1. So sánh: \(\left(26^{2018}+3^{2018}\right)^{2019}\)và \(\left(26^{2019}+3^{2019}\right)^{2018}\)
2. Tìm hai số \(a,b\inℕ^∗\)sao cho: \(a+2⋮b;b+3⋮a\)
So sánh :\(\left(26^{2018}+3^{2018}\right)^{2019}\)và \(\left(26^{2019}+3^{2019}\right)^{2018}\)
so sanh \(\left(26^{2018}+3^{2018}\right)^{2019}\)
và \(\left(26^{2019}+3^{2019}\right)^{2018}\)
Ta có: \(\left(26^{2018}+3^{2018}\right)^{2019}=26^{2018\cdot2019}+3^{2018\cdot2019}\left(1\right)\)
\(\left(26^{2019}+3^{2019}\right)^{2018}=26^{2019\cdot2018}+3^{2019\cdot2018}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left(26^{2018}+3^{2018}\right)^{2019}=\left(26^{2019}+3^{2019}\right)^{2018}\)
So sánh : \(\left(26^{2018}+3^{2018}\right)^{2019}\) và \(\left(26^{2019}+3^{2019}\right)^{2018}\)
\(A=\left(26^{2018}+3^{2018}\right)^{2019}\)
\(B=\left(26^{2019}+3^{2019}\right)^{2018}\)
\(B=\left(26^{2018}.26+3.3^{2018}\right)^{2018}< \left(26^{2018}.26+3^{2018}.26\right)^{2018}\)
\(B< \left(26^{2018}+3^{2018}\right)^{2018}.26^{2018}< \left(26^{2018}+3^{2018}\right)^{2018}.\left(26^{2018}+3^{2018}\right)\)
\(\Rightarrow B< \left(26^{2018}+3^{2018}\right)^{2019}\Rightarrow B< A\)
cho A+B+C=0 và AB+AC+BC=0
TÍNH M=\(\left(A-2018\right)^{2019}+\left(B-2018\right)^{2019}-\left(C+2018\right)^{2019}\)
Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(\Rightarrow a^2+b^2+c^2=\left(a+b+c\right)^2-2\left(ab+bc+ac\right)\)
hay \(a^2+b^2+c^2=0\Rightarrow a=b=c=0\)
Thay a = b = c = 0 vào M rồi tính như bình thường nha bạn!
Ta có :
\(a+b+c=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)^2=0\)
\(\Leftrightarrow\)\(a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Leftrightarrow\)\(a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow\)\(a^2+b^2+c^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}a^2=0\\b^2=0\\c^2=0\end{cases}\Leftrightarrow a=b=c=0}\)
\(\Rightarrow\)\(M=\left(a-2018\right)^{2019}+\left(b-2018\right)^{2019}-\left(c+2018\right)^{2019}\)
\(\Rightarrow\)\(M=-2018^{2019}-2018^{2019}-2018^{2019}\)
\(\Rightarrow\)\(M=-3.2018^{2019}\)
Chúc bạn học tốt ~
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). CMR:\(\dfrac{\left(a^{2018}+b^{2018}\right)^{2019}}{\left(c^{2018}+d^{2018}\right)^{2019}}=\dfrac{\left(a^{2019}-b^{2019}\right)^{2020}}{\left(c^{2019}+d^{2019}\right)^{2020}}\)
HELP ME!!!!!!! Mình cần gấp mai mình lộp bài rùi
Cứu mình với 9:00 sáng nay mình nộp bài rùi
a) Cho các số dương a,b,c,d; c khác d và \(\frac{a}{b}\)=\(\frac{c}{d}\). Chứng minh rằng : \(\frac{\left(a^{2018}+b^{2018}\right)^{2019}}{\left(c^{2018}+d^{2018}\right)^{2019}}\)=\(\frac{\left(a^{2019}-b^{2019}\right)^{2018}}{\left(c^{2019}-d^{2019}\right)^{2018}}\)
b) Cho biết |3x + 2y| + |5z - 7x| + \(\left(xy+yz+xz-500\right)^{2022}\)= 0 . Tính giá trị : \(A=\left(3x-y-z\right)^{2021}\)
Các bạn giải giúp mik nhé. Mik cần gấp lắm. Ai giải trc mik sẽ tick cho
Cho a,b,c thỏa mãn:\(a^2+b^2+c^2=ab+bc+ca\) và \(a^{2019}+b^{2019}+c^{2019}=3^{2020}\)
Tính \(A=\left(a-2\right)^{2017}+\left(b-3\right)^{2018}+\left(c-4\right)^{2019}\)
<=> \(2a^2+2b^2+2c^2=2ab+2bc+2ca< =>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0< =>\)
a=b=c => 32020 = 3.a2019 <=> 32019 = a2019 => a=b=c=3
A= 12017 + 02018 + (-1)2019 = 0
Cho \(A=1-\frac{2017}{2019}+\left(\frac{2017}{2019}\right)^2-\left(\frac{2017}{2019}\right)^3+...+\left(\frac{2017}{2019}\right)^{2018}\)
Chứng minh A không là số nguyên.
so sánh A=2018^2019 -1/2018^2019+1 và B = 2018^2019/2018^2019+2
Ta có: B = (2018 + 2019)/(2019 + 2020) = (2018 + 2019)/4039 = 2018/4039 + 2019/4039
Ta thấy : 2018/2019 > 2018/4039
2019/2020 > 2019/4039
=> 2018/2019 + 2019/2020 > 2018/4039 > 2019/4039
=> 2018/2019 + 2019/2020 > (2018 + 2019)/(2019 + 2020)
=> A > B