C/m nếu \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\) thì \(\dfrac{1}{x^{2003}}+\dfrac{1}{y^{2003}}+\dfrac{1}{z^{2003}}=\dfrac{1}{x^{2003}+y^{2003}+^{2003}}\)
K
Khách
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AH
Akai Haruma
Giáo viên
7 tháng 1 2019
Lời giải:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow \frac{x+y}{xy}+\frac{x+y}{z(x+y+z)}=0\)
\(\Leftrightarrow (x+y)\left(\frac{1}{xy}+\frac{1}{z(x+y+z)}\right)=0\)
\(\Leftrightarrow (x+y).\frac{z(x+y+z)+xy}{xyz(x+y+z)}=0\)
\(\Leftrightarrow (x+y).\frac{z(y+z)+x(z+y)}{xyz(x+y+z)}=0\)
\(\Leftrightarrow \frac{(x+y)(z+x)(z+y)}{xyz(x+y+z)}=0\Rightarrow (x+y)(y+z)(x+z)=0\)
\(\Rightarrow \left[\begin{matrix} x=-y\\ y=-z\\ z=-x\end{matrix}\right.\)
Không mất tổng quát, giả sử \(x=-y\):
\(\frac{1}{x^{2003}}+\frac{1}{y^{2003}}+\frac{1}{z^{2003}}=\frac{1}{(-y)^{2003}}+\frac{1}{y^{2003}}+\frac{1}{z^{2003}}=\frac{1}{z^{2003}}\)
\(\frac{1}{x^{2003}+y^{2003}+z^{2003}}=\frac{1}{(-y)^{2003}+y^{2003}+z^{2003}}=\frac{1}{z^{2003}}\)
Do đó: \(\frac{1}{x^{2003}}+\frac{1}{y^{2003}}+\frac{1}{z^{2003}}=\frac{1}{x^{2003}+y^{2003}+z^{2003}}\) (đpcm)
18 tháng 8 2017
\(\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)
=\(\left(\dfrac{x^2}{a^2}-\dfrac{x^2}{a^2+b^2+c^2}\right)+\left(\dfrac{y^2}{b^2}-\dfrac{y^2}{a^2+b^2+c^2}\right)\)+\(\left(\dfrac{z^2}{c^2}-\dfrac{z^2}{a^2+b^2+c^2}\right)=0\)
=\(x^2.\dfrac{b^2+c^2}{a^2+b^2+c^2}+y^2.\dfrac{a^2+c^2}{a^2+b^2+c^2}+z^2.\dfrac{a^2+b^2}{a^2+b^2+c^2}=0\)
Vì \(a,b,c\) \(\ne\)0 nên dấu "=" xảy ra khi \(x=y=z=0\)
\( \Rightarrow\)\(A=x^{2003}+y^{2003}+z^{2003}=0+0+0=0\)
Chúc Bạn Học Tốt !!!
NA
2
NM
23 tháng 8 2023
Ta có: \(2003^{2003}+1=2003^{2002+1}+1và2003^{2004}+1=2003^{2003+1}+1\)
\(\Rightarrow A>B\)
KT
1
25 tháng 5 2021
`(2-x)/2002-1=(1-x)/2003-x/2004`
`<=>(2-x)/2002-1+(x-1)/2003+x/2004=0`(chuyển vế)
`<=>(2-x)/2002+1+(x-1)/2003-1+x/2004-1=0`
`<=>(2004-x)/2002+(x-2004)/2003+(x-2004)/2004=0`
`<=>(x-2004)(1/2003+1/2004-1/2002)=0`
`<=>x=2004` do `1/2003+1/2004-1/2002 ne 0`
Vậy `x=2004`
12 tháng 7 2023
=>\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2001}{2003}\)
=>\(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2001}{4006}\)
=>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2001}{4006}\)
=>\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2001}{4006}\)
=>1/(x+1)=1/2-2001/4006=1/2003
=>x+1=2003
=>x=2002
9 tháng 9 2018
Bài 1.
Giải
a) Ta có: \(A=\dfrac{3n+9}{n-4}=\dfrac{3n-12+21}{n-4}=\dfrac{3\left(n-4\right)+21}{n-4}=3+\dfrac{21}{n-4}\)
Để \(A\in Z\) thì \(\dfrac{21}{n-4}\in Z\)
\(\Rightarrow21⋮\left(n-4\right)\)
\(\Rightarrow\left(n-4\right)\inƯ\left(21\right)\)
\(\Rightarrow\left(n-4\right)\in\left\{\pm1;\pm3;\pm7;\pm21\right\}\)
Ta có bẳng sau:
| \(n-4\) | \(-21\) | \(-7\) | \(-3\) | \(-1\) | \(1\) | \(3\) | \(7\) | \(21\) |
| \(n\) | \(-17\) | \(-3\) | \(1\) | \(3\) | \(5\) | \(7\) | \(11\) | \(25\) |
Vậy \(n\in\left\{-17;-3;1;3;5;7;11;25\right\}\) thì \(A\in Z.\)
b) Ta có: \(B=\dfrac{6n+5}{2n-1}=\dfrac{6n-3+8}{2n-1}=\dfrac{3\left(2n-1\right)+8}{2n-1}=3+\dfrac{8}{2n-1}\)
Để \(B\in Z\) thì \(\dfrac{8}{2n-1}\in Z\)
\(\Rightarrow8⋮\left(2n-1\right)\)
\(\Rightarrow\left(2n-1\right)\inƯ\left(8\right)\)
\(\Rightarrow\left(2n-1\right)\in\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
Ta có bảng sau:
| \(2n-1\) | \(-8\) | \(-4\) | \(-2\) | \(-1\) | \(1\) | \(2\) | \(4\) | \(8\) |
| \(2n\) | \(-7\) | \(-3\) | \(-1\) | \(0\) | \(2\) | \(3\) | \(5\) | \(9\) |
| \(n\) | \(\dfrac{-7}{2}\) | \(\dfrac{-3}{2}\) | \(\dfrac{-1}{2}\) | \(0\) | \(1\) | \(\dfrac{3}{2}\) | \(\dfrac{5}{2}\) | \(\dfrac{9}{2}\) |
Vậy \(n\in\left\{\dfrac{-7}{2};\dfrac{-3}{2};\dfrac{-1}{2};0;1;\dfrac{3}{2};\dfrac{5}{2};\dfrac{9}{2}\right\}\)
T
9 tháng 9 2018
Bạn Nguyen Thi Huyen giải bài 1 rồi nên mình giải tiếp các bài kia nhé!
Bài 2:
\(\dfrac{x-18}{2000}+\dfrac{x-17}{2001}=\dfrac{x-16}{2002}+\dfrac{x-15}{2003}\)
\(\Leftrightarrow\left(\dfrac{x-18}{2000}-1\right)+\left(\dfrac{x-17}{2001}-1\right)=\left(\dfrac{x-16}{2002}-1\right)+\left(\dfrac{x-15}{2003}-1\right)\)
\(\Leftrightarrow\dfrac{x-2018}{2000}+\dfrac{x-2018}{2001}=\dfrac{x-2018}{2002}+\dfrac{x-2018}{2003}\)
\(\Leftrightarrow\dfrac{x-2018}{2000}+\dfrac{x-2018}{2001}-\dfrac{x-2018}{2002}-\dfrac{x-2018}{2003}=0\)
\(\Leftrightarrow\left(x-2018\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
Dễ thấy \(\dfrac{1}{2000}>\dfrac{1}{2001}>\dfrac{1}{2002}>\dfrac{1}{2003}\) nên:
\(\dfrac{1}{2000}+\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}\ne0\). Do đó:
\(x-2018=0\Leftrightarrow x=2018\)
Bài 3:
a) \(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\Leftrightarrow\dfrac{20}{4x}+\dfrac{xy}{4x}=\dfrac{20+xy}{4x+4x}=\dfrac{20+xy}{8x}=\dfrac{1}{8}\)
Hoán vị ngoại tỉ ta có: \(\dfrac{20+xy}{8x}=\dfrac{1}{8}\Leftrightarrow\dfrac{8}{8x}=\dfrac{1}{x}=\dfrac{1}{8}\Leftrightarrow x=8\)
Thế x = 8 vào : \(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\) .Ta có: \(\dfrac{5}{8}+\dfrac{y}{4}=\dfrac{1}{8}\Leftrightarrow\dfrac{y}{4}=\dfrac{1}{8}-\dfrac{5}{8}=\dfrac{-2}{4}\). Ta có: \(\dfrac{y}{4}=\dfrac{-2}{4}\Leftrightarrow y=-2\)
Vậy: \(\left[{}\begin{matrix}x=8\\y=-2\end{matrix}\right.\)
b) \(\dfrac{1}{x}-\dfrac{2}{y}=\dfrac{3}{1}\Rightarrow\dfrac{y}{x}-2=\dfrac{3}{1}\) (hoán vị ngoại tỉ)
\(\Leftrightarrow\dfrac{y}{x}=\dfrac{5}{1}\). Suy ra nghiệm x,y có dạng \(\left[{}\begin{matrix}x=1k\\y=5k\end{matrix}\right.\left(k\in Z\right)\). Bằng các phép thử lại ta dễ dàng suy ra x,y vô nghiệm.
KT
1
M
2
HP
25 tháng 9 2021
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}=0\)
<=> \(\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)=0\)
<=> \(\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
<=> \(\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
<=> x + 2004 = 0
<=> x = -2004
(Bn nhớ thêm kết quả là 0 vào sau nữa nha)
\(\dfrac{xy+xz+yz}{xyz}=\dfrac{1}{x+y+z}\)
\(\left(xy+xz+yz\right)\left(x+y+z\right)=xyz\)
\(x^2y+xy^2+xyz+x^2z+xyz+xz^2+xyz+y^2z+z^2y=xyz\)
\(x^2\left(y+z\right)+xy\left(y+z\right)+xz\left(z+y\right)+yz\left(y+z\right)=0\)
\(\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]=0\)
\(\left(y+z\right)\left(x+z\right)\left(x+y\right)=0\)
\(\left[{}\begin{matrix}x=-y\\z=-x\\y=-z\end{matrix}\right.\)
\(\dfrac{1}{x^{2003}}+\dfrac{1}{y^{2003}}+\dfrac{1}{z^{2003}}=\dfrac{1}{z^{2003}}=\dfrac{1}{x^{2003}+y^{2003}+z^{2003}}\)