2013²⁰=.....3 ^2.10=.....9 ¹⁰=.......1
11²=......1
.....1 - .....1 = .....0
vì hiệu có tận cùng bằng 0 nên hiệu đó chia hết cho 10
tick mk nha

\(C=1.99+2.98+3.97+...+97.3+98.2+99.1\)

\(=1.99+2(99-1)+3(99-2)+...+97.(99-96)+98(99-97)+99(99-98)\)

\(=(1.99+2.99+3.99+...+97.99+98.99+99.99)-(1.2+2.3+...+96.97+97.98+98.99)\)

\(=99.(1+2+3+...+99)- (1.2+2.3+...+96.97+97.98+98.99)\)

\(=99.\frac{99.(99+1)}{2} - (1.2+2.3+...+96.97+97.98+98.99)\)

\(=490050 - (1.2+2.3+...+96.97+97.98+98.99)\)

Từ đây tính vế trong ngoặc ra rồi thực hiện trừ là oke

a) \(\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)

\(=\dfrac{\sqrt{x}\sqrt{y}\cdot\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)

\(=\sqrt{x}\sqrt{y}\)

\(=\sqrt{xy}\)

\(36\cdot\left(28+45\right)\cdot73\)

\(=36\cdot73\cdot73\)

\(=191844\)

Ta có:

+) \(\dfrac{106}{204}=\dfrac{\dfrac{21012}{19}}{21012}\)

+) \(\dfrac{107}{206}=\dfrac{10914}{21012}\)

Vì \(10914-\dfrac{21012}{19}=\dfrac{186354}{19}\) (số dương)

=> \(10914>\dfrac{21012}{19}\)

do đó: \(\dfrac{\dfrac{21012}{19}}{19}< \dfrac{10914}{21012}\)

hay \(\dfrac{106}{204}< \dfrac{107}{206}\)

Ko cần biết âu

100000000 con

tick đi nhé bn

Bài 1, 2, 3 trang 12 sgk sinh học 7 - loigiaihay.com

\(\left(\frac{3}{8}\right)^5=\frac{243}{8^5}>\frac{125}{8^5}>\frac{125}{9^5}=\frac{125}{\left(3^2\right)^5}=\frac{125}{3^{10}}>\frac{125}{3^{15}}=\frac{5^3}{\left(3^5\right)^3}=\left(\frac{5}{243}\right)^3\)

\(=>\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)