Tìm x biết
a,
\(|x-1|+|x-2|+|x-3|=2\)
b,
\(|3x+\frac{1}{2}|+|3x+\frac{1}{6}|+|3x+\frac{1}{12}|+|3x+\frac{1}{20}|+...+|3x+380|=58x\)
K
Khách
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Những câu hỏi liên quan
21 tháng 3 2020
a.ĐK: 2x2+1\(\ne0\) \(\forall x\)
Để phương trình bằng 0 thì 4x-8=0 ( Vì 2x2+1 >0 với mọi x)
\(\Leftrightarrow x=2\) (TM)
Vậy ...
b.ĐK: x-3\(\ne0\) \(\Leftrightarrow x\ne3\)
Để phương trình bằng 0 thì x2-x-6=0 (Vì x-3\(\ne0\))
\(\Leftrightarrow\left[{}\begin{matrix}x=2\:\left(TM\right)\\x=-3\:\left(TM\right)\end{matrix}\right.\)
Vậy ...
c. ĐK: x\(\ne\)2
\(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\Leftrightarrow\frac{x+5}{3\left(x-2\right)}-\frac{1}{2}=\frac{2x-3}{2\left(x-2\right)}\)
\(\Leftrightarrow\frac{2\left(x+5\right)-3\left(x-2\right)}{6\left(x-2\right)}=\frac{3\left(2x-3\right)}{6\left(x-2\right)}\)
\(\Leftrightarrow2x+10-3x+6=6x-9\) (x\(\ne\)2)
\(\Leftrightarrow x=\frac{25}{7}\left(TM\right)\)
Vậy ...
d. ĐK: \(x\ne\pm\frac{1}{3}\)
\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
\(\Leftrightarrow\frac{12}{1-9x^2}=\frac{\left(1-3x\right)^2-\left(1+3x\right)^2}{1-9x^2}\)
\(\Leftrightarrow12=1-6x+9x^2-1-6x-9x^2\) (\(x\ne\pm\frac{1}{3}\))
\(\Leftrightarrow x=-2\:\left(TM\right)\)
Vậy...
KN
5 tháng 3 2020
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
KN
5 tháng 3 2020
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
14 tháng 3 2020
a) \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
<=> 1 - x + 3(x + 1) = 2x + 3
<=> 1 - x + 3x + 3 = 2x + 3
<=> 1 - x + 3x + 3 - 2x = 3
<=> 4 = 3 (vô lý)
=> pt vô nghiệm
b) ĐKXĐ: \(x\ne1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
<=> (x - 2)(2 - x) - 5(x + 1)(2 - x) = 15(x - 2)
<=> 2x - x2 - 4 + 2x - 5x - 5x2 + 10 = 15x - 30
<=> -x + 4x2 - 14 = 15x - 30
<=> x - 4x2 + 14 = 15x - 30
<=> x - 4x2 + 14 + 15x - 30 = 0
<=> 16x - 4x2 - 16 = 0
<=> 4(4x - x2 - 4) = 0
<=> -x2 + 4x - 4 = 0
<=> x2 - 4x + 4 = 0
<=> (x - 2)2 = 0
<=> x - 2 = 0
<=> x = 2 (ktm)
=> pt vô nghiệm
c) xem bài 4 ở đây: Câu hỏi của gjfkm
d) ĐKXĐ: \(x\ne1;x\ne2;x\ne3\)
\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
<=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}+\frac{x+1}{\left(x-1\right)\left(x-3\right)}=\frac{2x+5}{\left(x-1\right)\left(x-3\right)}\)
<=> (x + 4)(x - 3) + (x + 1)(x - 2) = (2x + 5)(x - 2)
<=> x2 - 3x + 4x - 12 + x2 - 2x + x - 2 = 2x2 - 4x + 5x - 10
<=> 2x2 - 14 = 2x2 + x - 10
<=> 2x2 - 14 - 2x2 = x - 10
<=> -14 = x - 10
<=> -14 + 10 = x
<=> -4 = x
<=> x = -4
N
31 tháng 12 2018
a) \(=\frac{3x+2}{\left(3x+2\right).\left(3x-2\right)}-\frac{12x-8}{\left(3x+2\right).\left(3x-2\right)}-\frac{-3x+6}{\left(3x-2\right).\left(3x+2\right)}\)
\(b,\frac{x^2+1}{\left(x-1\right).\left(x^2+1\right)}-\frac{x.\left(x^2-1\right).\left(x-1\right)}{\left(x-1\right).\left(x^2+1\right)}.\left(\frac{1}{\left(x-1\right)^2}-\frac{1}{\left(x+1\right).\left(x-1\right)}\right)\)
p/s: hướng dấn cách tách thoy, tự làm nha~~lazy
31 tháng 12 2018
a )
\(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{3x-6}{4-9x^2}=0\)
\(\Leftrightarrow\frac{\left(3x+2\right)-4.\left(3x-2\right)}{9x^2-4}=\frac{3x-6}{4-9x^2}\) ( * )
Đkxđ : \(\hept{\begin{cases}9x^2-4\ne0\\4-9x^2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne\pm\sqrt{\frac{4}{9}}\\x\ne\pm\sqrt{\frac{4}{9}}\end{cases}}\Leftrightarrow x\ne\pm\frac{2}{3}\)
( * ) => \(\left(4-9x^2\right).\left[\left(3x+2\right)+\left(-12x+8\right)\right]=\left(9x^2-4\right).\left(3x-6\right)\)
\(\Leftrightarrow\left(4-9x^2\right).\left(-9x+10\right)=\left(9x^2-4\right).\left(3x-6\right)\)
\(\Leftrightarrow-36x+40+81x^3-90x^2=27x^3-54x^2-12x+24\)
\(\Leftrightarrow54x^3-36x^2-24x+16=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\left(loai\right)\\x=-\frac{2}{3}\left(loai\right)\end{cases}}\)
Vậy : phương trình vô nghiệm
NV
18 tháng 4 2017
\(1.\frac{7x-3}{x-1}=\frac{2}{3}\) ( \(x\ne1\))
\(\Leftrightarrow\frac{3\left(7x-1\right)}{3\left(x-1\right)}=\frac{2\left(x-1\right)}{3\left(x-1\right)}\)
\(\Rightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow19x=7\)
\(\Leftrightarrow x=\frac{7}{19}\)
\(2.\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\)
\(\Leftrightarrow\frac{\left(5x-1\right)\left(3x-1\right)}{\left(3x+2\right)\left(3x-1\right)}=\frac{\left(5x-7\right)\left(3x+2\right)}{\left(3x-1\right)\left(3x+2\right)}\)
\(\Rightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(\Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\)
\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)
\(\Leftrightarrow\left(15x^2-15x^2\right)+\left(-8x+11x\right)=-14-1\)
\(\Leftrightarrow3x=-15\)
\(\Leftrightarrow x=-5\)
\(3.\frac{1-x}{x+1}+3=\frac{2x+3}{3x-1}\)
\(\Leftrightarrow\frac{\left(1-x\right)\left(3x-1\right)}{\left(x+1\right)\left(3x-1\right)}+\frac{3\left(x+1\right)\left(3x-1\right)}{\left(x+1\right)\left(3x-1\right)}=\frac{\left(2x+3\right)\left(x+1\right)}{\left(3x-1\right)\left(0+1\right)}\)
\(\Rightarrow\left(1-x\right)\left(3x-1\right)+3\left(x+1\right)\left(3x-1\right)=\left(2x+3\right)\left(x+1\right)\)
\(\Leftrightarrow3x-1-3x^2+x+3\left(3x^2-x+3x-1\right)=2x^2+2x+3x+3\)
\(\Leftrightarrow3x-1-3x^2+x+9x^2-3x+9x-3=2x^2+2x+3x+3\)
\(\Leftrightarrow6x^2+10x-4=2x^2+5x+3\)
\(\Leftrightarrow\left(6x^2-2x^2\right)+\left(10x-5x\right)=7\)
\(\Leftrightarrow4x^2+5x-7=0\)
\(\Leftrightarrow\left(2x\right)^2+4x.\frac{5}{4}+\frac{16}{25}+\frac{191}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{5}{4}\right)^2-\frac{191}{25}=0\)
\(\left(2x+\frac{5}{4}\right)^2>0\)
\(\Rightarrow\left(2x+\frac{5}{4}\right)^2+\frac{191}{25}>0\)
=> PT vô nghiệm
\(4.\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)}{x^2-4}+\frac{\left(9x+4\right)\left(x-2\right)}{x^2-4}=\frac{2\left(3x-2\right)+1}{x^2-4}\)
\(\Rightarrow\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3\left(3x-2\right)+1\)
\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-2x+1\)
\(\Leftrightarrow3x^2-25x-6=3x^2-2x+1\)
\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(-25x+2x\right)+\left(-6-1\right)=0\)
\(\Leftrightarrow-23x-7=0\)
\(\Leftrightarrow-23x=7\)
\(\Leftrightarrow x=\frac{-7}{23}\)
\(5.\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
\(\Leftrightarrow\frac{\left(3x+2\right)^2}{9x^2-4}-\frac{6\left(3x-2\right)}{9x^2-4}=\frac{9x^2}{9x^2-4}\)
\(\Rightarrow\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)
\(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)
\(\Leftrightarrow\left(9x^2-9x^2\right)+\left(12x-18x\right)+\left(4+12\right)=0\)
\(\Leftrightarrow-6x+16=0\)
\(\Leftrightarrow-6x=-16\)
\(\Leftrightarrow x=\frac{16}{6}\)
\(6.1+\frac{1}{x+2}=\frac{12}{8-x^3}\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}+\frac{1\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}=\frac{12\left(x+2\right)}{\left(x+2\right)\left(8-x^3\right)}\)
\(\Rightarrow\left(x+2\right)\left(8-x^3\right)+1\left(8-x^3\right)=12\left(x+2\right)\)
\(\Leftrightarrow8x+x^4+16+2x^3+8-x^3=12x+24\)
\(\Leftrightarrow x^4+\left(2x^3-x^3\right)+\left(8x-12x\right)+\left(16-24\right)=0\)
\(\Leftrightarrow x^4+x^3-4x-8=0\)
\(\Leftrightarrow\left(x^4-4x\right)+\left(x^3-8\right)=0\)
Đến đấy mk tắc r xl bạn nhé
11 tháng 12 2019
\(a)=\frac{-2\left(x+3\right)}{x\left(1-3x\right)}.\frac{1-3x}{x\left(x+3\right)}\)
\(=\frac{-2}{x^2}\)
\(b)=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}\)
\(=x\left(x-3\right)\)
\(c)=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)
\(=\frac{\left(x+3\right).x}{x\left(x-1\right)\left(x+1\right)}-\frac{1.\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)-\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+3}{x+1}\)
# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha
Đặt bt trên là A nha
Đổi |x-1|=|1-x|
Suy ra A=|1-x|+x-2|+|x-3|
Áp dụng BĐTGTTĐ ta có
A=|1-x|+x-2|+|x-3|\(\ge\)|1-x+x-3|=2
Dấu = xảy ra khi \(\hept{\begin{cases}x-2=0\\1< x< 3\end{cases}}\)đồng thời xảy ra
Vậy x =2
b,
\(\left|3x+\frac{1}{2}\right|\ge0\)
\(\left|3x+\frac{1}{6}\right|\ge0\)
..........
\(\left|3x+380\right|\ge0\)
Suy ra đề bài \(\ge\)0
suy ra 58x \(\ge\)0
Suy ra \(3x+\frac{1}{2}+3x+\frac{1}{6}+......+3x+380=58x\)
Tự tính nhé hok tốt