( 13-x ).28=56
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\(a.ƯCLN\left(2^2;2\cdot3^5\right)=2\\ b.ƯCLN\left(3\cdot5^2;5^2\cdot7\right)=5^2=25\\ c.ƯCLN\left(2^2\cdot3;2^2\cdot3^2\cdot5;2^4\cdot11\right)=2^2=4\\ d.ƯCLN\left(2^2\cdot3\cdot5;3^2\cdot7;3\cdot5\cdot11\right)=3\)
S = \(\dfrac{3}{2}\)+ \(\dfrac{7}{6}\)+\(\dfrac{13}{12}\)+...+\(\dfrac{9901}{9900}\)
Xin giúp với!!!
\(S=\dfrac{3}{2}+\dfrac{7}{6}+\dfrac{13}{12}+...+\dfrac{9901}{9900}\)
\(=1+\dfrac{1}{2}+1+\dfrac{1}{6}+...+1+\dfrac{1}{9900}\)
\(=\left(1+1+1+...+1\right)+\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\)
\(=99+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=99+\left(1-\dfrac{1}{100}\right)=100-\dfrac{1}{100}=\dfrac{10000-1}{100}=\dfrac{9999}{100}\)
S = ( 1+\(\dfrac{1}{2}\) ) + ( 1 + \(\dfrac{1}{6}\) ) + .... + ( 1 + \(\dfrac{1}{9900}\) )
= 9900 + ( \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + ..... + \(\dfrac{1}{99.100}\) )
= 9900 + ( 1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + ..... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\) )
= 9900 + 1 - \(\dfrac{1}{100}\)
= 9901 - \(\dfrac{1}{100}\)
50-(20+40)
=50-60=-10
\(30+\left(31+69\right)-210\)
\(=30+100-210\)
\(=30-110=-80\)
Bài 7:
\(\dfrac{x-2}{5}=\dfrac{-2}{2y+1}\)
=>\(\left(x-2\right)\left(2y+1\right)=5\cdot\left(-2\right)=-10\)
mà 2y+1 lẻ
nên \(\left(x-2;2y+1\right)\in\left\{\left(10;-1\right);\left(-10;1\right);\left(2;-5\right);\left(-2;5\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(12;-1\right);\left(-8;0\right);\left(4;-3\right);\left(0;2\right)\right\}\)
Bài 6:
\(\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{70}+...+\dfrac{2}{x\left(x+3\right)}=\dfrac{101}{770}\)
=>\(\dfrac{2}{40}+\dfrac{2}{88}+\dfrac{2}{140}+...+\dfrac{2}{x\left(x+3\right)}=\dfrac{101}{770}\)
=>\(\dfrac{2}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{770}\)
=>\(\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{770}\)
=>\(\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{770}\)
=>\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{101}{770}:\dfrac{2}{3}=\dfrac{101}{770}\cdot\dfrac{3}{2}=\dfrac{303}{1540}\)
=>\(\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}=\dfrac{1}{308}\)
=>x+3=308
=>x=305
Bài 8:
a: \(\left(2x-1\right)^2+4>=4\forall x\)
=>\(B=\dfrac{20}{\left(2x-1\right)^2+4}< =\dfrac{20}{4}=5\forall x\)
Dấu '=' xảy ra khi 2x-1=0
=>\(x=\dfrac{1}{2}\)
b: \(x^2+1>=1\forall x\)
=>\(\left(x^2+1\right)^2>=1^2=1\forall x\)
=>\(\left(x^2+1\right)^2+5>=1+5=6\forall x\)
=>\(C=\dfrac{10}{\left(x^2+1\right)^2+5}< =\dfrac{10}{6}=\dfrac{5}{3}\forall x\)
Dấu '=' xảy ra khi x=0
a: KHi xét nghiệm viêm gan thì có 2 kết quả có thể xảy ra: Dương tính, Âm tính
b: Xác suất thực nghiệm là:
\(\dfrac{26}{230}=\dfrac{13}{115}\)
$2,5\times20,21\times5\times40\times0,2$
$=(2,5\times40)\times(5\times0,2)\times20,21$
$=100\times1\times20,21$
$=100\times20,21=2021$
\(2,5\cdot20,21\cdot5\cdot40\cdot0,2\\=\left(2,5\cdot40\right)\cdot20,21\cdot\left(5\cdot0,2\right)\\ =\left(2,5\cdot4\cdot10\right)\cdot20,21\cdot1\\ =100\cdot20,21\\ =2021\)
\(1-\left(4\dfrac{2}{5}+x-7\dfrac{2}{3}\right):15\dfrac{1}{3}=0\\ 1-\left(\dfrac{22}{5}+x-\dfrac{23}{3}\right):\dfrac{46}{3}=0\\ 1-\left(\dfrac{-49}{15}+x\right):\dfrac{46}{3}=0\\ \left(\dfrac{-49}{15}+x\right):\dfrac{46}{3}=1\\ -\dfrac{49}{15}+x=\dfrac{46}{3}\\ x=\dfrac{46}{3}+\dfrac{49}{15}\\ x=\dfrac{279}{15}=\dfrac{93}{5}\)
$1-\left(4.\frac25+x-\frac{7.2}{3}\right):15.\frac13=0$
$\Rightarrow \left(\frac85-\frac{14}{3}+x\right):15:3=1$
$\Rightarrow \left(-\frac{46}{15}+x\right):15=3$
$\Rightarrow -\frac{46}{15}+x=3.15$
$\Rightarrow -\frac{46}{15}+x=45$
$\Rightarrow x=45-\left(-\frac{46}{15}\right)=\frac{721}{15}$
\(\left(13-x\right)\cdot28=56\\ 13-x=56:28\\ 13-x=2\\ x=13-2\\ x=11\)
Vậy x=11
$\color{#6495ED}{\text{(13 - x).28 = 56}}$
$\color{#6495ED}{\text{(13 - x) = 56 : 28}}$
$\color{#6495ED}{\text{ 13 - x = 2}}$
$\color{#6495ED}{\text{ x = 13 - 2}}$
$\color{#6495ED}{\text{ x = 11}}$
Vậy \(x=11\)
$\color{#6495ED}{\text{4}}$$\color{#87CEFA}{\text{5}}$$\color{#ADD8E6}{\text{6}}$