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9 tháng 2 2021
Ta có : \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)
\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
\(\Leftrightarrow5x-200=0\)
\(\Leftrightarrow x=40\)
Vậy ...
9 tháng 2 2021
Ta có: \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)
\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
mà \(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}>0\)
nên 5x-200=0
\(\Leftrightarrow5x=200\)
hay x=40
Vậy: S={40}
IK
0
CM
11 tháng 9 2018
Ta có
56 x 2 – 45 y – 40 x y + 63 x = 56 x 2 + 63 x – 45 y + 40 x y = 7 x 8 x + 9 – 5 y 8 x + 9 = 7 x - 5 y 8 x + 9
Suy ra m = 8; n = 9
Đáp án cần chọn là: A
1 tháng 1 2018
\(pt\Leftrightarrow\frac{5x-150}{50}+\frac{5x-102}{49}+\frac{5x-56}{48}+\frac{5x-12}{47}+\frac{5x-16}{46}-14=0\)
\(\Leftrightarrow\frac{5x-150}{50}-1+\frac{5x-102}{49}-2+\frac{5x-56}{48}-3+\frac{5x-12}{47}-4+\frac{5x-16}{46}-4=0\)
\(\Leftrightarrow\frac{5x-200}{50}+\frac{5x-200}{49}+\frac{5x-200}{48}+\frac{5x-200}{47}+\frac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}\right)=0\)
Do \(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}\ne0\) nên \(5x-200=0\Rightarrow x=\frac{200}{5}=40\)
Vậy x= 40
KT
1 tháng 1 2018
\(\frac{5x-150}{50}+\frac{5x-102}{49}+\frac{5x-56}{48}+\frac{5x-12}{47}+\frac{5x-660}{46}=0\)
\(\Leftrightarrow\)\(\left(\frac{5x-150}{50}-1\right)+\left(\frac{5x-102}{49}-2\right)+\left(\frac{5x-56}{48}-3\right)+\left(\frac{5x-12}{47}-4\right)+\left(\frac{5x-660}{46}+10\right)=0\)
\(\Leftrightarrow\)\(\frac{5x-200}{50}+\frac{5x-200}{49}+\frac{5x-200}{48}+\frac{5x-200}{47}+\frac{5x-200}{46}=0\)
\(\Leftrightarrow\)\(\left(5x-200\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}\right)=0\)
\(\Leftrightarrow\)\(5x-200=0\)
\(\Leftrightarrow\)\(5x=200\)
\(\Leftrightarrow\)\(x=40\)
Vậy x = 40
TV
1
3 tháng 8 2017
\(4x^2-36x+56=0\)
\(\Rightarrow4\left(x^2-9x+14\right)=0\)
\(\Rightarrow4\left(x^2-7x-2x+14\right)=0\)
\(\Rightarrow4x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Rightarrow4\left(x-7\right)\left(x-2\right)=0\)
\(\Rightarrow\left(x-7\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-7=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow x=7;x=2\).
3 tháng 8 2017
\(4x^2-36x+56=0\)
\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot9+9^2-25=0\)
\(\Leftrightarrow\left(2x-9\right)^2-25=0\)
\(\Leftrightarrow\left(2x-9\right)^2=25\Leftrightarrow\left[{}\begin{matrix}2x-9=5\\2x-9=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=2\end{matrix}\right.\)
Vậy....................
DL
0
LS
2
\(0,0793m+\dfrac{56}{47}+100m=21\\ \Leftrightarrow0,0793m+100m=21-\dfrac{56}{47}\\ \Leftrightarrow100,0793m=\dfrac{931}{47}\\ \Rightarrow m\approx0,1979\left(2\right)\)