x(x-5)=0
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\(124\times\left(56+67\right)-24\times\left(67+56\right)\\ =124\times123-24\times123\\ =\left(124-24\right)\times123\\ =100\times123\\ =12300\)
124 x (56 + 67) - 24 x (67 + 56)
= (56 + 67) x (125 - 24)
= 123 x 100
= 12300
`x ∈ ƯC (150,35)`
Ta có:
`150 = 2 . 3 . 5^2`
`35 = 5 . 7 `
`=> UCLN(150;35) = 5`
Hay `x ∈ U(5) = {-5;-1;1;5}`
Vậy ....
`x ∈ BC (25,30,40)`
Ta có:
`25 = 5^2`
`30 = 2.3.5`
`40 = 2^3 . 5`
`=> BCNN = 5^2 . 2^3 . 3 = 600`
`=> x ∈ B(600) = {0;600;1200;....}`
Vậy ....
Chị My hơn Đạt số tuổi là:
10 - 7 = 3 (tuổi)
Sau bao nhiêu năm thì chị My vẫn hơn Đạt 3 tuổi
Hiệu số phần bằng nhau là:
5 - 4 = 1 (phần)
Năm đó tuổi của Đạt là:
`3:1 xx 4 = 12` (tuổi)
Sau số năm nữa là:
12 - 7 = 5 (năm)
ĐS: ...
\(A=\left(x+1\right)+\left(x+\dfrac{5}{45}\right)+\left(x+\dfrac{5}{117}\right)+\left(x+221\right)=10\\ \Rightarrow x+1+x+\dfrac{1}{9}+x+\dfrac{5}{117}+x+221=10\\ \Rightarrow4x+\left(1+\dfrac{1}{9}+\dfrac{5}{117}+221\right)=10\\ \Rightarrow4x+\dfrac{2888}{13}=10\\ \Rightarrow4x=10-\dfrac{2888}{13}\\ \Rightarrow4x=-\dfrac{2758}{13}\\ \Rightarrow x=-\dfrac{1379}{26}\)
\(1)-\dfrac{3}{7}+\dfrac{5}{13}-\dfrac{4}{7}+\dfrac{8}{13}\\ =\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)+\left(\dfrac{5}{13}+\dfrac{8}{13}\right)\\ =\dfrac{-7}{7}+\dfrac{13}{13}\\ =-1+1\\ =0\\ 2)-\dfrac{5}{14}-\dfrac{2}{14}+\dfrac{1}{8}+\dfrac{1}{8}\\ =\left(\dfrac{-5}{14}-\dfrac{2}{14}\right)+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)\\ =\dfrac{-7}{14}+\dfrac{2}{8}\\ =\dfrac{-1}{2}+\dfrac{1}{4}\\ =\dfrac{-1}{4}\\ 3)\dfrac{-5}{22}-1+\dfrac{3}{2}-\dfrac{6}{22}\\ =\left(\dfrac{-5}{22}-\dfrac{6}{22}\right)+\left(\dfrac{3}{2}-1\right)\\ =\dfrac{-11}{22}+\dfrac{1}{2}\\ =\dfrac{-1}{2}+\dfrac{1}{2}\\ =0\)
\(4,\dfrac{7}{16}+\dfrac{5}{9}+\dfrac{-3}{16}+\dfrac{-2}{9}\\ =\left(\dfrac{7}{16}-\dfrac{3}{16}\right)+\left(\dfrac{5}{9}-\dfrac{2}{9}\right)\\ =\dfrac{4}{16}+\dfrac{3}{9}\\ =\dfrac{1}{4}+\dfrac{1}{3}\\ =\dfrac{7}{12}\\ 5,\dfrac{2}{5}-\dfrac{3}{11}-\dfrac{7}{35}+\dfrac{14}{11}-\dfrac{1}{5}\\ =\left(-\dfrac{3}{11}+\dfrac{14}{11}\right)+\left(\dfrac{2}{5}-\dfrac{1}{5}\right)-\dfrac{7}{35}\\ =\dfrac{11}{11}+\dfrac{1}{5}-\dfrac{7}{35}\\ =1+\dfrac{1}{5}-\dfrac{1}{5}\\ =1\\ 6,\dfrac{3}{4}-\dfrac{3}{17}+\dfrac{-5}{6}+\dfrac{20}{17}-\dfrac{1}{4}\\ =\left(\dfrac{3}{4}-\dfrac{1}{4}\right)+\left(\dfrac{-3}{17}+\dfrac{20}{17}\right)+\dfrac{-5}{6}\\ =\dfrac{1}{2}+1+\dfrac{-5}{6}\\ =\dfrac{3}{2}-\dfrac{5}{6}\\ =\dfrac{9}{6}-\dfrac{5}{6}\\ =\dfrac{4}{6}=\dfrac{2}{3}\)
\(A=\left\{n^2\text{ }|\text{ }n\in N,\text{ }1\le n\le7\text{ }\right\}\)
Hoặc:
\(A=\left\{x\text{ }|\text{ }\text{x là số chính phương},\text{ }0< x< 50\right\}\)
\(7-\left(x-1\right)=15+3\left(x+1\right)\\ 7-x+1=15+3x+3\\ 8-x=18+3x\\ 3x+x=8-18\\ 4x=-10\\ x=-\dfrac{10}{4}\\ x=\dfrac{-5}{2}\)
Vậy: ...
\(x^2+2\left|x-\dfrac{1}{2}\right|=\left|x^2+2\right|\\ =>x^2+2\left|x-\dfrac{1}{2}\right|=x^2+2\left(x^2+2>0\right)\\ =>2\left|x-\dfrac{1}{2}\right|=2\\ =>\left|x-\dfrac{1}{2}\right|=1\\ TH1:x>\dfrac{1}{2}\\ =>x-\dfrac{1}{2}=1\\ =>x=1+\dfrac{1}{2}=\dfrac{3}{2}\\ TH2:x< =\dfrac{1}{2}\\ =>x-\dfrac{1}{2}=-1\\ =>x=\dfrac{1}{2}-1=-\dfrac{1}{2}\left(tm\right)\)
`x(x-5)=0`
TH1: `x=0`
TH2: `x-5=0`
`=>x=0+5`
`=>x=5`
Vậy: `x∈{0;5}`
Bổ sung cho @HT.Phong
Kết luận: \(x\) \(\in\) {0; 5}