Câu hỏi của Loveduda

Question

Tìm x:$\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10$

Nguyễn Huy Tú · Accepted Answer

$\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10$

$\Rightarrow\dfrac{x-241}{17}-1+\dfrac{x-220}{19}-2+\dfrac{x-195}{21}-3+\dfrac{x-166}{23}-4=0$

$\Rightarrow\dfrac{x-258}{17}+\dfrac{x-258}{19}+\dfrac{x-258}{21}+\dfrac{x-258}{23}=0$

$\Rightarrow\left(x-258\right)\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\right)=0$

Mà $\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\ne0$

$\Rightarrow x-258=0\Rightarrow x=258$

Vậy x = 258Câu trả lời: $\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10$

$\Rightarrow\dfrac{x-241}{17}-1+\dfrac{x-220}{19}-2+\dfrac{x-195}{21}-3+\dfrac{x-166}{23}-4=0$

$\Rightarrow\dfrac{x-258}{17}+\dfrac{x-258}{19}+\dfrac{x-258}{21}+\dfrac{x-258}{23}=0$

$\Rightarrow\left(x-258\right)\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\right)=0$

Mà $\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\ne0$

$\Rightarrow x-258=0\Rightarrow x=258$

Vậy x = 258

pu · Answer

x−24117+x−22019+x−19521+x−16623=10x−24117+x−22019+x−19521+x−16623=10

⇒x−24117−1+x−22019−2+x−19521−3+x−16623−4=0⇒x−24117−1+x−22019−2+x−19521−3+x−16623−4=0

⇒x−25817+x−25819+x−25821+x−25823=0⇒x−25817+x−25819+x−25821+x−25823=0

⇒(x−258)(117+119+121+123)=0⇒(x−258)(117+119+121+123)=0

Mà 117+119+121+123≠0117+119+121+123≠0

⇒x−258=0⇒x=258Câu trả lời: x−24117+x−22019+x−19521+x−16623=10x−24117+x−22019+x−19521+x−16623=10

⇒x−24117−1+x−22019−2+x−19521−3+x−16623−4=0⇒x−24117−1+x−22019−2+x−19521−3+x−16623−4=0

⇒x−25817+x−25819+x−25821+x−25823=0⇒x−25817+x−25819+x−25821+x−25823=0

⇒(x−258)(117+119+121+123)=0⇒(x−258)(117+119+121+123)=0

Mà 117+119+121+123≠0117+119+121+123≠0

⇒x−258=0⇒x=258