Tinh nhanh:
2017 2017 2017 x 2018 2018 2018 2018 /2018 2018 2018 x 2017 2017 2017 2017
Giúp mình nha!!!!!!!!!!!
Những câu hỏi liên quan
Tinh nhanh:
2017 2017 2017 x 2018 2018 2018 2018 /2018 2018 2018 x 2017 2017 2017 2017
( 2016 * 2017 + 2018 * 2 + 2015 ) : [ ( 2018 : 2017 - 2017 * 2015 ) + 2016 ]
2018 * 20182017 - 2017 * 20172018
2018 * 20182017 - 2017 * 20182018
Bai nay la bai tinh nhanh nha
Khó quá zậy
Nhưng mình bít kết quả rồi
đó là : -1,002482622
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
So sánh 2017^2016+2018/2017^2017+2018với 2017^2017+2018/2017^2018+2018
Ai kết bạn mình đi
A = 2017/2018 x 7/8 + 2017/2018 x 3/8 - 2017/2018 x 1/4
Ta có : A =\(\frac{2017}{2018}\)x \(\frac{7}{8}\)+ \(\frac{2017}{2018}\)x \(\frac{3}{8}\)- \(\frac{2017}{2018}\)x \(\frac{1}{4}\)
= \(\frac{2017}{2018}\) x ( \(\frac{7}{8}+\frac{3}{8}-\frac{1}{4}\))
= \(\frac{2017}{2018}\)x 1
=\(\frac{2017}{2018}\)
Vậy A= : \(\frac{2017}{2018}\)
Bài giải
\(A=\frac{2017}{2018}\text{ x }\frac{7}{8}+\frac{2017}{2018}\text{ x }\frac{3}{8}-\frac{2017}{2018}\text{ x }\frac{1}{4}\)
\(A=\frac{2017}{2018}\text{ x }\frac{1}{4}\left(\frac{7}{2}+\frac{3}{2}-1\right)=\frac{2017}{2018}\text{ x }\frac{1}{4}\text{ x }4==\frac{2017}{2018}\text{ x }1=\frac{2017}{2018}\)
CMR \(\left(2018^{2017}+2017^{2017}\right)^{2018}>\left(2018^{2018}+2017^{2018}\right)^{2017}\)
Có: \(\left(2018^{2018}+2017^{2018}\right)^{2017}< \left(2018^{2017}.2018+2017^{2017}.2018\right)^{2017}\)
\(=\left(2018^{2017}+2017^{2017}\right)^{2017}.2018^{2017}< \left(2018^{2017}+2017^{2017}\right)^{2017}.\left(2018^{2017}+2017^{2017}\right)\)
\(=\left(2018^{2017}+2017^{2017}\right)^{2018}\)
Cho hai số A = (2018^2017 + 2017^2017)^2018 ; B = (2018^2018 + 2017^2018)^2017. so sánh A và B
\(A=\left(2018^{2017}+2017^{2017}\right)^{2018}\) ; \(B=\left(2018^{2018}+2017^{2018}\right)^{2017}\)
Ta có:
\(B=\left(2018.2018^{2017}+2017.2017^{2017}\right)^{2017}\)
\(\Rightarrow B< \left(2018.2018^{2017}+2018.2017^{2017}\right)^{2017}\)
\(\Rightarrow B< \left(2018^{2017}+2017^{2017}\right)^{2017}.2018^{2017}\)
\(\Rightarrow B< \left(2018^{2017}+2017^{2017}\right)^{2017}.\left(2018^{2017}+2017^{2017}\right)\)
\(\Rightarrow B< \left(2018^{2017}+2017^{2017}\right)^{2018}=A\)
\(\Rightarrow B< A\)
Đúng 0
Bình luận (0)
Tính bằng cách thuận tiện:
A = 2017/2018 x 7/8 + 2017/2018 x 3/8 - 2017/2018 x 1/4
So sánh: A=(20182017 +20172017)2018 và B=(20182018+20172018)2017
A=2017^2018+1/2017^2018-3
B=2017^2018-1/2017^2018-5