A=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/16(1+2+3+4+...+16)
Những câu hỏi liên quan
a,1/2(x-4)-1/4(x-4/3)=2(x-1/2)
b, 3/4-(x-1/2)^2=-11/2
c,3/16+1/1/16.(x-2/3)^2=3/4
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`#040911`
a,
\(\dfrac{1}{2}\cdot\left(x-4\right)-\dfrac{1}{4}\cdot\left(x-\dfrac{4}{3}\right)=2\cdot\left(x-\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{1}{2}x-2-\dfrac{1}{4}x+\dfrac{1}{3}=2x-1\\\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{4}x-2x\right)=2-\dfrac{1}{3}-1\\ \Rightarrow-\dfrac{7}{4}x=\dfrac{2}{3}\\ \Rightarrow x=\dfrac{2}{3}\div\left(-\dfrac{7}{4}\right)\\ \Rightarrow x=-\dfrac{8}{21}\)
Vậy, \(x=-\dfrac{8}{21}\)
b,
\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)^2=-\dfrac{11}{2}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\left(-\dfrac{11}{2}\right)\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\pm\dfrac{5}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{5}{2}\\x-\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}+\dfrac{1}{2}\\x=-\dfrac{5}{2}+\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy, \(x\in\left\{-2;3\right\}\)
c,
\(\dfrac{3}{16}+1\dfrac{1}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\\ \Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\div\dfrac{17}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{17}\)
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c) \(\dfrac{3}{16}+\dfrac{1}{\dfrac{1}{16}}\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)
\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\)
\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\)
\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}:16\)
\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{256}=\left(\dfrac{3}{16}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{3}{16}\\x-\dfrac{2}{3}=-\dfrac{3}{16}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{16}+\dfrac{2}{3}\\x=-\dfrac{3}{16}+\dfrac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{41}{48}\\x=\dfrac{23}{48}\end{matrix}\right.\)
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Tính nhanh
A= 1+ 1/2 (1+2) +1/3 (1+2+3) +1/4 (1+2+3+4) +...+ 1/16 (1+2+3+4+...+16)
A=1+1/2x3+1/3X6+1/4X10+...+1/16X136
A=1+3/2+2+5/2+3+...+17/2
A=2/2+3/2+4/2+5/2+6/2+...+17/2
A=2+3+4+5+...+16+17/2
A=(2+17)x16:2/2
A=19x16:2/2
A=304:2/2
A=152/2
A=76
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Tính:
A= 1 + 1/2. (1+2) +1/3. (1+2+3) +1/4. (1+2+3+4) + .............. + 1/16. (1+2+3+4+........+15+16)
Tính nhanh A = 1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/16(1+2+3+...+16)
tính nhanh A= 1+1/2(1+2) +1/3(1+2+3)+1/4(1+2+3+4)+...+ 1/16(1+2+3+...+16)
ta có
A = \(1+\frac{1+2}{2}+\frac{1+2+3}{3}+\frac{1+2+3+4}{4}+......+\frac{1+2+3+\text{4 +....+16}}{16}\)
xét tổng S = 1+2+3+4+5+......+n = \(\frac{\left(n+1\right)n}{2}\) lấy \(\frac{S}{n}=\frac{\frac{\left(n+1\right)n}{2}}{n}=\frac{n+1}{2}\)
ta có
A=\(1+\frac{\frac{2\left(2+1\right)}{2}}{2}+\frac{\frac{3\left(3+1\right)}{2}}{3}+\frac{\frac{4\left(4+1\right)}{2}}{4}+\frac{\frac{5\left(5+1\right)}{2}}{5}+......+\frac{\frac{16\left(16+1\right)}{2}}{16}\)
A = \(1+\frac{1+2}{2}+\frac{1+3}{2}+\frac{1+4}{2}+\frac{1+5}{2}+......+\frac{1+16}{2}\)
A = \(1+\frac{1+2+1+3+1+\text{4+1+5+1+6+.....+1+16}}{2}\)
A = \(1+\frac{151}{2}\)
A = \(\frac{153}{2}\)
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Tính: A=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/16(1+2+3+...+16)
rút gọn biểu thức
a) A=16^8 -1/(2+1)(2^2+1)(2^4+1)(2^8+1(3^16+1)
b) B=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)/9^16-1
giúp mk vs ah mk đang cần gấp ah
a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)
b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)
\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)
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b tính nhanh A= 1+1/2(1+2) +1/3(1+2+3)+1/4(1+2+3+4)+...+ 1/16(1+2+3+...+16)
Tính nhanh:
A = 1+ 1/2(1+2) + 1/3+(1+2+3) +1/4(1+2+3+4)+.....+1/16(1+2+3+....+16)
