2018.20172017-2017.20182018=0

Giusp m với. Tìm số A biết. 121:A dư 10, 61 : A dư 10. giải thích cách làm theo lớp 6 b nhé.Thank

\(A=2018\left(20182017-2017\cdot10001\right)+2019\)

=20180000+2019

=20182019

A = 2018 . 20182017 - 2017 . 20182018 + 2019

= 2018 . 20182017 - 2017 . 10001 . 2018 + 2019

= 2018 . (20182017 - 20172017) + 2019

= 2018 . 10000 + 2019

= 20182019

Không bt có đúng ko nx

A=2018.20182017-2017.20182018+2019

A=2018.20182017-2017.2018.10001+2019

A=2018(20182017-20172017)+2019

A=2018.10000+2019

A=20180000+2019=20182019

chắc chắn đúng nhé :)

ĐKXĐ: a>0; a<>1

a: Sửa đề: \(M=\frac{a\cdot\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)

TA có: \(M=\frac{a\cdot\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)

\(=\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\frac{a-1}{\sqrt{a}}\cdot\frac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{a-1}\)

\(=\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}+\frac{a+1+a+1}{\sqrt{a}}=\frac{a+\sqrt{a}+1-a+\sqrt{a}-1+2a+2}{\sqrt{a}}=\frac{2a+2\sqrt{a}+2}{\sqrt{a}}\)

b: M=7

=>\(2a+2\sqrt{a}+2=7\sqrt{a}\)

=>\(2a-5\sqrt{a}+2=0\)

=>\(\left(\sqrt{a}-2\right)\left(2\sqrt{a}-1\right)=0\)

=>\(\left[\begin{array}{l}\sqrt{a}-2=0\\ 2\sqrt{a}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{a}=2\\ 2\sqrt{a}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}a=4\left(nhận\right)\\ a=\frac14\left(nhận\right)\end{array}\right.\)

c: M>6

=>\(\frac{2a+2\sqrt{a}+2}{\sqrt{a}}-6>0\)

=>\(\frac{2a-4\sqrt{a}+2}{\sqrt{a}}>0\)

=>\(\frac{2\left(\sqrt{a}-1\right)^2}{\sqrt{a}}>0\) (luôn đúng với mọi a thỏa mãn ĐKXĐ)