Bài 1: Cho \(\dfrac{a}{c}=\dfrac{c}{b}\) chứng minh rằng
a) \(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\) b) \(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)
Bài 2: Tìm x, y ∈ N biết: \(25-y^2=8\left(x-2009\right)^2\)
Bài 3:
a) \(A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^3.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
B) CMR: Với mọi số nguyên dương n thì: \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
Bài 4: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
Bài 5:
a) Số A được chia thành 3 số tỉ lệ theo \(\dfrac{2}{5}:\dfrac{3}{4}:\dfrac{1}{6}\). Biết rằng tổng các bình phương của ba số đó = 24309. Tìm số A
b) Cho \(\dfrac{a}{c}=\dfrac{c}{b}\). CMR: \(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\)
Bài 4:
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)
=>\(\left(x-7\right)^{x+1}\cdot\left[\left(x-7\right)^{10}-1\right]=0\)
=>\(\left[{}\begin{matrix}x-7=0\\x-7=-1\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=6\\x=8\end{matrix}\right.\)
Bài 3:
a: Sửa đề: \(A=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
\(=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^5\left(3+1\right)}+\dfrac{5^{10}\cdot7^3\left(7-1\right)}{5^9\cdot7^3\left(1+2^3\right)}\)
\(=\dfrac{1}{3}\cdot\dfrac{2}{4}+\dfrac{5\cdot6}{9}\)
\(=\dfrac{1}{6}+5\cdot\dfrac{2}{3}=\dfrac{1}{6}+5\cdot\dfrac{4}{6}=\dfrac{21}{6}=\dfrac{7}{2}\)
b: \(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=\left(3^n\cdot9+3^n\right)-\left(2^n\cdot4+2^n\right)\)
\(=3^n\cdot10-2^n\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot10\)
\(=10\cdot\left(3^n-2^{n-1}\right)⋮10\)
