K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

27 tháng 7 2019

\(pt\Leftrightarrow\frac{4+x}{3}-\frac{x+1}{4}+\frac{x-3}{4}=\frac{6-2x}{2}\)

\(\Leftrightarrow\frac{4+x}{3}-1=3-x\)

\(\Leftrightarrow4+x-3=9-3x\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=2\)

Vậy pt có nghiệm duy nhất x=2

20 tháng 7 2019

\(\text{a) }\frac{6}{x-4}-\frac{x}{x+2}=\frac{6}{x-4}.\frac{x}{x+2}\)

\(ĐKXĐ:x\ne-2;x\ne4\)

\(MTC:\left(x-4\right)\left(x+2\right)\)

\(\Leftrightarrow\frac{6\left(x+2\right)}{\left(x-4\right)\left(x+2\right)}-\frac{x\left(x-4\right)}{\left(x-4\right)\left(x+2\right)}=\frac{6x}{\left(x-4\right)\left(x+2\right)}\)

\(\Rightarrow6\left(x+2\right)-x\left(x-4\right)=6x\)

\(\Leftrightarrow6x+12-x^2+4x=6x\)

\(\Leftrightarrow6x+12-x^2+4x-6x=0\)

\(\Leftrightarrow-x^2+4x+12=0\)

\(\Leftrightarrow-\left(x^2-4x-12\right)=0\)

\(\Leftrightarrow x^2-4x-12=0\)

\(\Leftrightarrow x^2+2x-6x-12=0\)

\(\Leftrightarrow x\left(x+2\right)-6\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-6\right)=0\)

\(\Leftrightarrow x=-2\left(\text{loại}\right)\text{ hoặc }x=6\left(\text{nhận}\right)\)

Vậy \(S=\left\{6\right\}\)

\(\text{b) }\frac{2x+3}{2x-1}=\frac{x-3}{x+5}\)

\(ĐKXĐ:x\ne\frac{1}{2};x\ne-5\)

\(\Leftrightarrow\left(2x+3\right)\left(x+5\right)=\left(2x-1\right)\left(x-3\right)\left[\text{Tỉ lệ thức}\right]\)

\(\Leftrightarrow2x^2+10x+3x+15=2x^2-6x-x+3\)

\(\Leftrightarrow2x^2+13x+15=2x^2-7x+3\)

\(\Leftrightarrow2x^2+13x-2x^2+7x=3-15\)

\(\Leftrightarrow20x=-12\)

\(\Leftrightarrow x=\frac{-12}{20}=\frac{-3}{5}\)

Vậy \(S=\left\{\frac{-3}{5}\right\}\)

11 tháng 6 2017

1)

a) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}< =>\frac{2\left(x+5\right)}{2\left(3x-6\right)}-\frac{3x-6}{2\left(3x-6\right)}=\frac{3\left(2x-3\right)}{3\left(2x-4\right)}.\)

(đk:x khác \(\frac{1}{2}\))

\(\frac{2x+10}{6x-12}-\frac{3x-6}{6x-12}=\frac{6x-9}{6x-12}< =>2x+10-3x+6=6x-9< =>x=\frac{25}{7}\)

Vậy x=\(\frac{25}{7}\)

b) /7-2x/=x-3 \(x\ge\frac{7}{2}\)

(đk \(x\ge3,\frac{7}{2}< =>x\ge\frac{7}{2}\))

\(\Rightarrow\orbr{\begin{cases}7-2x=x-3\\7-2x=-\left(x-3\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\left(< \frac{7}{2}\Rightarrow l\right)\\x=4\left(tm\right)\end{cases}}}\)

Vậy x=4

2)

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

\(\Leftrightarrow\frac{30\left(x-1\right)}{60}+\frac{20\left(x-2\right)}{60}+\frac{15\left(x-3\right)}{60}-\frac{12\left(x-4\right)}{60}-\frac{10\left(x-5\right)}{60}>0\)

\(\Leftrightarrow30x-30+20x-40+15x-45-12x+48-10x+50>0\Leftrightarrow43x-17>0\Leftrightarrow x>\frac{17}{43}\)

23 tháng 3 2019

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

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

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

\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0

\(x-1=0\)

\(x=1\)