\(\Leftrightarrow\left(-x-\frac{4}{7}\right)-\frac{53}{12}=\frac{-5}{6}\)
\(\Leftrightarrow-x-\frac{4}{7}=\frac{43}{12}\)
\(\Leftrightarrow-x=\frac{349}{84}\)
\(\Leftrightarrow x=-\frac{349}{84}\)
Tính A=\(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(11^4+\frac{1}{4}\right)}{\frac{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(12^4+\frac{1}{4}\right)}{ }}\)
giải hộ mk phương trình, thanks:
a) \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
b) \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)
giải phương trình
a,\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{9\cdot10}\right)\left(x-1\right)+\frac{1}{10}x=x-\frac{9}{10}\)
b,\(\frac{x+1}{1}+\frac{2x+3}{3}+\frac{3x+5}{5}+\frac{20x+39}{39}=22+\frac{4}{3}+\frac{6}{5}+\frac{40}{39}\)
c,(x-20)+(x-19)+(x-18)+...+100+101=101
1.Giải phương trình: \(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
2.Giải phương trình: \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
Tìm x :a) \(\frac{x-214}{86}+\frac{x-132}{84}+\frac{x-54}{82}+\frac{x-20}{80}=10\)
b) \(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)
c) \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
GIẢI CÁC PHƯƠNG TRÌNH SAU :
Bài 1:
a, \(x^2-4+3>=0\)
b, \(x^4-4x^2+3< 0\)
Bài 2 :
a, \(\frac{7x+5}{5}-x=\frac{\left|3x-5\right|}{2}\)
b, \(x-\frac{\left|3x-2\right|}{5}=3-\frac{2x-5}{3}\)
c, \(\left|x-1\right|+\left|x-2\right|=1\)
\(\text{Giải phương trình:}\)
\(a,\frac{5-x}{4x^2-8x}+\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\)
\(b,\frac{x-49}{50}+\frac{x-50}{49}=\frac{49}{x-50}+\frac{50}{x-49}\)
\(c,\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{1}{x+3}\)
Giải các phương trình:
a) \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
b) \(\frac{3}{\left(x-1\right)\left(x-2\right)}+\frac{2}{\left(x-3\right)\left(x-1\right)}=\frac{1}{\left(x-2\right)\left(x-3\right)}\)
c) \(1+\frac{1}{x+2}=\frac{12}{8+x^3}\)
d) \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{\left(x-3\right)\left(x+3\right)}\)
Rút gọn : \(\left[\left(x^3-1-\frac{7-x^3}{3+x^3}\right).\frac{4}{x^5+3x^2}\right]:\left[\frac{3x^6-12}{x^9+6x^6+9x^3}.\frac{x}{3x^3+6}\right]\)
