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MN
27 tháng 7 2018
\(\dfrac{x-18}{x+4}=\dfrac{x-17}{x+16}\)
⇒ (x-18) . (x+16) = (x-17) . (x+4)
x(x+16) -18(x+16) = x(x+4) - 17(x+4)
\(x^2+16x-18x-288=x^2+4x-17x-68\)
\(x^2+16-18x-x^2-4x+17x=-68+288\)
11x=220
x= 220 : 11
x = 20
27 tháng 7 2018
\(\dfrac{x-18}{x+4}=\dfrac{x-17}{x+16}\)
\(\Leftrightarrow\left(x-18\right)\left(x+16\right)=\left(x-17\right)\left(x+4\right)\)
\(\Leftrightarrow x^2-2x-288=x^2-13x-68\)
\(\Leftrightarrow220=11x\)
\(\Leftrightarrow x=20\)
\(\)
30 tháng 8 2017
\(\dfrac{x-1}{50}+\dfrac{x-2}{49}=\dfrac{x-3}{48}+\dfrac{x-4}{47}\)
\(\Rightarrow\dfrac{x-1}{50}-1+\dfrac{x-2}{49}-1=\dfrac{x-3}{48}-1+\dfrac{x-4}{47}-1\)
\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}=\dfrac{x-51}{48}+\dfrac{x-51}{47}\)
\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}-\dfrac{x-51}{48}-\dfrac{x-51}{47}=0\)
\(\Rightarrow\left(x-51\right)\left(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\right)=0\)
Vì \(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\ne0\) nên \(x-51=0\Rightarrow x=51\)
\(\dfrac{x+25}{6}+\dfrac{x+20}{11}+\dfrac{x+16}{15}+3=0\)
\(\Rightarrow\dfrac{x+25}{6}+1+\dfrac{x+20}{11}+1+\dfrac{x+16}{15}+1=0\)
\(\Rightarrow\dfrac{x+31}{6}+\dfrac{x+31}{11}+\dfrac{x+31}{15}=0\)
\(\Rightarrow\left(x+31\right)\left(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\right)=0\)
Vì \(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\ne0\) nên \(x+31=0\Rightarrow x=-31\)
\(\dfrac{x-15}{6}+\dfrac{x-10}{11}=\dfrac{x-3}{18}+\dfrac{x-7}{14}\)
\(\Rightarrow\dfrac{x-15}{6}-1+\dfrac{x-10}{11}-1=\dfrac{x-3}{18}-1+\dfrac{x-7}{14}-1\)
\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}=\dfrac{x-21}{18}+\dfrac{x-21}{14}\)
\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}-\dfrac{x-21}{18}-\dfrac{x-21}{14}=0\)
\(\Rightarrow\left(x-21\right)\left(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\right)=0\)
Vì \(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\ne0\) nên \(x-21=0\Rightarrow x=21\)
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.
9 tháng 8 2017
\(\dfrac{x}{4}=\dfrac{18}{x+1};x^2+1=72\)
\(\)\(\Rightarrow x\left(x+1\right)=18.4\)
\(\Rightarrow x^2+x=72\)
\(\Rightarrow\left\{{}\begin{matrix}x^2+1=72\\x^2+x=72\end{matrix}\right.\)
\(\Rightarrow x^2+1=x^2+x\)
\(\Rightarrow x=1\)
11 tháng 8 2017
thanks, bn cs thể làm giúp mik phần b nữa đc hông?
12 tháng 11 2016
Đặt \(\frac{x-18}{x+4}=\frac{x-17}{x+16}=k\)
Suy ra: \(x-18=k\left(x+4\right)\Rightarrow x=\frac{4k+18}{1-k}\left(1\right)\\ x-17=k\left(x+16\right)\Rightarrow x=\frac{16k+17}{1-k}\left(2\right)\)
Từ (1) và (2) ta được: \(4k+18=16k+17,\) suy ra \(k=\frac{1}{12},x=20\)
TT
12 tháng 11 2016
Hôm nay cô giao bài nhìu, tui đăng nhìu, vất vả nhìu cho chú đấy
1
1
26 tháng 10 2017
\(\dfrac{x-18}{x+4}=\dfrac{x-17}{x+16}\Leftrightarrow\left(x-18\right)\left(x+16\right)=\left(x+4\right)\left(x-17\right)\)
\(\Leftrightarrow x^2+16x-18x-288=x^2-17x+4x-68\)
\(\Leftrightarrow x^2+16x-18x-288-\left(x^2-17x+4x-68\right)=0\)
\(\Leftrightarrow x^2+16x-18x-288-x^2+17x-4x+68=0\)\(\Leftrightarrow11x-220=0\Leftrightarrow11x=220\Leftrightarrow x=\dfrac{220}{11}=20\)
vậy \(x=20\)
3 tháng 8 2018
a) 27x : 3x = 9
(27 : 3)x = 9
9x = 91
x = 1
b) 25 : 5x =5
5x = 25 : 5
5x = 51
x = 1
c) 2 : (x + 2)2 = \(\dfrac{1}{18}\)
(x + 2)2 = 2 : \(\dfrac{1}{18}\)
(x + 2)2 = 36
\(\Rightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
d) (5x - 1)2 = \(\dfrac{36}{49}\)
(5x - 1)2 = \(\left(\dfrac{6}{7}\right)^2\)
Bạn làm tiếp nha, mình có việc bận :v
22 tháng 9 2018
3, Tìm x, biết
\(d,\dfrac{-16}{x}=\dfrac{x}{-4}=>x^2=\left(-16\right).\left(-4\right)=>x^2=64\)
\(=>x=8\) hay \(x=-8\)
\(e,\dfrac{x}{-2}=\dfrac{\dfrac{8}{25}}{-x}=>-x^2=-2.\dfrac{8}{5}=\dfrac{-16}{25}\)
\(=>-x^2=0,64=>x=0,8\)
\(g,\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(=>x^2=\left(-15\right).\left(-60\right)\)\(=>x^2=900=>x=30\) hay \(x=-30\)
22 tháng 9 2018
d) \(\dfrac{-16}{x}=\dfrac{x}{-4}\)
= 16 . 4 = x.x
= 64 = \(x^2\)
= \(8^2=x^2\)
vậy x = 8
e)\(\dfrac{x}{-2}=\dfrac{8}{\dfrac{25}{-x}}\)
= -2 . \(\dfrac{8}{25}\) = -x . x
= -0,64 = \(-x^2\)
= 0,64 = \(x^2\)
0,8\(^2=x^2\)
vậy x = 0,8
g) \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
= -15 . -60 = x.x
= 900 = \(x^2\)
30 \(^2=x^2\)
vậy x = 30
21 tháng 6 2017
\(\left|x+\dfrac{11}{17}\right|\ge0\)
\(\left|x+\dfrac{2}{17}\right|\ge0\)
\(\left|x+\dfrac{4}{17}\right|\ge0\)
\(\Leftrightarrow\left|x+\dfrac{11}{17}\right|+\left|x+\dfrac{2}{17}\right|+\left|x+\dfrac{4}{17}\right|\ge0\)
\(\Leftrightarrow x+\dfrac{11}{17}+x+\dfrac{2}{17}+x+\dfrac{4}{17}=4x\)
\(3x+\left(\dfrac{11}{17}+\dfrac{2}{17}+\dfrac{4}{17}\right)=4x\)
\(3x+1=4x\Leftrightarrow4x-3x=1\Leftrightarrow x=1\)
Vậy...
21 tháng 6 2017
Ta có: \(\left\{{}\begin{matrix}\left|x+\dfrac{11}{17}\right|\ge0\\\left|x+\dfrac{2}{17}\right|\ge0\\\left|x+\dfrac{4}{17}\right|\ge0\end{matrix}\right.\Leftrightarrow\left|x+\dfrac{11}{17}\right|+\left|x+\dfrac{2}{17}\right|+\left|x+\dfrac{4}{17}\right|\ge0\)
\(\Leftrightarrow4x\ge0\Leftrightarrow x\ge0\)
\(\Leftrightarrow x+\dfrac{11}{17}+x+\dfrac{2}{17}+x+\dfrac{4}{17}=4x\)
\(\Leftrightarrow3x+1=4x\)
\(\Leftrightarrow x=1\)
Vậy x = 1
Đặt \(\dfrac{x-18}{x+4}=\dfrac{x-17}{x+16}=k\)
\(\Rightarrow x-18=k.\left(x+4\right)\Rightarrow x=\dfrac{4k+18}{1-k}\left(1\right)\)
\(x-17=k.\left(x+16\right)\Rightarrow x=\dfrac{16k+17}{1-k}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow4k+18=16k+17\Rightarrow-12k=-1\Rightarrow k=\dfrac{1}{12}\)
\(\Rightarrow x=\dfrac{4.\dfrac{1}{12}+18}{1-\dfrac{1}{12}}=\dfrac{\dfrac{55}{3}}{\dfrac{11}{12}}=20\)
Vậy x = 20
\(\dfrac{x-18}{x+4}=\dfrac{x-17}{x+16}\)
(x\(-\)18).(x+16)=(x-17).(x+4)
x\(^2\)-18x+16x-18.16=x\(^2\)-17x+4x-4.17
x\(^2\)-18x+16x-288=x\(^2\)-17x+4x-68
x\(^2\)-18x+16x-x\(^2\)+17x-4x=-68+288
11x=220
x=220/11
x=20