c/ m à

Gọi số thứ nhất là a

số thứ 2 là b

Theo bài ra ta có \(\left\{{}\begin{matrix}a=5k+3\left(k\in Z\right)\\b=10n+7\left(n\in N\right)\end{matrix}\right.\)

Suy ra \(a^2+b^2=\left(5k+3\right)^2+\left(10n+7\right)^2\)

=\(25k^2+30k+9+100n^2+140n+49\)

=\(25k^2+30k+100n^2+140n+58\)

Vì \(\left\{{}\begin{matrix}25k^2⋮5\\30k⋮5\\100n^2⋮5\\140n⋮5\end{matrix}\right.\)

Mà 58 chia 5 dư 3

Vậy tổng bình phương của hai số này chia cho 5 dư 3

a) \(\dfrac{1}{3}.\left(-\dfrac{4}{5}\right)+\dfrac{1}{3}.\left(-\dfrac{6}{5}\right)\)

=\(\dfrac{1}{3}.\left[\left(-\dfrac{4}{5}\right)+\left(-\dfrac{6}{5}\right)\right]\)

=\(\dfrac{1}{3}.\left(-\dfrac{10}{5}\right)\)

=\(\dfrac{1}{3}.\left(-2\right)\)

=\(-\dfrac{2}{3}\)

\(\left|x-3\right|=x\)

Suy ra x-3=x hoặc x-3=-x

+) x-3=x

3=0(Vô lí)

+) x-3=-x

2x=3

x=\(\dfrac{3}{2}\)

a) \(\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)\)

=\(a^3+b^3+\left(a^3-b^3\right)\)

=\(a^3+b^3+a^3-b^3\)

=\(2a^3\)

b) \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

=\(\left(a+b\right)\left(a^2-2ab+b^2-ab\right)\)

=\(\left(a+b\right)\left[\left(a^2-2ab+b^2\right)-ab\right]\)

=\(\left(a+b\right)\left[\left(a-b\right)^2-ab\right]\)

\(5^5-5^4+5^3=5^3.\left(5^2-5+1\right)=5^3.\left(25-5+1\right)\)

=\(5^3.21\)

Vì 21 chia hết cho 7

Suy ra \(5^3.21⋮7\)

Vậy \(5^5-5^4+5^3⋮7\)

Vì \(\widehat{O1}=\widehat{O2}\)

Mà \(\widehat{O1}+\widehat{O2}=180^0\)

Suy ra \(\widehat{O1}=\widehat{O2}=90^0\)

Suy ra xx' vuông góc với yy'

\(3^{27}=3^{25}.3^2=3^{25}.9\)

\(30.3^{24}=3.10.3^{24}=3^{25}.10\)

Vì \(\left\{{}\begin{matrix}3^{25}=3^{25}\\9< 10\end{matrix}\right.\)

Suy ra \(3^{25}.9< 3^{25}.10\)

Suy ra \(3^{27}< 30.3^{24}\)

đề