Tính nhanh:
a)\(\frac{2018x2019-2017}{2017x2018+2019}\) b)\(\frac{2018x2018}{2017x2019+1}\)
so sánh phân số 2018x2018/2017x2019 và 1
A.2018x2018/2017x2019 = 1
B.2018X2018/2017X2019 > 1
C.2018x2018/2017x2019 <1
D.Không so sánh được
Ta có :
2018 x 2018 = ( 2017 + 1 ) x ( 2019 - 1 )
= ( 2017 + 1 ) x 2019 - ( 2017 + 1 )
= 2017 x 2019 + 2019 - 2017 - 1
= 2017 x 2019 + 1 > 2017 x 2019
\(\Rightarrow\frac{2018\times2018}{2017\times2019}=\frac{2017\times2019+1}{2017\times2019}=1+\frac{1}{2017\times2019}>1\)
Vậy ta chọn B
~~Học tốt~~
các bn chọ hộ mình, đúng mình k cho
\(\frac{2016+2017x2018}{2017x2019-1}\)
\(\frac{2016+2017.2018}{2017.2019-1}\)
\(=\frac{\left(2016+1\right)+2017.2018-1}{2017.2019-1}\)
\(=\frac{2017+2017.2018-1}{2017.2019-1}\)
\(=\frac{2017.\left(1+2018\right)-1}{2017.2019-1}\)
\(=\frac{2017.2019-1}{2017.2019-1}=1\)
\(\frac{2016+2017\times2018}{2017\times2019-1}\)
\(=\frac{2016+2017\times2018}{2017\times\left(2018+1\right)-1}\)
\(=\frac{2016+2017\times2018}{2017\times2018+2017-1}\)
\(=\frac{2016+2017\times2018}{2017\times2018+2016}\)
\(=1\)
__CHÚC BN HOK TỐT__
\(\frac{2016+2017x2018}{2017x2019-1}\) rút gọn phân số
\(\frac{2016+2017.2018}{2017.2019-1}\)
= \(\frac{2016+2017.2018}{2017.2018+2017-1}\)
= \(\frac{2016+2017.2018}{2017.2018+2016}\)
= 1
\(\frac{2017x2018+1}{2019+2016x2018}\)Tính nhanh
Ta có :
\(\frac{2017\times2018+1}{2019+2016\times2018}\)
\(=\frac{2017\times2018+1}{1+2018+2016\times2018}\)
\(=\frac{2017\times2018+1}{1+2018\times\left(2016+1\right)}\)
\(=\frac{2017\times2018+1}{1+2018\times2017}\)
\(=1\)
\(\frac{2017.2018+1}{2019+2016.2018}\)
\(=\frac{2017.2018+1}{1+2018+2016.2018}\)
\(=\frac{2017.(2018+1)}{(1+2018).\left(2016+1\right)}\)
\(=\frac{2017.2019}{2019.2017}\)
\(=\frac{1}{1}=1\)
\(B=\dfrac{2017x2018+1000}{2018x2018-1018}\) help meeeeeeeeeeeeeeeeeeeeeee
\(B=\dfrac{2017\times2018+1000}{2017\times2018+2018-1018}\\ B=\dfrac{2017\times2018+1000}{2017\times2018+1000}\\ B=1\)
so sánh 2 phân số
A = \(\frac{2017x2018-1}{2017x2018-2}\)và B = \(\frac{2017}{2018}\)
Ta có:
\(A=\frac{2017\cdot2018-1}{2017\cdot2018-2}\)
\(A=\frac{2017\cdot2018-2+1}{2017\cdot2018-2}\)
\(A=\frac{2017\cdot2018-2}{2017\cdot2018-2}+\frac{1}{2017\cdot2018-2}\)
\(A=1+\frac{1}{2017\cdot2018-2}\)
Ta có phân số trung gian là 1. Ta có:
\(A>1\) ; \(B< 1\)
\(\Rightarrow A>1>B\)
\(\Rightarrow A>B\)
Vậy A>B
Chúc em học tốt!
\(\Rightarrow\text{❤️✔✨♕✨✔️❤ }\Leftarrow\)
\(\text{Ta có :}\)
\(A=\frac{2017\cdot2018-1}{2017\cdot2018-2}=\frac{4070305}{4070304}=1\frac{1}{4070304}\)
\(B=\frac{2017}{2018}\)
\(\text{Vì : }1\frac{1}{4070304}>1\text{ mà }\frac{2017}{2018}< 1\text{ nên }1\frac{1}{4070304}>\frac{2017}{2018}\)
\(\Rightarrow A>B\)
tínhnhanh
2018x2018-2017x2019
\(2018\cdot2018-2017\cdot2019\)
\(=2018\cdot2018-2017\left(2018-1\right)\)
\(=2018\cdot2018-2017\cdot2018-2017\cdot1\)
\(=2018\left(2018-2017\right)-2017\cdot1\)
\(=2018\cdot1-2017\cdot1\)
\(=1\)
\(2018\cdot2018-2017\cdot2019\)
\(=(2019-2018)\cdot(2018-2017)\)
\(=1\)
\(2018\times2018-\left(2018-1\right)\times\left(2018+1\right)=2018\times2018-2018\times2018-2018+2018-1\)
\(=-1\)
A=2018x2019-1
2017x2018+807x5
A=4074342-1=4074341
2017x2018=4070306+4035=4074341
Cho a(b+c)2+b(c+a)2+c(a+b)2=4abc và a2017+b2017+c2017=1
Tính A=\(\frac{1}{a^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}\)
Ta có a(b+c)^2 +b(c+a)^2+c(a+b)^2 =4abc
ab^2+ac^2+2abc+ba^2bc^2+2abc+ca^2+cb^2+2abc=4abc
ab^2+ac^2+bc^2+ba^2+cb^2+ca^2+2abc=0
(ab^2+abc)+(ac^2+abc)+(bc^2+cb^2)+(a^2b+a^2c)=0
ab(b+c)+ac(b+c)+bc(b+c)+a^2(b+c)=0
(b+c)(ab+ac+bc+a^2)=0
(b+c)(a+b)(a+c)=0
*th1:b+c=0=> b=-c
=> b^2017 +c^2017 =0
mà a^2017 +b^2017 +c^2017=1
=>a^2017=1 => a=1
thay vào A rồi dc A=1
các th khác tương tự