/95) = 0
<=> 300-x = 0 ( vì 1/99+1/97+1/95 > 0 )
<=> x=300
Tk mk nha
\(pt\Leftrightarrow\frac{29-x}{21}+1+\frac{27-x}{23}+1+...=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
Do \(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}>0\) nên 50 - x = 0 hay x = 50.
pt<=>29-x/21+1+27-x/23+1+...=0
<=>50-x/21+50-x/23+50-x/25+50-x/27+50-x/29=0
<=>(50-x).(1/21+1/23+1/25+1/27+1/29)=0
Do 1/21+1/23+1/25+1/27+1/29>0 nên 50-x=0 hay x=50