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\(A=75\left(4^{1993}+4^{1992}+...+4^2+5\right)+31\)
\(=25\left(4-1\right)\left(4^{1993}+4^{1992}+...+4^2+4+1\right)+31\)
\(=25\left(4^{1994}+4^{1993}+...+4^3+4^2+4-4^{1993}-....-4-1\right)+31\)
\(=25.\left(4^{1994}-1\right)+31\)
\(=25.4^{1994}-25+31\)
\(=25.4^{1994}+6\)
Bài giải
\(A=75\cdot\left(4^{1993}+4^{1992}+...+4^2+4\right)+31\)
Đặt \(B=4^{1993}+4^{1992}+...+4^2+4\)
\(B=4+4^2+...+4^{1992}+4^{1993}\)
\(4B=4^2+4^3+...+4^{1993}+4^{1994}\)
\(4B-B=3B=4^{1994}-4\)
\(B=\frac{4^{1994}-4}{3}\)
Thay \(B=\frac{4^{1994}-4}{3}\) vào biểu thức ta có :
\(A=75\cdot\frac{4^{1994}-4}{3}+31\)
\(B=25\cdot3\cdot\frac{4^{1994}-4}{3}+31\)
\(B=25\cdot\left(4^{1994}-4\right)+31\)
Võ Thị Thảo Minh
em hãy sử dụng đẳng thức này để rút gọn :
a2 - b2 = (a - b)(a + b)
\(A=25.3\left(4^{1975}+4^{1974}+...+4^2+4+1\right)+25\)
\(=25\left(4-1\right)\left(4^{1975}+4^{1974}+...+4^2+4+1\right)+25\)
Áp dụng hằng đẳng thức, ta có : \(A=25\left(4^{1976}-1\right)+25=25.4^{1976}\)
Vậy \(A⋮4^{1976}\)
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=\frac{5^{32}-1}{2}\)
\(5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=5\left(4x^2-4x+1\right)+4\left(x^2+3x-x-3\right)-2\left(25-30x+9x^2\right)\)
\(=20x^2-20x+5+4x^2+12x-4x-12-50+60x-18x^2\)
\(=6x^2+48x-57\)
\(5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=5\left(4x^2-4x+1\right)+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=5\left(4x^2-4x+1\right)+4\left(x-1\right)\left(x+3\right)-2\left(25-3x+9x^2\right)\)
\(=20x^2-20x+5+4\left(x-1\right)\left(x+3\right)-2\left(25-30x+9x^2\right)\)
\(=20x^2-20x+5+4x^2+8x-12-50+60x-18x^2\)
\(=6x^2+48x-57\)
Ta có \(A=75\left(4^{1993}+4^{1992}+....+4+1\right)+25\)
\(\Leftrightarrow A=25\left(4-1\right)\left(4^{1993}+4^{1992}+...+4+1\right)+25\)
Vận dụng hằng đẳng thức
\(a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+...+b^{n-1}\right)\)
Ta có
\(\left(4-1\right)\left(4^{1993}+4^{1992}+...+4+1\right)=4^{1994}-1\)
\(\Rightarrow A=25\left(4^{1994}-1\right)+25\)
\(\Leftrightarrow A=25\cdot4^{1994}\)
Vậy \(A=25\cdot4^{1994}\)