Tính A=(-1).(-1)^2.(-1)^3....(-1)^2016
TÍNH GIÁ TRỊ BIỂU THỨC A=2016+(2016/1+2)+(2016/1+2+3)+....+(2016/1+2+3+4+...+2016)
Tính A= 1/20+1/30+1/42+1/56+...+1/990
Tính S= 1/6+1/66+1/176+1/336+...+1/496*501
So sánh A và B :
A=2016^2016+2/2016^2016-1 và B=2016^2016/2016^2016-3
A = 1/2 + 1/3 +1/4 +.....+1/2016 + 1/2017 B = 2016/1 + 2015/2 + ......+ 2/2015 + 1/2016 . Tính B/A
\(\frac{B}{A}=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{\frac{2017}{1}+\frac{2017}{2}+...+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}=2017\div\frac{1}{2017}=4068289\)
Bài 3: Tính
A = ( 1 + 2 ) . 1/2 + ( 1 + 2 + 3 ) . 1/3 + ... + ( 1 + 2 + 3 +...+ 2016) . 1/2016
\(A=\left(1+2\right).\frac{1}{2}+\left(1+2+3\right).\frac{1}{3}+...+\left(1+2+3+...+2016\right).\frac{1}{2016}\)
\(A=\left(1+2\right).2:2.\frac{1}{2}+\left(1+3\right).3:2.\frac{1}{3}+...+\left(1+2016\right).2016:2.\frac{1}{2016}\)
\(A=3:2+4:2+...+2017:2\)
\(A=3.\frac{1}{2}+4.\frac{1}{2}+...+2017.\frac{1}{2}\)
\(A=\frac{1}{2}.\left(3+4+...+2017\right)\)
\(A=\frac{1}{2}.\left(3+2017\right).2015:2\)
\(A=\frac{1}{2}.2020.2015.\frac{1}{2}\)
\(A=505.2015=1017575\)
Cho A = 1/2 + 1/3 + 1/4 + ... + 1/2017 B = 1/2016 + 2/2015 +3/2014+ ...+ 2015/2 + 2016/1 Tính B : A
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
tính a: [1-1/2*2]*[1-1/3*3]*[1-1/4*4]*[1-1/5*5]*......*[1-1/2015*2015]*[1-1/2016*2016]
Bài 1: Cho a,b,c thỏa mãn (a+b-c)/c=(b+c-a)/a=(c+a-b)/b
tính P=(1+b/a)*(1+c/b)*(1+a/c)
Bài 2: Cho a+b+c=0
tính B=((a^2+b^2-c^2)*(b^2+c^2-a^2)*(c^2+a^2-b^2))/(10*a^2*b^2*c^2)
Bài 3: cho a^3*b^3+b^3*c^3+c^3*a^3=3*a^3*b^3*c^3
tính M(1+a/b)*(1+b/c)*(1+c/a)
Bài 4: cho 3 số a,b,c TM a*b*c=2016
tính P=2016*a/(a*b+2016*a+2016) + b/(b*c+b+2016) + c/(a*c+c+1)
Bài 5: cho a+b+c=0
tính Q=1/(a^2+b^2-c^2) + 1/(b^2+c^2-a^2) + 1/(a^2+c^2-b^2)
Cho A=1/2+1/3+1/4+...+1/2016+1/2017
B=2016/1+2015/2+2014/3+....+2/2015+1/2016
tính B/A
mọi người giải giúp em em xin cảm ơn ạ :))
Câu 1
a) Chứng tỏ rằng 1/3 - 1/3^2 + 1/3^3 - 1/3^4 + 1/3^5 - 1/3^6 < 1/4
b) Cho A= 2015^2016 + 2016^2015 x 2015 và B= 1 + 2^2 + 3^2 + ......+2016^2. Tính AB có chia hết cho 5 không? Vì sao?
Tính A=(1/1009+1/1010+...+1/2016+1/2017)(1-1/2+1/3+1/4+...+1/2015+1/2016)