giải phương trình
\(\frac{6}{x-5}+\frac{2}{x-8}=\frac{18}{x^2-13x+40}-1\)
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14 tháng 3 2017
3/(x^2-13x+40)+2/(x^2-8x+15)+1/(x^2-5x+6)+6/5+0
3/(x-8)(x-5)+2/(x-5)(x-3)+1/(x-3)(x-2)+6/5=0
1/(x-8)-1/(x-5)+1/(x-5)-1/(x-3)+1/(x-3)-1/(x-2)+6/5=0
1/(x-8)-1/(x-2)+6/5=0
ban tu giai tiep nhan
m^2x+2x=5-3mx
m^2x+3mx+2x=5
x(m^2+3m+2)=5
khi 0x=5 thi pt vo nghiem
m^2+3m+2=0
(m+1)(m+2)=0
m=-1 hoac m=-2
LN
3
TN
5 tháng 6 2017
phân tích mẫu thành nhân tử r áp dụng \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) sau đó rút gọn quy đồng
5 tháng 6 2017
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\) \(\left(ĐKXĐ:x\ne0;x\ne-4;x\ne-5;x\ne-6;x\ne-7\right)\)
\(\Leftrightarrow\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}+\frac{1}{x^2+6x+7x+42}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{\left(x+6\right)\left(x+7\right)+\left(x+4\right)\left(x+7\right)+\left(x+4\right)\left(x+5\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{\left(x^2+13x+42\right)+\left(x^2+11x+28\right)+\left(x^2+9x+20\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{x^2+13x+42+x^2+11x+28+x^2+9x+20}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3x^2+33x+90}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3\left(x^2+11x+30\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)=18.3\left(x^2+11x+30\right)\)
\(\Leftrightarrow\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)=54\left(x+5\right)\left(x+6\right)\)
\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=54\)
\(\Leftrightarrow x^2+11x+28-54=0\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow x^2+13x-2x-26=0\)
\(\Leftrightarrow x\left(x+13\right)-2\left(x+13\right)=0\)
\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+13=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-13\left(tm\right)\\x=2\left(tm\right)\end{cases}}\)
7 tháng 2 2018
\(\dfrac{2}{x^2-x-6}+\dfrac{x+1}{x^2+x-12}=\dfrac{x}{x^2+6x+8}\)
\(\Leftrightarrow\dfrac{2}{\left(x-3\right)\left(x+2\right)}+\dfrac{x+1}{\left(x-3\right)\left(x+4\right)}=\dfrac{x}{\left(x+2\right)\left(x+4\right)}\)
=> 2(x+4)+(x+1)(x+2)=x(x-3)
⇔2x+8+x2+2x+x+2=x2-3x
⇔x2+5x+10=x2-3x
⇔x2-x2+5x+3x=-10
⇔8x=-10
\(\Leftrightarrow\dfrac{-5}{4}\)
Vậy S={-\(\dfrac{5}{4}\)}
NT
1
27 tháng 12 2019
Câu hỏi của Phạm Tiến Dũng new - Toán lớp 9 - Học toán với OnlineMath
LD
1
NN
23 tháng 4 2017
Ta có:
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}\) \(=\frac{1}{18}\)
\(\Leftrightarrow\)\(\frac{1}{\left(x+4\right)\left(x+5\right)}\) \(+\frac{1}{\left(x+5\right)\left(x+6\right)}\) \(+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}\) \(=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow x^2+11x-26=0\Leftrightarrow\hept{\begin{cases}x_1=2\\x_2=-13\end{cases}}\)
Vậy nghiệm của phương trình là {2;-13}
KN
13 tháng 2 2020
\(ĐKXĐ:x\ne-4;x\ne-5;x\ne-6;x\ne-7\)
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{3}{54}\)
\(\Rightarrow\left(x+4\right)\left(x+7\right)=54\)
\(\Leftrightarrow x^2+11x+28=54\)
\(\Leftrightarrow x^2+11x-26=0\)
Ta có \(\Delta=11^2+4.26=225,\sqrt{\Delta}=15\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-11+15}{2}=2\\x=\frac{-11-15}{2}=-13\end{cases}}\)
Vậy tập nghiệm S = {2;-13}
\(\frac{6}{x-5}+\frac{2}{x-8}=\frac{18}{x^2-13x+40}-1\)
điều kiện: \(x\ne5;8\)
\(\frac{6\left(x-8\right)+2\left(x-5\right)}{\left(x-5\right)\left(x-8\right)}-\frac{18}{x^2-13x+40}+1=0\)
\(\frac{6x-48+2x-10}{\left(x-5\right)\left(x-8\right)}-\frac{18}{x^2-8x-5x+40}+1=0\)
\(\frac{8x-58}{\left(x-5\right)\left(x-8\right)}-\frac{18}{x\left(x-8\right)-5\left(x-8\right)}+1=0\)
\(\frac{8x-58}{\left(x-5\right)\left(x-8\right)}-\frac{18}{\left(x-5\right)\left(x-8\right)}+\frac{\left(x-5\right)\left(x-8\right)}{\left(x-5\right)\left(x-8\right)}=0\)
\(\frac{8x-58-18+x^2-13x+40}{\left(x-5\right)\left(x-8\right)}=0\)
\(\frac{x^2-5x-36}{\left(x-5\right)\left(x-8\right)}=0\)
=> \(x^2-5x-36=0\)
\(x^2+4x-9x-36=0\)
\(x\left(x+4\right)-9\left(x+4\right)=0\)
\(\left(x-9\right)\left(x+4\right)=0\)
Vậy x - 9 = 0 hoặc x + 4 = 0
hay x = 9 (thỏa mãn điều kiện) hoặc x = -4 (thỏa mãn điều kiện)
vậy...
\(\frac{6}{x-5}+\frac{2}{x-8}=\frac{18}{x^2-13x+40}-1\)
ĐKXĐ: \(x\ne5,8\)
\(\Leftrightarrow\frac{6}{x-5}+\frac{2}{x-8}=\frac{18}{\left(x-5\right)\left(x-8\right)}-1\)
\(\Leftrightarrow6\left(x-8\right)+2\left(x-5\right)=18-\left(x-5\right)\left(x-8\right)\)
\(\Leftrightarrow8x-58=-22-x^2+13x\)
\(\Leftrightarrow8x-58+22+x^2-13x=0\)
\(\Leftrightarrow-5x-36+x^2=0\)
\(\Leftrightarrow\left(x-9\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-9=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=-4\end{cases}}\)
Vậy: phương trình có tập nghiệm là: S = {9; -4}