Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(52\cdot31\cdot x=4929\\ \Leftrightarrow1612x=4929\\ \Leftrightarrow x=\dfrac{159}{52}\)
\(x\cdot1999=1999\cdot199,8\\ \Leftrightarrow x=199,8\)
( X . 0,25 + 1999).2000 = (53 + 1999).2000
=> X . 0,25 +1999 = 53+1999
=> X . 0,25 = 53
=> X = 53 : 0,25 = 53 . \(\frac{1}{4}\)
=> X = \(\frac{53}{4}\)
(x.0,25+1999) . 2000=(53+1999) . 2000
(x.0,25+1999) . 2000= 2052 . 2000
(x.0,25+1999) . 2000= 4104000
x.0,25+1999 = 4104000 : 2000
x.0,25+1999 = 2052
x.0,25 = 2052-1999
x.0,25 = 53
x = 53:0.25
x = 212
Vậy x=212.
Mình ko chép lai đề bài nhé:
=(1000+999) x 177 + (1000+999) x 118 -999 x 59
=1000 x (177+118) + 999 x (177+118-59)
=1000 x 295 + 999 x 236
=295000+235764
=530764
Chúc bạn học tốt nha
=(1000+999) x 177 + (1000+999) x 118 -999 x 59
=1000 x (177+118) + 999 x (177+118-59)
=1000 x 295 + 999 x 236
=530764
\(A=\dfrac{1999^{1999}+1}{1999^{1998}+1}\)
\(\dfrac{1}{1999}A=\dfrac{1999^{1999}+1}{1999^{1999}+1999}\)
\(\dfrac{1}{1999}A=\dfrac{1999^{1999}}{1999^{1999}}-\dfrac{1998}{1999^{1999}+1999}\)
\(\dfrac{1}{1999}A=1-\dfrac{1998}{1999^{1999}+1999}\)
\(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}\)
\(\dfrac{1}{1999}B=\dfrac{1999^{2000}+1}{1999^{2000}+1999}\)
\(\dfrac{1}{1999}B=\dfrac{1999^{2000}}{1999^{2000}}-\dfrac{1998}{1999^{2000}+1999}\)
\(\dfrac{1}{1999}B=1-\dfrac{1998}{1999^{2000}+1999}\)
Vì \(\dfrac{1998}{1999^{1999}+1999}>\dfrac{1998}{1999^{2000}+1999}=>\dfrac{1}{1999}A< \dfrac{1}{1999}B=>A< B\)
\(A=\dfrac{1999^{1999}+1}{1999^{1998}+1}=\dfrac{\left(1999^{1999}+1\right)^2}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)
\(A=\dfrac{\left(1999^{1999}\right)^2+2.1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\left(1\right)\)
\(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}=\dfrac{\left(1999^{2000}+1\right)\left(1999^{1998}+1\right)}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)
\(B=\dfrac{\left(1999.1999^{1999}+1\right)\left(\dfrac{1}{1999}.1999^{1999}+1\right)}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)
\(B=\dfrac{\left(1999^{1999}\right)^2+1999.1999^{1999}+\dfrac{1}{1999}.1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)
\(B=\dfrac{\left(1999^{1999}\right)^2+\left(1999+\dfrac{1}{1999}\right).1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\left(2\right)\)
mà \(\left(1999+\dfrac{1}{1999}\right)>2\)
\(\left(1\right).\left(2\right)\Rightarrow A< B\)
Sửa lại :
\(=\frac{1999.2000+1999-1}{1998+1999.2000}.\frac{7}{5}=\frac{1999.2000+1998}{1998+1999.2000}.\frac{7}{5}=1.\frac{7}{5}=\frac{7}{5}\)
\(\frac{1999.2000-1}{1998+1999.2000}.\frac{7}{5}\)
= \(\frac{-1}{1998}\) . \(\frac{7}{5}\)
= \(\frac{-7}{9990}\)
Xx1999=1999x199,8
Xx1999=399400,2
x=399400,2:1999
x=199,8
Vì x nhân 1999=1999 nhân 199,8 nên x = 199,8
k mk nha!!!cảm on bạn rất nhiều