Tính nhanh:
M \(=\dfrac{345x6789+3456x345-245x345}{690}\)x\(\left(1-\dfrac{1}{100}\right)\)x\(\left(1-\dfrac{1}{99}\right)\)x...x\(\left(1-\dfrac{1}{3}\right)\)x\(\left(1-\dfrac{1}{2}\right)\)
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10 tháng 4 2021 lúc 22:28
Tìm x∈Z, biết:a) left(x-20right)+left(x-19right)+left(x-18right)+...+99+100100b) 213-x.left(dfrac{1}{2}+dfrac{1}{2^2}+dfrac{1}{2^3}+...+dfrac{1}{2^{2020}}right):left(1-dfrac{1}{2^{2020}}right)13
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\(Tìm\) \(x\)∈\(Z\)\(,\) \(biết\)\(:\)
\(a\)) \(\left(x-20\right)+\left(x-19\right)+\left(x-18\right)+...+99+100=100\)
\(b\)) \(213-x.\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}\right):\left(1-\dfrac{1}{2^{2020}}\right)=13\)
a) Quy luật là gì ??
b)
Đặt
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2020}}\\\Rightarrow2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2019}}\\ \Rightarrow2A-A=1-\dfrac{1}{2^{2020}}\Rightarrow A=1-\dfrac{1}{2^{2020}}\)
Suy ra , phương trình trở thành :
213 -x =13
<=> x=200
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Tính:
a) dfrac{x^2}{left(x-yright)left(x-zright)}+dfrac{y^2}{left(y-zright)left(y-xright)}+dfrac{z^2}{left(z-xright)left(z-yright)}
b) dfrac{x^2-yz}{left(x+yright)left(x+zright)}+dfrac{y^2-zx}{left(y+zright)left(y+xright)}+dfrac{z^2-xy}{left(z+xright)left(z+yright)}
c) dfrac{1}{xleft(x-yright)left(x-zright)}+dfrac{1}{yleft(y-xright)left(y-zright)}+dfrac{1}{zleft(z-xright)left(z-yright)}
d) dfrac{1}{xleft(x+1right)}+dfrac{1}{left(x+1right)left(x+2right)}+dfrac{1}{left(x+2right)left(x+3right)}...
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Tính:
a) \(\dfrac{x^2}{\left(x-y\right)\left(x-z\right)}+\dfrac{y^2}{\left(y-z\right)\left(y-x\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)}\)
b) \(\dfrac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+\dfrac{y^2-zx}{\left(y+z\right)\left(y+x\right)}+\dfrac{z^2-xy}{\left(z+x\right)\left(z+y\right)}\)
c) \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}+\dfrac{1}{y\left(y-x\right)\left(y-z\right)}+\dfrac{1}{z\left(z-x\right)\left(z-y\right)}\)
d) \(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)
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d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
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f, x^2-x+25x^2-2.dfrac{1}{2}.x+left(dfrac{1}{2}right)^2-left(dfrac{1}{2}right)^2+25left(x-dfrac{1}{2}right)^2+dfrac{99}{4}Vì left(x-dfrac{1}{2}right)^2 ≥ 0 nên left(x-dfrac{1}{2}right)^2+dfrac{99}{4}gedfrac{99}{4} với mọi xDấu xảy ra ⇔ x-dfrac{1}{2}0Leftrightarrow xdfrac{1}{2}Vậy GTNN của đa thức là dfrac{99}{4} tại xdfrac{1}{2}
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f, \(x^2-x+25\)
\(=x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2+25\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{99}{4}\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\) ≥ 0 nên \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{99}{4}\ge\dfrac{99}{4}\) với mọi x
Dấu "=" xảy ra ⇔ \(x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy GTNN của đa thức là \(\dfrac{99}{4}\) tại \(x=\dfrac{1}{2}\)
Làm tính nhanh (làm kĩ giúp ạ)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+5}-\dfrac{1}{x+6}\)
=1/x-1/x+6
\(=\dfrac{x+6-x}{x\left(x+6\right)}=\dfrac{6}{x\left(x+6\right)}\)
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Đố :
Đố em tính nhanh được tổng sau :
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}\)
Tìm x, biết \(\left|x+\dfrac{1}{1\cdot2}\right|+\left|x+\dfrac{1}{2\cdot3}\right|+\left|x+\dfrac{1}{3\cdot4}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|=100x\)
Vì \(\left|x+\dfrac{1}{1\cdot2}\right|+\left|x+\dfrac{1}{2\cdot3}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|\ge0\forall x\)
\(\Rightarrow100x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+\dfrac{1}{1\cdot2}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|=x+\dfrac{1}{1\cdot2}+...+x+\dfrac{1}{99\cdot100}\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1\cdot2}+...+\dfrac{1}{99\cdot100}\right)=100x\)
\(\Rightarrow99x+\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)=100x\)
\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}=x\)
\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=x\)
\(\Rightarrow x=1-\dfrac{1}{100}=\dfrac{99}{100}\)
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tính nhanh
\(\left(1-\dfrac{1}{2}\right)x\left(1-\dfrac{1}{3}\right)x\left(1-\dfrac{1}{4}\right)x\left(1-\dfrac{1}{5}\right)=\)
làm ơn giúp mik với ạ ^^
\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)
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\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\times\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{4}{4}-\dfrac{1}{4}\right)\times\left(\dfrac{5}{5}-\dfrac{1}{5}\right)\)
\(=\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\)
\(=\dfrac{1\times2\times3\times4}{2\times3\times4\times5}\)
\(=\dfrac{1}{5}\)
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Cần giúp nhanh vsBài 1. Tìm xa) left|x+dfrac{7}{4}right|dfrac{1}{2}b) left|2x+1right|-dfrac{2}{5}dfrac{1}{3}c) 3x.left(x+dfrac{2}{3}right)0d) x+dfrac{1}{3}dfrac{2}{5}-left(-dfrac{1}{3}right)Bài 2. Tính nhanhAdfrac{1}{100}-dfrac{1}{100.99}-dfrac{1}{99.98}-dfrac{1}{98.97}-....-dfrac{1}{3.2}-dfrac{1}{2.1}
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Cần giúp nhanh vs
Bài 1. Tìm x
a) \(\left|x+\dfrac{7}{4}\right|=\dfrac{1}{2}\)
b) \(\left|2x+1\right|-\dfrac{2}{5}=\dfrac{1}{3}\)
c) \(3x.\left(x+\dfrac{2}{3}\right)=0\)
d) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)\)
Bài 2. Tính nhanh
\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
Bài 1:
a.
$|x+\frac{7}{4}|=\frac{1}{2}$
\(\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.\)
b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$
$|2x+1|=\frac{11}{15}$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.\)
c.
$3x(x+\frac{2}{3})=0$
\(\Leftrightarrow \left[\begin{matrix} 3x=0\\ x+\frac{2}{3}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\frac{-3}{2}\end{matrix}\right.\)
d.
$x+\frac{1}{3}=\frac{2}{5}-(\frac{-1}{3})=\frac{2}{5}+\frac{1}{3}$
$\Leftrightarrow x=\frac{2}{5}$
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Bài 2:
$\frac{1}{100}-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}$
$=\frac{99}{100}$
$\Rightarrow A=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}=\frac{-49}{50}$
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Bài 1:
a) Ta có: \(\left|x+\dfrac{7}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{7}{4}=\dfrac{1}{2}\\x+\dfrac{7}{4}=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)
b) Ta có: \(\left|2x+1\right|-\dfrac{2}{5}=\dfrac{1}{3}\)
\(\Leftrightarrow\left|2x+1\right|=\dfrac{11}{15}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{11}{15}\\2x+1=\dfrac{-11}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-4}{15}\\2x=\dfrac{-26}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{15}\\x=\dfrac{-13}{15}\end{matrix}\right.\)
c) Ta có: \(3x\left(x+\dfrac{2}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)
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Thực hiện phép tính:
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)
ĐKXĐ: \(x\ne0;-1;-2;...;-100\)
Ta có: \(\dfrac{1}{x\left(x+1\right)}=\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}=\dfrac{1}{x}-\dfrac{1}{x+1}\)
Tương tự, ta có biểu thức đã cho bằng:
\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+99}-\dfrac{1}{x+100}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+100}\)
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