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\(A=\dfrac{-5x}{21}+\dfrac{-5y}{21}+\dfrac{-5z}{21}\)
\(A=\dfrac{-5x+\left(-5y\right)}{21}+\dfrac{-5z}{21}\)
\(A=\dfrac{-5\cdot\left(x+y\right)}{21}+\dfrac{-5z}{21}\)
\(A=\dfrac{-5\cdot\left(-z\right)}{21}+\dfrac{-5z}{21}\)
\(A=\dfrac{5z}{21}+\dfrac{-5z}{21}\)
\(A=\dfrac{5z+\left(-5z\right)}{21}=\dfrac{0}{21}=0\)
Vậy \(A=0\)
\(\dfrac{-5x}{21}+\dfrac{-5y}{21}+\dfrac{-5z}{21}=\dfrac{-5x-5y-5z}{21}\)
= \(\dfrac{-5\left(x+y\right)-5z}{21}=\dfrac{-5\left(-z\right)-5z}{21}=\dfrac{5z-5z}{21}=\dfrac{0}{21}=0\)
bài 3:
a, đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)
A/D tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
Đặt \(S=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1008}\right)\)
\(=\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\)
Nên:
\(A=\left(\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right):\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)\(=\left(\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right):\left(\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)\)\(\Rightarrow A=1\)
Vậy A = 1
Chúc bạn học tốt!!
Ta có; \(\dfrac{5x}{3}:\dfrac{10x^2+5x}{21}=\dfrac{5x}{3}.\dfrac{21}{10x^2+5x}=\dfrac{\left(5x\right)21}{3.5x.\left(2x+1\right)}=\dfrac{7}{2x+1}\)
là số nguyên.
Do đó \(7⋮2x+1\Leftrightarrow2x+1\in U\left(7\right)=\left\{-7;-1;1;7\right\}\)
Ta có bảng:
| 2x + 1 | -7 | -1 | 1 | 7 |
| 2x | -8 | -2 | 0 | 6 |
| x | -4 | -1 | 0 | 3 |
| KL | TM | TM | TM | TM |
Vậy \(x\in\left\{-4;-1;0;3\right\}\).
Đặt \(A=\frac{5x}{3}:\frac{10x^2+5x}{21}\)
Ta có:\(A=\frac{5x}{3}:\frac{10x^2+5x}{21}\)
\(A=\frac{5x}{3}.\frac{21}{5x\left(2x+1\right)}\)
\(A=\frac{7}{2x+1}\left(ĐKXĐ:x\ne\frac{1}{2}\right)\)
Để A nguyên thì 7 phải chia hết cho 2x+1
Hay \(\left(2x+1\right)\inƯ\left(7\right)\)
Vậy Ư(7) là:[1,-1,7,-7]
Do đó ta có bảng sau:
| 2x+1 | -7 | -1 | 1 | 7 |
| 2x | -8 | -2 | 0 | 6 |
| x | -4 | -1 | 0 | 3 |
Vậy để A ngyên thì \(x\in\left[-4;-1;0;3\right]\)
a) \(-\dfrac{2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(\Rightarrow-\dfrac{2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{2}{3}x+\dfrac{2}{3}x\)
\(\Rightarrow\dfrac{1}{2}=\dfrac{4}{3}x\)
\(\Rightarrow x=\dfrac{1}{2}:\dfrac{4}{3}=\dfrac{3}{8}\)
Vậy \(x=\dfrac{3}{8}\).