(1+1/3). (1+1/8). (1+1/15). .. (1+1/24) =100/101x
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1. cho f(x)=x8-101x7+101x6-101x5+...+101x2-101x+25. tính f(100)
f(x)=x8-101x7+101x6-101x5+...+101x2-101x+25
=x8-(100+1)x7+(100+1)x6-(100+1)x5+...+(100+1)x2-(100+1)x+25
f(100 ) hay x= 100
Thay 100 = x ,có :
=x8-(x+1)x7+(x+1)x6-(x+1)x5+...+(x+1)x2-(x+1)x+25
= x8 - x8 - x7+ x7 + x6 - x6 - x5 + x5 + .......................+ x3 + x2 - x2 + x + 25
= x+ 25
f(100 0 = 100 + 25 = 125
Vậy f(100 ) =125
Câu 2:
\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
Có \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge0\)
do đó phương trình ban đầu tương đương với:
\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)
\(\Leftrightarrow100x+\left(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\right)=101x\)
\(\Leftrightarrow x=\frac{100.101}{2.101}=50\)
A = \(1\dfrac{13}{15}.25.\dfrac{1}{100}.3+\left(\dfrac{8}{15}-\dfrac{78}{60}\right)\div1\dfrac{23}{24}\)
A = \(\dfrac{1.15+13}{15}.25.\dfrac{1}{100}.3+\left(\dfrac{32}{60}-\dfrac{78}{60}\right)\div\dfrac{1.24+23}{24}\)
A = \(\dfrac{28}{15}.25.\dfrac{1}{100}.3+\left(-\dfrac{23}{30}\right)\div\dfrac{47}{24}\)
A = \(\left(\dfrac{28}{15}.25\right).\left(\dfrac{1}{100}.3\right)+\left(-\dfrac{23}{30}\right)\div\dfrac{47}{24}\)
A = \(\left(\dfrac{28.25}{15}\right).\left(\dfrac{1.3}{100}\right)+\left(-\dfrac{23}{30}\right)\div\dfrac{47}{24}\)
A = \(\dfrac{140}{3}.\dfrac{3}{100}+\left(-\dfrac{23}{30}\right)\div\dfrac{47}{24}\)
A = \(\dfrac{7}{5}+\left(-\dfrac{92}{235}\right)=\dfrac{237}{235}\)
\(A=1\dfrac{13}{15}.25.\dfrac{1}{100}.3+\left(\dfrac{8}{15}-\dfrac{78}{60}\right):1\dfrac{23}{24}\\ =\dfrac{28}{15}.25.\dfrac{1}{100}.3-\dfrac{23}{30}:\dfrac{47}{24}\\ =\dfrac{7}{5}-\dfrac{23}{30}.\dfrac{24}{47}\\ =\dfrac{7}{5}-\dfrac{92}{235}\\ =\dfrac{7.47}{235}-\dfrac{92}{235}\\ =\dfrac{237}{235}=1\dfrac{2}{235}\)