Lời giải:
PT $\Leftrightarrow \frac{x-342}{15}-1+\frac{x-323}{17}-2+\frac{x-300}{19}-3+\frac{x-273}{21}-4=0$
$\Leftrightarrow \frac{x-357}{15}+\frac{x-357}{17}+\frac{x-357}{19}+\frac{x-357}{21}=0$
$(x-357)\left(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\right)=0$
Dễ thấy: $\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\neq 0$
$\Rightarrow x-357=0$
$\Rightarrow x=357$
Giải các phương trình:
\(a,2x^4+3x^3+8x^2+6x+5=0\)
b, \(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)
x-342/15+x-323/17+x-300/19+x-273/21=10
x-342/15 x-323/17 x-300/19 x-273/21=10
Giải phương trình
\(\frac{x-342}{15}\)+ \(\frac{x-323}{17}\) + \(\frac{x-300}{19}\) + \(\frac{x-273}{21}\) =0
giải phương trình sau: (x-342)/17+(x-323)/19+(x-300)/19+(x+275)/21=10
\(\frac{x-342}{15}\) + \(\frac{x-323}{17}\) + \(\frac{x-300}{19}\) + \(\frac{x-273}{21}\) =10
giải pt
Đề: Giải
a)\(\frac{3x-2}{9x+1}<\frac{1}{3}\)
b)\(\frac{x-342}{15}+\frac{x-323}{17}+\frac{x-300}{19}+\frac{x-273}{21}=10\)
Giải các phương trình sau:
\(j.\dfrac{1}{x-1}-\dfrac{7}{x-2}=\dfrac{1}{\left(x-1\right)\left(2-x\right)}\)
\(k.\dfrac{2x+19}{5x^2-5}-\dfrac{17}{x^2-1}=\dfrac{3}{1-x}\)
\(l.\dfrac{1}{x-1}-\dfrac{2x^2+5}{x^3-1}=\dfrac{4}{x^2+x+1}\)
Giải pt sau:
\(\dfrac{2032-x}{25}+\dfrac{2053-x}{23}+\dfrac{2070-x}{21}+\dfrac{2083-x}{19}=10\)
