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Ta có:
A=(100^2 -99^2)+(98^2 - 97^2)+(96^2 - 95^2)+.........+(2^2 - 1)
=(100-99)(100+99) + (98-97)(98+97) + (96-95)(96+95)+........+(2-1)(2+1)
=100+99+98+97+......+2+1=5050
Ở đây mình nhóm các hạng tử rồi AD hằng đẳng thức A^2 - B^2 = (A-B)(A+B)
\(A=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+...+2+1\)
\(=\left(100+1\right).\frac{100-1}{2}=\frac{101.99}{2}=\frac{9999}{2}\)
(99-98)(99+98)+(97-96)(97+96)+...+(3-2)(3+2)+1
=99+98+97+96+...+3+2+1
=((99+1).99)/2
=4950
A=(992-982)+(972-962)+...+(32-22) +1
A= (99-98)(99+98)+(97-96)(97+96)+...+(3-2)(3+2)+1
A= 197+193+...+5+1
A=((197-1):4+1):2 . (197+1)
A=25.198
(đến bước này thì tự tính nha bạn)
a) \(413.\left(413-26\right)+169=413^2-2.13.413+13^2=\left(413-13\right)^2=160000\)
b) \(\left(625^2+3\right).\left(25^4-3\right)-5^{16}+10\)
\(=\left(5^8+3\right)\left(5^8-3\right)-5^{16}+10\)
\(=5^{16}-9-5^{16}+10=1\)
c) \(\frac{41^2+39^2+8^2.39}{41^2-39^2}=\frac{\left(41+39\right)^2}{\left(41-39\right)\left(41+39\right)}=\frac{41+39}{41-39}=\frac{80}{2}=40\)
\(1+2+....+2^{99}=2\left(1+2+....+2^{99}\right)-1-2-....-2^{99}=2^{100}-1\)
\(\Rightarrow2^{100}-\left(1+2+....+2^{99}\right)=2^{100}-\left(2^{100}-1\right)=1\)
Đặt biểu thức đã cho là A
\(\Rightarrow A=2^{100}-\left(2^{99}+2^{98}+2^{97}+......+2^2+2+1\right)\)
Đặt \(B=2^{99}+2^{98}+2^{97}+.......+2^2+2+1\)
\(\Rightarrow2B=2^{100}+2^{99}+2^{98}+.........+2^3+2^2+2\)
\(\Rightarrow2B-B=B=2^{100}-1\)
\(\Rightarrow A=2^{100}-B=2^{100}-\left(2^{100}-1\right)=2^{100}-2^{100}+1=1\)
111-98+113-96+115-94+...+207-2
=(111+113+115+...+207)-(2+4+...+96+98)
=\(\dfrac{\left(207-11\right):2+1\cdot\left(11+207\right)}{2}\)-\(\dfrac{\left(98-2\right):2+1\cdot\left(2+98\right)}{2}\)
=10791-2450=8341.
a \(\dfrac{1}{x-y}+\dfrac{2}{x+y}+\dfrac{3x}{y^2-x^2}\)
\(=\dfrac{x+y+2x-2y-3x}{\left(x-y\right)\left(x+y\right)}=\dfrac{-y}{\left(x-y\right)\left(x+y\right)}\)
b: \(\dfrac{1}{x-2}+\dfrac{1}{x+2}-\dfrac{4x-4}{x^2-4}\)
\(=\dfrac{x+2+x-2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{-2x+4}{\left(x-2\right)\left(x+2\right)}\)
=-2/x+2
c: \(\dfrac{x+1}{x+3}-\dfrac{x-1}{3-x}+\dfrac{2x-2x^2}{x^2-9}\)
\(=\dfrac{\left(x+1\right)\left(x-3\right)+\left(x-1\right)\left(x+3\right)+2x-2x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{x^2-2x-3+x^2+2x-3+2x-2x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{2x-6}{\left(x+3\right)\left(x-3\right)}=\dfrac{2}{x+3}\)
a) Ta có: \(413^2+213^2-326\cdot213\)
\(=413^2-2\cdot413\cdot213+213^2+106500\)
\(=200^2+106500\)
\(=40000+106500\)
\(=146500\)