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\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{n}{2^2}+...+\frac{100}{2^{100}}\)
\(2A=2\left(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{100}{2^{100}}\right)=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{n-1}{2^n}+...+\frac{100}{2^{99}}\)
\(2A-A=A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}-\frac{100}{2^{100}}\)
Đặt \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2B=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(\Rightarrow2B-B=B=2-\frac{1}{2^{99}}\)
\(\Rightarrow A=2-\frac{1}{2^{99}}-\frac{100}{2^{100}}< 2\)
ta có : \(a^8+b^8-a^6b^2-a^2b^6\ne\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)\)
và \(a^2b^2\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)\) cũng có thể âm
\(\Rightarrow\) sai
Câu 1:
\(a^3+a^2b-ab^2-b^3\)
\(=a^2\left(a+b\right)-b^2\left(a+b\right)\)
\(=\left(a+b\right)\left(a^2-b^2\right)\)
\(=\left(a+b\right)\left(a-b\right)\left(a+b\right)\)
\(=\left(a+b\right)^2\left(a-b\right)\)
Câu 2:
\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)+bc^3-a^3b+a^3c-b^3c\)
\(=a\left(b-c\right)\left(b^2+bc+c^2\right)-a^3\left(b-c\right)-bc\left(b-c\right)\left(b+c\right)\)
\(=\left(b-c\right)\left(ab^2+abc+c^2a-a^3-b^2c-bc^2\right)\)
\(=\left(b-c\right)\left[a\left(c-a\right)\left(c+a\right)-b^2\left(c-a\right)-bc\left(c-a\right)\right]\)
\(=\left(b-c\right)\left(c-a\right)\left(ca+a^2-b^2-bc\right)\)
\(=\left(b-c\right)\left(c-a\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)
A=(992-982)+(972-962)+.....+(32-22)+1=((98+1)2-982)+......+((2+1)2-22)+1
=(2.98+1)+(2.96+1)+....+(2.2+1)+1=50+4.(1+2+...+48+49)=50.4.(49.50/2)=50.4.49.25=245000
2a=2(2^2+2^3+2^4+...+2^100)
2a=2^3+2^4+2^5+2^101
2a-a=(2^3+2^4+...+2^101)-(2^2+2^3+...+2^100)
a=2^101-2^2
còn lại tự tính nhé