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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)
Ta có: \(A=\left(\dfrac{x-2}{x+2}+\dfrac{x}{x-2}+\dfrac{2x+4}{4-x^2}\right)\cdot\left(x+\dfrac{5}{x-3}\right)\)
\(=\left(\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}+\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x+4}{\left(x-2\right)\left(x+2\right)}\right)\cdot\left(\dfrac{x\left(x-3\right)+5}{\left(x-3\right)}\right)\)
\(=\dfrac{x^2-4x+4+x^2+2x-2x-4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2-3x+5}{x-3}\)
\(=\dfrac{2x^2-4x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2-3x+5}{x-3}\)
\(=\dfrac{2x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2-3x+5}{x-3}\)
\(=\dfrac{2x\left(x^2-3x+5\right)}{\left(x+2\right)\left(x-3\right)}\)
\(\left(3x-4\right)^2-2\left(3x-4\right)\left(x-4\right)+\left(x-4\right)^2\)
\(=\left(3x-4-x+4\right)^2\)
\(=4x^2\)
\(\left(3x-4\right)^2+\left(4-x\right)^2-2\left(3x-4\right)\left(x-4\right)=\left(3x-4\right)^2-2\left(3x-4\right)\left(x-4\right)+\left(x-4\right)^2=\left(3x-4-x+4\right)^2=\left(2x\right)^2=4x^2\)
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 \(A=25\cdot4^{1994}\)