1+1/3+1/6+1/10+...+2/(x.x+1)=1994/1993
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\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{\frac{x\left(x+1\right)}{2}}=1\frac{1993}{1991}\)
\(\Leftrightarrow\left(1\cdot\frac{1}{2}\right)+\left(\frac{1}{3}\cdot\frac{1}{2}\right)+\left(\frac{1}{6}\cdot\frac{1}{2}\right)+....+\left(\frac{1}{\frac{x\left(x+1\right)}{2}}\cdot\frac{1}{2}\right)=1\frac{1993}{1991}\div2\)
\(\Leftrightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{x\left(x+1\right)}=\frac{1992}{1991}\)
\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1991}\)
\(\Leftrightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1992}{1991}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{1992}{1991}\)
\(\Leftrightarrow\frac{1}{x+1}=-\frac{1}{1991}\)
\(\Leftrightarrow x=-1992\)
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{x\left(x+1\right):2}=1\frac{1994}{1993}\)
\(< =>1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{3987}{1993}\)
\(< =>1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{3987}{1993}\)
\(< =>1+2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{3987}{1993}\)
\(< =>1+2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{3987}{1993}< =>2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{3987}{1993}-1=\frac{1994}{1993}\)
\(< =>\frac{1}{2}-\frac{1}{x+1}=\frac{1994}{1993}:2=\frac{997}{1993}< =>\frac{1}{x+1}=\frac{1}{2}-\frac{997}{1993}=-\frac{1}{3986}\)
<=>x=-3987
\(\Rightarrow\frac{1}{2}.\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}\right)=\frac{1}{2}.1\frac{1994}{1993}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}=\frac{3987}{3986}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3987}{3986}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{3987}{3986}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{-3986}\)
=> x + 1 = -3986
=> x = -3987
a/ A = B
vì \(\frac{10^{1993}+10}{10^{1993}+1}=1\)và \(\frac{10^{1994}+10}{10^{1994}+1}=1\)
Học tốt
A = B
vì \(\frac{10^{1993}+10}{10^{1993}+1}=10\) và \(\frac{10^{1994}+10}{10^{1994}+1}=10\)
học tốt
\(A=\frac{10^{1993}+10}{10^{1993}+1}\)
\(=\frac{10^{1993}+1+9}{10^{1993}+1}\)
\(=\frac{10^{1993}+1}{10^{1993}+1}+\frac{9}{10^{1993}+1}\)
\(=1+\frac{9}{10^{1993}+1}\)( 1 )
\(B=\frac{10^{1994}+10}{10^{1994}+1}\)
\(=\frac{10^{1994}+1+9}{10^{1994}+1}\)
\(=\frac{10^{1994}+1}{10^{1994}+1}+\frac{9}{10^{1994}+1}\)
\(=1+\frac{9}{10^{1994}+1}\)( 2 )
Vì \(\frac{9}{10^{1993}+1}>\frac{9}{10^{1994}+1}\)( 3 )
Từ ( 1 )( 2 )( 3 )\(\Rightarrow1+\frac{9}{10^{1993}+1}>1+\frac{9}{10^{1994}+1}\)
\(\Rightarrow A>B\)
1-2-3+4+5-6-7+8+...+1993-1994
=(1-2-3+4)+(5-6-7+8)+...+(1993-1994)
=0+0...+-1
=-1
\(\left(1-2-3+4\right)+\left(5-6-7+8\right)+...+\left(1993-1994\right)\)
\(=0+0+....+\left(-1\right)\)
\(=\left(-1\right)\)
Ta có 1-2-3+4+5-6-7+8+...+1993-1994
=(1-2-3+4)+(5-6-7+8)+...+(1889-1990-1991+1992)+1993-1994
= 0 + +0 +.....+ 0 + 1993-1994
= -1