1250 cm

Ta có : \(3^4=\overline{...1}\)

<=>  \(\left(3^4\right)^{502}=\overline{...1}\)

<=> \(\left(3^4\right)^{502}\cdot3=\overline{...3}\)

<=> \(3^{2009}=\overline{...3}\)(1)

Và \(7^8=\overline{...1}\)

<=> \(\left(7^8\right)^{251}=\overline{...1}\)

<=> \(7^{2008}\cdot7^2=\overline{...9}\)

<=> \(7^{2010}=\overline{...9}\)(2)

Và \(13^4=\overline{...1}\)

<=> \(\left(13^4\right)^{502}=\overline{...1}\)

<=> \(\left(13^4\right)^{502}\cdot13^3=\overline{...7}\)(3)

Từ (1)(2)(3)=> b= \(3^{2009}\cdot7^{2010}\cdot13^{2011}=\overline{...3}\cdot\overline{...7}\cdot\overline{...9}=\overline{...9}\)

Vậy chữ số hàng đơn vị của b là 9.

Ta có: A=\(\left(2x+3y\right)\left(5x-2y\right)=10x^2-4xy+15xy-6y^2\)

              = \(10x^2+11xy-6y^2\)

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Đề sai , thay n=1 ko ra .

* Ta có: A= \(-x^2-4x+15=-\left(x^2+4x+4\right)+19\)

                    =  \(-\left(x+2\right)^2+19=19-\left(x+2\right)^2\le19\)

Dấu = xảy ra khi x=-2

Vậy max A=19 khi x=-2

* Ta có: B=\(-3x^2-4x+8=-3\left(x^2+2\cdot\frac{2}{3}+\frac{16}{36}\right)+\frac{28}{3}\)

                 =  \(-3\left(x+\frac{2}{3}\right)^2+\frac{28}{3}=\frac{28}{3}-3\left(x+\frac{2}{3}\right)^2\le\frac{28}{3}\)

Dấu = xảy ra khi x=\(-\frac{2}{3}\)

Vậy max B= \(\frac{28}{3}\)khi x=\(-\frac{2}{3}\)

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