1 + 1 = ?
giup mk nhe
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\(2x^2-6x-1=0\)
Theo Vi - ét, ta có :
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{6}{2}=3\\x_1x_2=\dfrac{c}{a}=-\dfrac{1}{2}\end{matrix}\right.\)
Ta có :
\(A=\dfrac{x_1-2}{x_2-1}+\dfrac{x_2-2}{x_1-1}\)
\(=\dfrac{\left(x_1-2\right)\left(x_1-1\right)+\left(x_2-2\right)\left(x_2-1\right)}{\left(x_2-1\right)\left(x_1-1\right)}\)
\(=\dfrac{x_1^2-x_1-2x_1+2+x_2^2-x_2-2x_2+2}{x_1x_2-\left(x_1+x_2\right)+1}\)
\(=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-3\left(x_1+x_2\right)+4}{x_1x_2-\left(x_1+x_2\right)+1}\)
\(=\dfrac{3^2-2.\left(-\dfrac{1}{2}\right)-3.3+4}{-\dfrac{1}{2}-3+1}\)
\(=-2\)
3S= 1.2.(3-0)+ 2.3.(4-1)+...+ n.(n+1).[(n+2)-(n-1)]
=[1.2.3+ 2.3.4+...+ (n-1)n(n+1)+ n(n+1)(n+2)]- [0.1.2+ 1.2.3+...+(n-1)n(n+1)]
=n(n+1)(n+2)
=>S
Ta có: (a-1)2=(a-1)(a-1)
=a(a-1)-1(a-1)
=(a2-a)-(a-12)
=a2-2a+1
\(S1=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(S1=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(S1=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(S1=1-\frac{1}{10}\)
\(S1=\frac{9}{10}\)
CHÚC BN HC GIỎI !!!!!!!!!! TỨ DIỆP THẢO
S=\(\frac{1}{1.2}+\frac{1}{2.3}+...............+\frac{1}{9.10}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...............+\frac{1}{9}-\frac{1}{10}\)
=\(1-\frac{1}{10}\)
=\(\frac{9}{10}\)
Trả lời :
1 + 1 = 1 x 2 = 2
~ Hok tốt ~
1 + 1 = 2
HỌC TỐT