Giải phương trình:\(\dfrac{x-29}{1970}+\dfrac{x-27}{1972}+\dfrac{x-25}{1974}+\dfrac{x-23}{1976}+\dfrac{x-21}{1978}+\dfrac{x-19}{1980}=\dfrac{x-1970}{29}+\dfrac{x-1972}{27}+\dfrac{x-1974}{25}+\dfrac{x-1976}{23}+\dfrac{x-1978}{21}+\dfrac{x-1980}{19}\)
\(\text{Giải phương trình sau(biến đổi đặc biệt):}\)
\(E=\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-21}{1978}+\frac{x-19}{1980}=\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}+\frac{x-1978}{21}+\frac{x-1980}{19}\)
E = ( x - 29 ) / 1970 + ( x - 27 ) / 1972 + ( x - 25 ) / 1974 + ( x - 23 ) / 1976 + ( x - 21 ) / 1978 + ( x - 19 ) / 1980 = ( x - 1970 ) / 29 + ( x - 1972 ) / 27 + ( x - 1974 ) / 25 + ( x - 1976 ) / 23 + ( x - 1978 ) / 21 + ( x - 1980 ) / 19
( Trừ từng số hạng cho 1 ra như sau )
E = (x - 1999)/ 1970 + ( x - 1999 ) / 1972 + ( x - 1999) / 1974 + ( x - 1999)/ 1976 + ( x -1999) / 1978 + ( x - 1999)/ 1980 = ( x - 1999)/29 + ( x - 1999) / 27 + ( x - 1999 ) / 25 + ( x - 1999) / 23 + ( x - 1999)/21 + ( x - 1999) / 19
< = > ( x - 1999 ) / 1970 + ( x - 1999 ) / 1972 + ( x - 1999 ) / 1974 + ( x - 1999) / 1976 + ( x - 1999) / 1978 + ( x - 1999) / 1980 - ( x - 1999) / 29 - ( x - 1999)/ 27 - ( 1 - 1999) / 25 - ( x-1999) / 23 - ( x - 1999) / 21 - ( x - 1999) / 19 = 0 ( chuyển vế )
< = > ( x - 1999 ) ( 1/1970 + 1/ 1972 + 1/1974 + 1/1976 + 1/1978 + 1/1980 - 1/29 - 1/27 - 1/25 - 1/23 - 1/21 - 1/19) = 0
Vì ( 1/1970 + 1/1972 + 1/1974 + 1/1976 + 1/1978 + 1/1980 - 1/29 -1/27 - 1/25 - 123 - 1/21 - 1/19 ) khác 0 nên để đẳng thức bằng 0 thì bắt buộc x - 1999 = 0
< = > x = 0 + 1999 = 1999
Vậy tập nghiệm của phương trình là S = { 1999 }
Giải các phương trình sau
1, \(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-21}{1978}+\frac{x-19}{1980}=\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1976}{23}+\frac{x-1978}{21}+\frac{x-1980}{19}\)
2, \(\frac{x-b-c}{a}+\frac{x-a-c}{b}+\frac{x-a-b}{c}=3\)
Bạn nào biết giúp mk với ạ . Mk cảm ơn
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
\(\frac{x-b-c}{a}-1+\frac{x-a-c}{b}-1+\frac{x-a-b}{c}-1=0\)
\(\Leftrightarrow\frac{x-\left(a+b+c\right)}{a}+\frac{x-\left(a+b+c\right)}{b}+\frac{x-\left(a+b+c\right)}{c}=0\)
\(\Leftrightarrow\left(x-\left(a+b+c\right)\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)
\(\Rightarrow x=a+b+c\)
\(\dfrac{29-x}{21}\)+\(\dfrac{27-x}{23}\)+\(\dfrac{25-x}{25}\)+\(\dfrac{23-x}{27}\)+\(\dfrac{21-x}{29}\)=\(\dfrac{(29-x+1}{21}\)+\(\dfrac{(27-x+1)}{23}\)+\(\dfrac{(25-x+1)}{25}\)+\(\dfrac{(23-x+1)}{21}\)=-5 +5
GIẢI nốt hộ mình với ạ
mọi người ơi giúp mình câu giải phương trình này với:
\(\dfrac{x-5}{1990}+\dfrac{x-15}{1980}+\dfrac{x-25}{1970}+\dfrac{x-1990}{5}+\dfrac{x-1980}{15}+\dfrac{x-1970}{25}=0\)
\(\Leftrightarrow\left(\dfrac{x-5}{1990}-1\right)+\left(\dfrac{x-15}{1980}-1\right)+\left(\dfrac{x-25}{1970}-1\right)\\ +\left(\dfrac{x-1990}{5}-1\right)+\left(\dfrac{x-1980}{15}-1\right)+\left(\dfrac{x-1970}{25}-1\right)=0\\ \Leftrightarrow\dfrac{x-1995}{1990}+\dfrac{x-1995}{1980}+\dfrac{x-1995}{1970}+\dfrac{x-1995}{5}\\ +\dfrac{n-1995}{15}+\dfrac{n-1995}{25}=0\\ \Rightarrow\left(x-1995\right)\left(\dfrac{1}{1990}+\dfrac{1}{1980}+\dfrac{1}{1970}+\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}\right)=0\)
\(\Rightarrow x-1995=0\\ \Rightarrow x=1995\)
X-29/1970 + x-27/1972 + x-25/1974 + x-23/1976 + x-1970/29 + x-1972/27 - 6 =0
Các bạn giải chi tiết giúp mình nhesss. Thanks ạ 😘😘😘
\(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-1970}{29}+\frac{x-1972}{27}-6=0\)
\(\Leftrightarrow\)\(\frac{x-29}{1970}-1+\frac{x-27}{1972}-1+\frac{x-25}{1974}-1+\frac{x-23}{1976}-1+\frac{x-1970}{29}-1+\frac{x-1972}{27}-1=0\) \(\Leftrightarrow\) \(\frac{x-1999}{1970}+\frac{x-1999}{1972}+\frac{x-1999}{1974}+\frac{x-1999}{1976}+\frac{x-1970}{29}+\frac{x-1999}{27}=0\)
\(\Leftrightarrow\)\(\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}\right)=0\)
Vì \(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}\ne0\)
\(\Rightarrow\)\(x-1999=0\)
\(\Leftrightarrow\)\(x=1999\)
Vậy...
Cảm ơn bạn nhiều nha. Lần sau mình có bài j khó nữa nhớ giúp mình vs nhá😊😊😊
giải phương trình:
\(\dfrac{x-5}{1990}+\dfrac{x-15}{1980}+\dfrac{x-25}{1970}=\dfrac{x-1990}{5}+\dfrac{x-1980}{15}+\dfrac{x-1970}{25}\)
Ta có: \(\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}=\frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1970}{25}\)
\(\Leftrightarrow\)\(\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}-3=\frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1970}{25}-3\)
\(\Leftrightarrow\)\(\frac{x-5}{1990}-1+\frac{x-15}{1980}-1+\frac{x-25}{1970}-1=\frac{x-1990}{5}-1+\frac{x-1980}{15}-1+\frac{x-1970}{25}-1\)\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}=\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}\)
\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}-\frac{x-1995}{5}-\frac{x-1995}{15}-\frac{x-1995}{25}=0\)
\(\Leftrightarrow\)\(\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}-\frac{1}{15}-\frac{1}{25}\right)=0\)
\(\Leftrightarrow\)\(x-1995=0\)
\(\Leftrightarrow\)\(x=1995\)
Giải các phương trình sau
a)\(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
b)\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=-5\)
a) \(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
\(\Leftrightarrow\dfrac{x-5}{100}-1+\dfrac{x-4}{101}-1+\dfrac{x-3}{102}-1=\dfrac{x-100}{5}-1+\dfrac{x-101}{4}-1+\dfrac{x-102}{3}-1\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left(x-105\right)=0;\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}\right)\ne0\)
\(\Leftrightarrow x=105\)
b) \(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=-5\)
\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{29}+\dfrac{1}{27}+\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}\right)=0\)
\(\Leftrightarrow50-x=0;\left(\dfrac{1}{29}+\dfrac{1}{27}+\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}\right)\ne0\)
\(\Leftrightarrow x=50\)
giả phương trình:
\(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}=4\)
\(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}-8=0\)
a) Ta có: \(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}=4\)
\(\Leftrightarrow\frac{x-91}{37}-1+\frac{x-86}{42}-1+\frac{x-78}{50}-1+\frac{x-49}{79}-1=0\)
\(\Leftrightarrow\frac{x-91-37}{37}+\frac{x-86-42}{42}+\frac{x-78-50}{50}+\frac{x-49-79}{79}=0\)
\(\Leftrightarrow\frac{x-128}{37}+\frac{x-128}{42}+\frac{x-128}{50}+\frac{x-128}{79}=0\)
\(\Leftrightarrow\left(x-128\right)\left(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\right)=0\)
Vì \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}>0\)
nên x-128=0
hay x=128
Vậy: x=128
b) Ta có: \(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}-8=0\)
\(\Leftrightarrow\frac{x-29}{1970}-1+\frac{x-27}{1972}-1+\frac{x-25}{1974}-1+\frac{x-23}{1976}-1+\frac{x-1970}{29}-1+\frac{x-1972}{27}-1+\frac{x-1974}{25}-1+\frac{x-1976}{23}-1=0\)
\(\Leftrightarrow\frac{x-29-1970}{1970}+\frac{x-27-1972}{1972}+\frac{x-25-1974}{1974}+\frac{x-23-1976}{1976}+\frac{x-1970-29}{29}+\frac{x-1972-27}{27}+\frac{x-1974-25}{25}+\frac{x-1976-23}{23}=0\)
\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}\right)=0\)
Vì \(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}>0\)
nên x-1999=0
hay x=1999
Vậy: x=1999
a) Ta có \(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}\)=4
<=>\(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}-4=0\)
<=>\(\frac{x-91}{37}-1+\frac{x-86}{42}-1+\frac{x-78}{50}-1+\frac{x-49}{79}-1=0\)
<=>\(\frac{x-128}{37}+\frac{x-128}{42}+\frac{x-128}{50}+\frac{x-128}{79}=0\)
<=>(x-128)\(\left(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\right)=0\)
Vì \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}>0\)=>x-128=0<=>x=128
b)Tương tự
<=>x-128=0
<=>x=128
Chú ý \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\)>0
b)tương tự
\(\dfrac{29-x}{21}\)+\(\dfrac{27-x}{23}\)+\(\dfrac{25-x}{25}\)+\(\dfrac{23-x}{27}\)=-4
\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}\ne0\right)=0\Leftrightarrow x=50\)
\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}=-4\\ \Leftrightarrow\left(\dfrac{29-x}{21}+1\right)+\left(\dfrac{27-x}{23}+1\right)+\left(\dfrac{25-x}{25}+1\right)+\left(\dfrac{23-x}{27}+1\right)=0\\ \Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}=0\\ \Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}\right)=0\\ \Leftrightarrow50-x=0\left(vì.\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}\ne0\right)\\ \Leftrightarrow x=50\)