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Ở ngoặc đầu tiên của A thì mỗi số đều chia hết cho 2(dựa vào cơ số).

Vế tiếp theo thì toàn số lẻ lũy thừa lên chia 2 dư 1,mà có 4 số nên chia hết cho 2.

Vậy hiệu của chúng,tức A chia hết cho 2.

2006 là số chẵn lũy thừa lên chia hết cho 2 còn số kia lẻ nên chia 2 dư 1.

Vậy chia 2 dư 1.

Chúc em học tốt^^

Bài 2:Tìm x biết

\\(\\left(4x+3\\right)^3+\\left(5-7x\\right)^3+\\left(3x-8\\right)^3=0\\)

\\(\\Leftrightarrow\\left[\\left(4x\\right)^3+3.\\left(4x\\right)^2.3+3.4x.3^2+3^3\\right]+\\left[5^3-3.5^2.7x+3.5.\\left(7x\\right)^2-\\left(7x\\right)^3\\right]+\\left[\\left(3x\\right)^3-3.\\left(3x\\right)^2.8+3.3x.8^2-8^3\\right]=0\\)

\\(\\Leftrightarrow64x^3+144x^2+108x+27+125-525x+735x^2-343x^3+27x^3-216x^2+576x-512=0\\)

\\(\\Leftrightarrow-252x^3+663x^2+159x-360=0\\)

\\(\\Leftrightarrow3\\left(-84x^3+221x^2+53x-120\\right)=0\\)

Bài 2: Đặt \(4x+3=a;5-7x=b;3x-8=c\Rightarrow a+b+c=0\)

Kết hợp với đề bài ta có \(\left\{{}\begin{matrix}a^3+b^3+c^3=0\\a+b+c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^3+b^3+c^3-3abc+3abc=0\\a+b+c=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc=0\left(1\right)\\a+b+c=0\left(2\right)\end{matrix}\right.\)

Thay (2) vào (1) suy ra \(3abc=0\Leftrightarrow a=0\text{hoặc }b=0\text{hoặc }c=0\)

+) a = 0 suy ra \(x=-\frac{3}{4}\)

+) b = 0 suy ra \(x=\frac{5}{7}\)

+) c = 0 suy ra \(x=\frac{8}{3}\)

Vậy...

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3+25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3+5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)

\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1+7\right)}{5^9.7^3\left(1+2^3\right)}\)

\(=\frac{2}{12}-\frac{5.8}{9}=\frac{1}{6}-\frac{40}{9}=\frac{-77}{18}\)

b ) 3ⁿ⁺² - 2ⁿ⁺² + 3ⁿ - 2ⁿ

= ( 3ⁿ⁺² + 3ⁿ ) - ( 2ⁿ⁺² + 2ⁿ )

= 3ⁿ ( 3² + 1 ) - 2ⁿ ( 2² + 1 )

= 3ⁿ.10 - 2^n-1.2.5

= 3ⁿ.10 - 2^n-1.10

= ( 3ⁿ - 2^n-1 ).10 chia hết cho 10 ( đpcm )

\(A=8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)

\(A=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)

\(A=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}\)

\(A=\dfrac{28}{7}-\dfrac{31}{9}\)

\(A=4-\dfrac{31}{9}\)

\(A=\dfrac{4}{1}-\dfrac{31}{9}\)

\(A=\dfrac{36}{9}-\dfrac{31}{9}\)

\(\Rightarrow A=\dfrac{5}{9}\)

\(B=\left(10\dfrac{2}{9}+2\dfrac{3}{5}\right)-6\dfrac{2}{9}\)

\(B=\left(\dfrac{92}{9}+\dfrac{13}{5}\right)-\dfrac{56}{9}\)

\(B=\dfrac{92}{9}-\dfrac{56}{9}+\dfrac{13}{5}\)

\(B=\dfrac{36}{9}+\dfrac{13}{5}\)

\(B=4+\dfrac{13}{5}\)

\(B=\dfrac{20}{5}+\dfrac{13}{5}\)

\(\Rightarrow B=\dfrac{33}{5}\)

A = \(8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)

A = \(8\dfrac{2}{7}-3\dfrac{4}{9}-4\dfrac{2}{7}\)\

A = \(\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)

A = \(\left(8-4\right)-3\dfrac{4}{9}\)

A = \(4-3\dfrac{4}{9}\)

A = \(\dfrac{36}{9}-\dfrac{31}{9}\)

A = \(\dfrac{5}{9}\).

B = \(\left(10\dfrac{2}{9}+2\dfrac{3}{5}\right)-6\dfrac{2}{9}\)

B = \(10\dfrac{2}{9}+2\dfrac{3}{5}-6\dfrac{2}{9}\)

B = \(\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)

B = \(4+2\dfrac{3}{5}\)

B = \(\dfrac{20}{5}+\dfrac{13}{5}\)

B = \(\dfrac{33}{5}\).

A= \(8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)=8\dfrac{2}{7}-3\dfrac{4}{9}-4\dfrac{2}{7}=8\dfrac{2}{7}-4\dfrac{2}{7}-3\dfrac{4}{9}=4-3\dfrac{4}{9}=3\dfrac{9}{9}-3\dfrac{4}{9}=\dfrac{5}{9}\)

\(\frac{A}{n}=\frac{4n+4}{n}=4+\frac{4}{n}\)
\(\Rightarrow n\in U\left(4\right)\)
Lập bảng tiếp nhé!
\(\frac{B}{n}=\frac{5n+6}{n}=5+\frac{6}{n}\)
Lập bảng

\(2.\)
a)\(\left(\frac{3}{29}-\frac{1}{5}\right)\cdot\frac{29}{3}=\frac{3}{29}\cdot\frac{29}{3}-\frac{1}{5}\cdot\frac{29}{3}=1-\left(1+\frac{14}{15}\right)=1-1-\frac{14}{15}=\frac{14}{15}\)
b)\(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}=\frac{5}{9}\cdot\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)