Giải phương trình sau:
\(x+\frac{x+1}{2}+\frac{x+2}{3}+\frac{x+3}{4}=1\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
1 tháng 3 2020
\(x-\frac{\frac{x}{2}-\frac{3+x}{4}}{2}=3-\frac{\left(1-\frac{6-x}{3}\right).\frac{1}{2}}{2}\)
\(\Leftrightarrow2x-\frac{x}{2}+\frac{3+x}{4}=6-\frac{1}{2}+\frac{6-x}{6}\)
\(\Leftrightarrow24x-6x+9+3x=72-6+12-2x\)
\(\Leftrightarrow23x=69\)
\(\Leftrightarrow x=3\)
Vậy nghiệm của pt x=3
TH
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
\(x-1=0\)
\(x=1\)
DD
1
21 tháng 4 2017
phương trình tương đương với 1+\(\frac{1}{x}+1+\frac{1}{x+3}\)=1+\(\frac{1}{x+1}+1+\frac{1}{x+2}\)\(\Leftrightarrow\frac{1}{x}+\frac{1}{x+3}=\frac{1}{x+2}+\frac{1}{x+1}\)
\(\Leftrightarrow\frac{2x+3}{x\left(x+3\right)}=\frac{2x+3}{\left(x+1\right)\left(x+2\right)}\)\(\Leftrightarrow\left(2x+3\right)\left(\frac{1}{x\left(x+3\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}\right)\)=0
\(\Leftrightarrow\left(2x+3\right)\left(\frac{\left(x+1\right)\left(x+2\right)-x\left(x+3\right)}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(\frac{2}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\right)=0\)\(\Leftrightarrow2x+3=0\Leftrightarrow x=\frac{-3}{2}\)
S
subjects
CTVHS
18 tháng 9
\(a,\frac{x}{x-3}-\frac{6}{x^2-9}=\frac{1}{x+3}\) (đkxđ: x khác 3, -3)
\(\frac{x\left(x+3\right)-6}{\left(x-3\right)\left(x+3\right)}=\frac{1}{x+3}\)
\(x\left(x+3\right)-6=x-3\)
\(x^2+2x-3=0\)
\(\left(x+3\right)\left(x-1\right)=0\)
\(\Longrightarrow\left[\begin{array}{l}x=-3\left(L\right)\\ x=1\left(N\right)\end{array}\right.\)
\(b,\frac{x^2}{x-2}+\frac{x}{1-x}=\frac{4}{x^2-3x+2}\) (đkxđ: \(x\ne1,x\ne2)\)
\(\frac{x^2}{x-2}-\frac{x}{x-1}=\frac{4}{\left(x-1\right)\left(x-2\right)}\)
\(\frac{x^2\left(x-1\right)-x\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=\frac{4}{\left(x-1\right)\left(x-2\right)}\)
\(x^2\left(x-1\right)-x\left(x-2\right)=4\)
\(x^3-x^2-x^2+2x=4\)
\(x^3-2x^2+2x-4=0\)
\(\left(x^3-2x^2\right)+\left(2x-4\right)=0\)
\(x^2\left(x-2\right)+2\left(x-2\right)=0\)
\(\left(x-2\right)\left(x^2+2\right)=0\)
vì \(x^2+2>0\forall x\) ⇒ x - 2 = 0
⇒ x = 2 (ko thoả mãn)
vậy phương trình vô nghiệm
x=-0,44
\(x+\frac{x+1}{2}+\frac{x+2}{3}+\frac{x+3}{4}=1\)
\(\Rightarrow\frac{12x}{12}+\frac{6x+6}{12}+\frac{4x+8}{12}+\frac{3x+9}{12}=\frac{12}{12}\)
\(\Rightarrow25x+23=12\)
\(\Rightarrow x=\frac{-11}{25}\)