giai pt sau
\(\frac{|x+3|}{4}-\frac{|x-4|}{9}=\frac{1}{2}-\frac{x+5}{36}..\)
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G
2
5 tháng 7 2018
\(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{1}{x^2+2x-3}=1.\)
\(ĐK:\hept{\begin{cases}x-1\ne0\\x+3\ne\\x^2+2x-3\ne0\end{cases}0}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne\Leftrightarrow-3\end{cases}}\)
\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)+4-x^2-2x+3=0\)
\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5+4-x^2-2x+3=0\)
\(\Leftrightarrow3x+9=0\)
\(\Leftrightarrow3x=-9\Leftrightarrow x=-3\) (loại)
Vậy pt vô No
27 tháng 2 2020
1/ \(\frac{3\left(x+3\right)}{4}+\frac{1}{2}=\frac{5x+9}{3}-\frac{7x-9}{4}\)
=> \(\frac{9\left(x+3\right)}{12}+\frac{6}{12}=\frac{4\left(5x+9\right)}{12}-\frac{3\left(7x-9\right)}{12}\)
=> \(9\left(x+3\right)+6=4\left(5x+9\right)-3\left(7x-9\right)\)
=> \(9x+27+6=20x+36-21x+27\)
=> \(9x-20x+21x=27-27-6+36\)
=> \(10x=30\)
=> \(x=3\)
Vậy phương trình có tập nghiệm là \(S=\left\{3\right\}\)
2.Ta có : \(\frac{2x-3}{3}-\frac{x-3}{6}=\frac{4x+3}{5}-17\)
=> \(\frac{10\left(2x-3\right)}{30}-\frac{5\left(x-3\right)}{30}=\frac{6\left(4x+3\right)}{30}-\frac{510}{30}\)
=> \(10\left(2x-3\right)-5\left(x-3\right)=6\left(4x+3\right)-510\)
=> \(20x-30-5x+15=24x+18-510\)
=> \(20x-5x-24x=18-510+30-15\)
=> \(-9x=-477\)
=> \(x=53\)
Vậy phương trình có tập nghiệm là \(S=\left\{53\right\}\)
3/ Ta có : \(\frac{5x-1}{6}+\frac{2\left(x+4\right)}{9}=\frac{7x-5}{15}+x-1\)
=> \(\frac{30\left(5x-1\right)}{180}+\frac{40\left(x+4\right)}{180}=\frac{12\left(7x-5\right)}{180}+\frac{180x}{180}-\frac{180}{180}\)
=> \(30\left(5x-1\right)+40\left(x+4\right)=12\left(7x-5\right)+180x-180\)
=> \(150x-30+40x+160=84x-60+180x-180\)
=> \(150x+40x-180x-84x=-60-180-160+30\)
=> \(-74x=-370\)
=> \(x=5\)
Vậy phương trình có tập nghiệm là \(S=\left\{5\right\}\)
28 tháng 2 2020
a) ĐKXĐ: x≠0
Ta có: \(\frac{9}{x}+2=-6\)
⇔\(\frac{9}{x}+2+6=0\)
⇔\(\frac{9}{x}+8=0\)
⇔\(\frac{9}{x}+\frac{8x}{x}=0\)
⇔9+8x=0
⇔8x=-9
hay \(x=-\frac{9}{8}\)
Vậy: \(x=-\frac{9}{8}\)
b) ĐKXĐ: x≠0;x≠-1;x≠-3
Ta có: \(\frac{7}{x+1}+\frac{-18x}{x\left(x^2+4x+3\right)}=\frac{-4}{x+3}\)
⇔\(\frac{7}{x+1}+\frac{-18x}{x\left(x+1\right)\left(x+3\right)}-\frac{-4}{x+3}=0\)
⇔\(\frac{7x\left(x+3\right)}{\left(x+1\right)\cdot x\cdot\left(x+3\right)}+\frac{-18x}{\left(x+1\right)\cdot x\cdot\left(x+3\right)}-\frac{-4x\left(x+1\right)}{\left(x+3\right)\cdot x\cdot\left(x+1\right)}=0\)
⇔\(7x^2+21x-18x+4x\left(x+1\right)=0\)
\(\Leftrightarrow7x^2+21x-18x+4x^2+4x=0\)
⇔\(11x^2+7x=0\)
\(\Leftrightarrow x\left(11x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\11x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\11x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\frac{-7}{11}\end{matrix}\right.\)
Vậy: \(x=\frac{-7}{11}\)
c) ĐKXĐ: x≠1; x≠-3
Ta có: \(\frac{3x-1}{x-1}-1=\frac{2x+5}{x+3}+\frac{4}{x^2-2x+3}\)
⇔\(\frac{3x-1}{x-1}-1-\frac{2x+5}{x+3}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
⇔\(\frac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x+5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
⇔\(\left(3x-1\right)\left(x+3\right)-\left(x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)-4=0\)
\(\Leftrightarrow3x^2+9x-x-3-\left(x^2+3x-x-3\right)-\left(2x^2-2x+5x-5\right)-4=0\)
\(\Leftrightarrow3x^2+8x-3-\left(x^2+2x-3\right)-\left(2x^2+3x-5\right)-4=0\)
\(\Leftrightarrow3x^2+8x-3-x^2-2x+3-2x^2-3x+5-4=0\)
\(\Leftrightarrow3x+1=0\)
\(\Leftrightarrow3x=-1\)
hay \(x=\frac{-1}{3}\)
Vậy: \(x=\frac{-1}{3}\)
1 tháng 6 2020
\(\frac{3x+2}{x+4}+\frac{2x+1}{x-2}=5-\frac{x-32}{x^2+2x-8}\)
\(\Leftrightarrow\) \(\frac{\left(3x+2\right)\left(x-2\right)}{\left(x+4\right)\left(x-2\right)}+\frac{\left(2x+1\right)\left(x+4\right)}{\left(x+4\right)\left(x-2\right)}=\frac{5\left(x+4\right)\left(x-2\right)}{\left(x+4\right)\left(x-2\right)}-\frac{x-32}{\left(x+4\right)\left(x-2\right)}\)
\(\Rightarrow\) (3x + 2)(x - 2) + (2x + 1)(x + 4) = 5(x + 4)(x - 2) - x + 32
\(\Leftrightarrow\) 3x2 - 6x + 2x - 4 + 2x2 + 8x + x + 4 = 5x2 - 10x + 20x - 40 - x + 32
\(\Leftrightarrow\) 5x2 + 5x = 5x2 + 9x - 8
\(\Leftrightarrow\) 5x2 + 5x - 5x2 - 9x + 8 = 0
\(\Leftrightarrow\) -4x + 8 = 0
\(\Leftrightarrow\) x - 2 = 0
\(\Leftrightarrow\) x = 2
Vậy S = {2}
\(\frac{x+2m}{x+3}+\frac{x-m}{x-3}=\frac{mx\left(x+1\right)}{x^2-9}\) (đkxđ: x \(\ne\) \(\pm\) 3)
\(\Leftrightarrow\) \(\frac{\left(x+2m\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-m\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{mx\left(x+1\right)}{\left(x+3\right)\left(x-3\right)}\)
\(\Rightarrow\) (x + 2m)(x - 3) + (x - m)(x + 3) = mx(x + 1)
\(\Leftrightarrow\) x2 - 3x + 2mx - 6m + x2 + 3x - mx - 3m - mx2 - mx = 0
\(\Leftrightarrow\) (2 - m)x2 - 9m = 0
Thay m = 1 ta được:
(2 - 1)x2 - 9 . 1 = 0
\(\Leftrightarrow\) x2 - 9 = 0
\(\Leftrightarrow\) (x - 3)(x + 3) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(KTM\right)\\x=-3\left(KTM\right)\end{matrix}\right.\)
Vậy S = \(\varnothing\)
Thay m = 2 ta được:
(2 - 2)x2 - 9 . 2 = 0
\(\Leftrightarrow\) -18 = 0
\(\Rightarrow\) Pt vô nghiệm
Vậy S = \(\varnothing\)
Chúc bn học tốt!!
12 tháng 2 2019
\(ĐKXĐ:x\ne0;-2;-4;-6;-8\)\(\frac{1}{x\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+8\right)}=\frac{4}{105}\)
\(\Leftrightarrow\frac{2}{x\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{2}{\left(x+4\right)\left(x+6\right)}+\frac{2}{\left(x+6\right)\left(x+8\right)}=\frac{8}{105}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+4}+...+\frac{1}{x+6}-\frac{1}{x+8}=\frac{8}{105}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+8}=\frac{8}{105}\)
Quy đồng làm nốt
8 tháng 11 2016
Thực hiện các phép đổi tương đương , ta đưa ( 1 ) về dạng :
\(\frac{x+4}{2x^2-5x+2}-\frac{x+4}{2x^2-7x+3}=0\)
\(\Leftrightarrow\left(x+4\right)\left(\frac{1}{2x^2-5x+2}-\frac{1}{2x^2-7x+3}\right)=0\)
\(\Leftrightarrow\frac{\left(x+4\right)\left(1-2x\right)}{\left(2x^2-5x+2\right)\left(2x^2-7x+3\right)}=0\)
\(\Leftrightarrow\left(x+4\right)\left(1-2x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-4\\x=\frac{1}{2}\end{array}\right.\)
Thữ vào mẫu thức : Với \(x=\frac{1}{2}\) thì \(2x^2-5x+2=0\)
Với \(x=-4\) thì \(\left(2x^2-5x+2\right)\left(2x^2-7x+3\right)\ne0\)
Vậy phương trình ( 1 ) là cho nghiệm duy nhất là \(x=-4\)
\(9\text{|}x+3\text{|}-4\text{|}x-4\text{|}=18-x+5.\) ( quy đồng) mẫu chung là 36
phá dấu bừa nhé
TH1 : \(9\left(-x-3\right)-4\left(-x+4\right)=18-x+5\)
\(-9x-27+4x-16=18-x+5\)
\(-4x=18+5+27+16=66\)
\(x=\frac{66}{-4}\)
TH2: \(9\left(x+3\right)-4\left(x-4\right)=18-x+5\) ( quy đồng ) mẫu chung là 36
\(9x+27-4x+16=18-x+5\)
\(6x=18+5-27-16=-20.\)
\(x=-\frac{20}{6}\)
p/s làm bừa nhé đừng chửi
phá dấu trị tuyệt đối ra có bị làm sao k ?