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10 tháng 7 2017

ĐK \(0\le x\le4\)

\(\Leftrightarrow\frac{\left(4-x\right)!x!}{24}-\frac{\left(5-x\right)\left(4-x\right)!x!}{120}=\frac{\left(6-x\right)\left(5-x\right)\left(4-x\right)!x!}{720}\)

\(\Leftrightarrow\left(4-x\right)!x!\left[\frac{1}{24}-\frac{5-x}{120}-\frac{\left(6-x\right)\left(5-x\right)}{720}\right]=0\)

\(\frac{\Leftrightarrow1}{24}-\frac{5-x}{120}-\frac{\left(6-x\right)\left(5-x\right)}{720}=0\)do \(\left(4-x\right)!x!\ne0\forall x\)

\(\Leftrightarrow\frac{30-6\left(5-x\right)-\left(30-11x+x^2\right)}{720}=0\Leftrightarrow30-30+6x-30+11x-x^2=0\)

\(\Leftrightarrow x^2-17x+30=0\Rightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=15\left(l\right)\end{cases}}\)

Vậy x=2

27 tháng 5 2020

\(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}=\frac{3\left(x-3\right)\left(3-x\right)}{12}\)

\(\Leftrightarrow12\left(x-3\right)-2\left(x-3\right)\left(2x-5\right)=3\left(x-3\right)\left(3-x\right)\)

\(\Leftrightarrow12\left(x-3\right)-2\left(x-3\right)\left(2x-5\right)-3\left(x-3\right)\left(3-x\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(13-x\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x-3=0\\13-x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\x=13\end{cases}}}\)

Vậy tập nghiệm của phương trình trên là:\(S=\left\{3;13\right\}\)

 #hoktot<3# 

\(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

\(\frac{12\left(x-3\right)}{12}-\frac{\left(x-3\right)\left(2x-5\right)2}{12}=\frac{\left(x-3\right)\left(3-x\right)3}{12}\)

Khử mẫu : \(12\left(x-3\right)-\left(x-3\right)\left(2x-5\right)2=\left(x-3\right)\left(3-x\right)3\)

\(34x-66-4x^2=18x-3x^2-27\)

\(34x-66-4x^2-18x+3x^2+27=0\)

\(16x-39-x^2=0\)

Phân tích nốt nhé ! 

11 tháng 2 2019

\(\Leftrightarrow\frac{12\left(x-3\right)-2\left[\left(x-3\right)\left(2x-5\right)\right]}{12}=\frac{3\left[\left(x-3\right)\left(3-x\right)\right]}{12}\)

\(\Leftrightarrow12x-36-2\left(2x^2-5x-6x+15\right)=3\left(3x-x^2-9+3x\right)\)

\(\Leftrightarrow12x-36-4x^2+10x+12x-30=9x-3x^2-27+9x\)

\(\Leftrightarrow-x^2+34x-39=0\)

\(\Leftrightarrow-\left(x-17\right)^2-250=0\)

\(\Leftrightarrow-\left(x-17\right)^2=250\left(voly\right)\)

vay \(S=\varnothing\)

tớ ko bt lm abc , tớ lm d thôi nha , thứ lỗi 

\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)

\(\frac{3x+13}{2x^2+x-6}=\frac{5}{x-6}+\frac{7}{1-2x}\)

\(\frac{3x+13}{\left(x+2\right)\left(2x-3\right)}=\frac{3x+37}{\left(x-6\right)\left(2x-1\right)}\)

\(\frac{10-9x}{-4x^3+32x^2-51x+18}=0\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{10}{9}\end{cases}}\)

25 tháng 2 2020

giup minh voi cac bạn

8 tháng 1 2020

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

4 tháng 4 2020

ĐK: \(x\notin\left\{-2,-3,-4,-5,-6\right\}\)

\(\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{15}\)

\(\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{15}\)\(\Leftrightarrow\frac{x+6-x-2}{\left(x+6\right)\left(x+2\right)}=\frac{1}{15}\) \(\Leftrightarrow\frac{4}{x^2+8x+12}=\frac{1}{15}\)

\(\Leftrightarrow x^2+8x+12=60\Leftrightarrow x^2+8x-48=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-12\end{matrix}\right.\) (tm)