Lời giải:
Xét tử số:
$1+(1+2)+(1+2+3)+...+(1+2+3+....98)$
$=\underbrace{(1+1+.....+1)}_{98}+\underbrace{(2+2+...+2)}_{97}+....+\underbrace{97+97}_{2}+98$
$=1.98+2.97+3.96+...+98.1$
Do đó: $B=\frac{1.98+2.97+3.96+..+98.1}{1.98+2.97+3.96+...+98.1}=1$
Câu 1: Chứng minh rằng:
\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+\(\frac{1}{5^2}\)+\(\frac{1}{6^2}\).....+\(\frac{1}{100^2}\)<\(\frac{1}{2}\)
Câu 2: Rút gọn biểu thức:
A=\(\left(\frac{1}{2}+1\right)\).\(\left(\frac{1}{3}+1\right)\).\(\left(\frac{1}{4}+ 1\right)\)....\(\left(\frac{1}{98}+1\right)\).\(\left(\frac{1}{99}+1\right)\)
\(Mn giúp êm với ạ\) ω
Mơn mn nhìu❤
Tính giá trị của các biểu thức sau :
\(A=27+46+79+34+53\)
\(B=-377-\left(98-277\right)\)
\(C=-1,7.2,3+1,7.\left(-3,7\right)-1,7.3-0,17:0,1\)
\(D=2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{3}{5}.2,75+\left(-1,2\right):\dfrac{4}{11}\)
\(E=\dfrac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
1, Chứng minh các biểu thức sau không phụ thuộc vào biến :
\(A=\left(x+3\right)^3-\left(x+9\right).\left(x^2-27\right)\)
\(B=\left(x+y\right).\left(x^2-x.y+y^2\right)+\left(x-y\right).\left(x^2+x.y+y^2\right)-2.\left(x^3-9\right)\)
2, Cho \(\left(a^2+b^2\right).\left(x^2-y^2\right)=\left(ã.+b.y\right)\)
CMR : \(\frac{a}{x}=\frac{b}{y}\)
Xin giải giúp:
1. \(\left(x-\frac{5}{24}\right)\frac{18}{7}=-\frac{12}{7}\)
2. \(\frac{3}{4}+\frac{1}{4}\left(x-1\right)=\frac{1}{2}\)
3. \(\left(4x-\frac{1}{2}\right)\left(\frac{x}{3}-\frac{1}{5}\right)=0\)
Cảm ơn nhiều!!
Thực hiện phép tính sau:
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{10^2}\right)\).
1/ \(\left(\frac{7}{8}-\frac{3}{4}\right).1\frac{1}{3}-\frac{2}{7}.\left(3,5\right)^2\)
2/\(1,25:\left(10,3-9,8\right)-\frac{3}{4}\)
1.tính giá trị biểu thức
a) \(\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)
b) \(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right).\left(1+\dfrac{1}{9.11}\right)\)
2. Chứng tỏ:
\(\dfrac{1}{201}+\dfrac{1}{202}+.........+\dfrac{1}{399}+\dfrac{1}{400}\)>\(\dfrac{1}{2}\)
Tính giá trị biểu thức sau:
\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{4}-1\right)\cdot...\cdot\left(\dfrac{1}{99}-1\right)\cdot\left(\dfrac{1}{100}-1\right)\)
a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16\frac{2}{3}\right)=0\)
b) \(\left(\frac{x}{3}-5\frac{1}{4}\right)^2-\frac{-2}{5}=1\frac{1}{25}\)
c) \(\frac{17-x}{12}=\frac{3}{17-x}\)
d) \(1\frac{1}{3}-25\%\left(x-\frac{8}{3}\right)+2x=1,6:\frac{3}{5}\)
