\(\frac{74-x}{26}+\frac{75-x}{25}+\frac{76-x}{24}+\frac{77-x}{23}+\frac{78-x}{22}=-5\)
Tìm X
giải phương trình sau
\(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=0\)
Sửa đề: \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)Ta có: \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)
\(\Leftrightarrow\dfrac{74-x}{26}+1+\dfrac{75-x}{25}+1+\dfrac{76-x}{24}+1+\dfrac{77-x}{23}+1+\dfrac{78-x}{22}+1=0\)
\(\Leftrightarrow\dfrac{100-x}{26}+\dfrac{100-x}{25}+\dfrac{100-x}{24}+\dfrac{100-x}{23}+\dfrac{100-x}{22}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\right)=0\)
mà \(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}>0\)
nên 100-x=0
hay x=100
Vậy: S={100}
Ta có : \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)
\(\Leftrightarrow\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}+5=0\)
\(\Leftrightarrow\dfrac{74-x}{26}+1+\dfrac{75-x}{25}+1+\dfrac{76-x}{24}+1+\dfrac{77-x}{23}+1+\dfrac{78-x}{22}+1=0\)
\(\Leftrightarrow\dfrac{100-x}{26}+\dfrac{100-x}{25}+\dfrac{100-x}{24}+\dfrac{100-x}{23}+\dfrac{100-x}{22}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\right)=0\)
Thấy : \(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\ne0\)
\(\Rightarrow100-x=0\)
\(\Leftrightarrow x=100\)
Vậy ...
\(\frac{x+1}{77}+\frac{x+2}{76}=\frac{x+3}{75}+\frac{x+4}{74}\)
Giải phương trình
<=>\(\left(\frac{x+1}{77}+1\right)+\left(\frac{x+2}{76}+1\right)=\left(\frac{x+3}{75}+1\right)+\left(\frac{x+4}{74}+1\right)\)
<=> \(\frac{x+1+77}{77}+\frac{x+2+76}{76}=\frac{x+3+75}{75}+\frac{x+4+74}{74}\)
<=> \(\frac{x+78}{77}+\frac{x+78}{76}=\frac{x+78}{75}+\frac{x+78}{74}\)
<=> \(\frac{x+78}{77}+\frac{x+78}{76}-\frac{x+78}{75}-\frac{x+78}{74}\)
<=> \(\left(x+78\right)\left(\frac{1}{77}+\frac{1}{76}-\frac{1}{75}-\frac{1}{74}\right)\)
Vì \(\frac{1}{77}+\frac{1}{76}-\frac{1}{75}-\frac{1}{74}\ne0\) nên phương trình trên <=> x + 78 = 0
<=> x = -78
Tập nghiệm của phương trình trên là S= \(\left\{-78\right\}\)
Chúc bạn học tốt !
a) \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(\Leftrightarrow\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
\(\Leftrightarrow\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
\(\Leftrightarrow x-23=0\)( vì \(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\ne0\))
\(\Leftrightarrow x=23\)
Vậy nghiệm của pt x=23
\(\frac{x-25}{1979}-\frac{x-24}{1980}-\frac{x-23}{1981}-\frac{x-22}{1982}=\frac{x-1979}{25}-\frac{x-1980}{24}-\frac{x-1981}{23}-\frac{x-1982}{22}\)
Bạn cộng mỗi vế cho 4 trong đó mỗi phần tử cộng với 1 = -1954(hình như vậy) thì x = 2004
\(\frac{x-25}{1979}-\frac{x-24}{1980}-\frac{x-23}{1981}-\frac{x-22}{1982}=\frac{x-1979}{25}-\frac{x-1980}{24}-\frac{x-1981}{23}-\frac{x-1982}{22}\)
Cộng mỗi vế với 4, trong đó mỗi phần tử sẽ cộng với 1. Hình như kết quả là -1954 thì phải đúng ko? Mk mới chỉ mò kết quả thôi nhưng cách làm thì mk biết! Nếu có gì thắc mắc thì nhắn tin cho mk giải cho!
Tìm x, biết:
\(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)
\(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)
\(\Leftrightarrow\left[\frac{x-12}{77}-1\right]+\left[\frac{x-11}{78}-1\right]=\left[\frac{x-74}{15}-1\right]-\left[\frac{x-73}{16}-1\right]\)
\(\Leftrightarrow\frac{x-12-77}{77}+\frac{x-11-78}{78}=\frac{x-74-15}{15}+\frac{x-73-16}{16}\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}=0\)
\(\Leftrightarrow\left[x-89\right]\cdot\left[\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right]=0\)
\(\Leftrightarrow x-89=0\)
\(\Leftrightarrow x=89\)
Vậy x = 89
mik ko biết
mà nhìn như toán nâng cao ý
\(\frac{x-25}{1979}-\frac{x-24}{1980}-\frac{x-23}{1981}-\frac{x-22}{1982}=\frac{x-1979}{25}-\frac{x-1980}{24}-\frac{x-1981}{23}-\frac{x-1982}{22}\)
giúp tôi với nếu đúng cho 3 tick
P/s: Chuyển tất cả các hạng tử sang 1 vế rồi cộng thêm 1 vào các vế có dấu (+) đằng trước, cộng thêm -1 vào các hạng tử có dấu (-) phía trước rồi đặt nhân tử chung ra ngoài ta được:
\(Pt\Leftrightarrow\left(x-2004\right)\left(\frac{1}{1979}-\frac{1}{1980}-\frac{1}{1981}-\frac{1}{1982}-\frac{1}{25}+\frac{1}{24}+\frac{1}{23}+\frac{1}{22}\right)=0\)
\(\Leftrightarrow x-2004=0\)
\(\Rightarrow x=2004\)
Vậy x = 2004
https://olm.vn/hoi-dap/detail/263823966145.html?pos=616279814817
Tìm x biết \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
\(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
=> x-23=0
x=0+23
x=23. Vậy x=23
Chúc bạn học tốt!^_^
\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
=>( x-13)(\(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\) = 0
ta thấy 1/24>1/25>1/26>1/27 => 1/24+1/25 - 1/ 26 - 1/17 > 0
=> x -13 = -
=> x=13
\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
Giải:
Ta có: \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(\Leftrightarrow\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
\(\Leftrightarrow\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)\)
\(\Leftrightarrow x-23=0\) (Vì \(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\) ≠ 0)
\(\Leftrightarrow x=23\)
Vậy nghiệm của phương trình là x = 23.
Chúc bạn học tốt@@
\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\Leftrightarrow\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\Leftrightarrow x-23=0\Leftrightarrow x=23\)
Vậy $x=23$