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\(a)4(15-18)-(3-5).(-3)^2\)
\(=4.\left(-3\right)-2.9\)
\(=6\)
\(b)(-4-11):(-3)\)
\(=-15:\left(-3\right)\)
\(=5\)
\(c)143-(-57)-(-2)^3.(-3)^0\)
\(=143+57+8.1\)
\(=208\)
\(d,\left(-5\right).\left[\left(90-2.5\right)-10^2+10:\left|-2\right|\right]\)
\(=-5.\left[80-100+10:2\right]\)
\(=-5.-15\)
\(=75\)
$\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}=\frac{2}{3.5}+\frac{2}{5.7}+$\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}$
=$\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{11}-\frac{1}{13}$
=$\frac{1}{3}-\frac{1}{13}=\frac{10}{39}$
Vậy....
\(S=\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}+\dfrac{2}{143}\)
\(S=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)
\(S=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)
\(S=\dfrac{1}{3}-\dfrac{1}{13}\)
\(S=\dfrac{13}{39}-\dfrac{3}{39}\)
\(S=\dfrac{10}{39}\)
Vậy \(S=\dfrac{10}{39}\)
a)\(\dfrac{x}{15}=\dfrac{3}{y}\Rightarrow x.y=45\)=1.45=3.15=5.9
=> x=1,y=45
x=45,y=1
x=3,y=15
x=15,y=3
x=5,y=9
x=9,y=5.
b) làm tương tự (x+1)(2y-5)=143=1.143=11.13
* x+1=1,2y-5=143 => x=0;y=74
*x+1=143,2y-5=1 => x=142;y=3
*x+1=11,2y-5=13 =>x=10;y=9
*x+1=13,2y-5=11 => x=12, y=8
x.y-x-y=2 <=> y(x-1)=2+x
=> \(y=\dfrac{2+x}{x-1}=\dfrac{x-1+3}{x-1}=1+\dfrac{3}{x-1}\)
x,y thuộc Z=> x-1 là ước của 3 {1;3;-1;-3}
x-1=1=>x=2=>y=4.
x-1=-1=>x=0=>y=-2
x-1=3=>x=4=>y=2
x-1=-3=>x=-2=>y=0
a) \(A=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.10}+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{100}\right)+\dfrac{1}{143}=\dfrac{1}{2}.\dfrac{99}{100}+\dfrac{1}{143}=\dfrac{99}{200}+\dfrac{1}{143}=\dfrac{99.143+200.1}{200.143}=\dfrac{14157+200}{28600}=\dfrac{14357}{28600}\)
b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=14950\)
\(\Rightarrow x+x+...+x+\left(1+2+...+99\right)=14950\)
\(\Rightarrow100x+\left(\left(99+1\right):2\right).99:2=14950\)
\(\Rightarrow100x+2475=14950\Rightarrow100x=12475\Rightarrow x=\dfrac{12475}{100}=\dfrac{499}{4}\)
a) (x-2)(x+3) <0 => x-2 và x+3 phải trái dấu
=> x-2<0 và x+3>0
hoặc x-2>0 và x+3<0
=> x<2 và x>-3 => -3<x<2
hoặc x>2 và x<-3 ( vô lý ) ( loại )
=> x \(\in\) { -2;-1;0;1 }
Đúng 100%, tích nha, please!!
\(\Rightarrow2x=\dfrac{143}{2}-15=\dfrac{113}{2}\\ \Rightarrow x=\dfrac{113}{2}\cdot\dfrac{1}{2}=\dfrac{113}{4}\)
2.x + 15 = 143 : 2
2.x + 15 = 71,5
2.x = 71,5 - 15
2.x = 56,5
x = 56,5 : 2
x = 28,25