\(\frac{149-x}{25}\)+\(\frac{170-x}{23}\)+\(\frac{187-x}{21}\)+\(\frac{200-x}{19}\)=10
giải phương trình sau : (149-x)/25+(170-x)/23+(187-x)/21+(200-x)/19=10
\(\frac{149-x}{25}+\frac{170-x}{23}+\frac{187-x}{21}+\frac{200-x}{19}=10\)
\(\Rightarrow\frac{149-x}{25}-1+\frac{170-x}{23}-2+\frac{187-x}{21}-3+\frac{200-x}{19}-4=0\)
\(\Rightarrow\frac{124-x}{25}+\frac{124-x}{23}+\frac{124-x}{21}+\frac{124-x}{19}=0\)
\(\Rightarrow\left(124-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)
Mà \(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}>0\Rightarrow x-124=0\Rightarrow x=124\)
Giải PT\(\frac{x-29}{1986}+\frac{x-27}{1988}+\frac{x-25}{1990}+\frac{x-23}{1992}+\frac{x-21}{1994}+\frac{x-19}{1996}=\frac{x-1986}{29}+\frac{x-1988}{27}+\frac{x-1990}{25}+\frac{x-1992}{23}+\frac{x-1994}{21}+\frac{x-1996}{19}\)
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
Giải phương trình
\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-170}{23}=10\)
giúp mình nhen mấy bạn
Ta có: \(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-170}{23}=10\)
\(\Rightarrow\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-170}{23}-10=0\)
\(\Leftrightarrow\left(\frac{x-241}{17}-1\right)+\left(\frac{x-220}{19}-2\right)+\left(\frac{x-195}{21}-3\right)+\left(\frac{x-170}{23}-4\right)=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=0\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=0\)
\(\Leftrightarrow\left(x-258\right)=0\)
\(\Rightarrow x=258\)
Vậy x=258
\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10\)
Mk chi p bang 123 vi bam may tinh, con ck jai thi hk p!
Tìm x biết :
\(\frac{2032-x}{25}+\frac{2053-x}{23}+\frac{2070-x}{21}+\frac{2083-x}{19}-10=0\)
X sẽ bằng 2007 vì:
2032-x/25+2053-x/21+2070-x/21+2038-x/19 = 10 ( vì đỏi vế số 10 nên = 0+10=10)
10= 1+2+3+4 (Có 4 phân số thì mỗi phân số tương ứng lần lượt la 1 ,2 ,3 ,4)
Vậy x =2007
Chúc bạn học giỏi
=>\(\left(\frac{2032-x}{25}-1\right)+\left(\frac{2053-x}{23}-2\right)+\left(\frac{2070-x}{21}-3\right)+\left(\frac{2083-x}{19}-4\right)=0\)
=>\(\frac{2007-x}{25}+\frac{2007-x}{23}+\frac{2007-x}{21}+\frac{2007-x}{19}=0\)
=>\(\left(2007-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)
Vì \(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\ne0\)
=> 2007 - x = 0 => x = 2007
GIÚP MK VS Ạ!!!
\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}-\frac{199-x}{19}=10\)
\(\text{Giải phương trình sau(biến đổi đặc biệt):}\)
\(E=\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-21}{1978}+\frac{x-19}{1980}=\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}+\frac{x-1978}{21}+\frac{x-1980}{19}\)
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 }
Câu cuối r , giúp mk vs nha
\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}-\frac{199-x}{19}=10\)
Tìm x biết: \(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-170}{22}=10\)
=> \(\frac{x-241}{17}-1+\frac{x-220}{19}-2+\frac{x-195}{21}-3+\frac{x-170}{22}-4=0\)
<=> \(\left(\frac{x-241}{17}-\frac{17}{17}\right)+\left(\frac{x-220}{19}-\frac{38}{19}\right)+\left(\frac{x-195}{21}-\frac{63}{21}\right)+\left(\frac{x-170}{22}-\frac{88}{22}\right)=0\)
<=> \(\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{22}=0\)
<=> \(\left(x-258\right).\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{22}\right)=0\)
<=> x - 258 = 0 do \(\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{22}\right)\ne0\)
=> x = 258
10=1+2+3+4
X=241+17x1=258
X=220+19x2=258
X=195+21x3=258
X=170+22x4=258.