\(\left(x-\frac{2x}{x+2}\right)^2\)+\(\frac{4x^2}{x+2}\)=12
\(\left(\frac{x^2}{x+2}\right)^2\)+4.\(\frac{x^2}{x+2}\)-12 =0
Đặt
\(\frac{x^2}{x+2}\)=t
đưa về phương trình bậc 2 theo ẩn t.
\(\left(x-\frac{2x}{x+2}\right)^2\)+\(\frac{4x^2}{x+2}\)=12
\(\left(\frac{x^2}{x+2}\right)^2\)+4.\(\frac{x^2}{x+2}\)-12 =0
Đặt
\(\frac{x^2}{x+2}\)=t
đưa về phương trình bậc 2 theo ẩn t.
Giải PT
\(x^2+\frac{4x^2}{\left(x+2\right)^2}=12\)
Giải các pt sau
a, \(\frac{6}{x^2+2}+\frac{7}{x^2+3}+\frac{12}{x^2+8}-\frac{3x^2+16}{x^2+10}=1\)
b,\(2x\left(8x-1\right)^2\left(4x-1\right)=9\)
giải các pt
\(a,\frac{2x-13}{2x-16}+\frac{2\left(x-6\right)}{x-8}=\frac{7}{8}+\frac{2\left(5x-39\right)}{3x-24}\)
\(b,x\left(x-2\right)\left(x-1\right)\left(x+1\right)=24\)
\(c,x^4+2x^3+5x^2+4x-12=0\)
Giải pt sau :
\(\left(3x-2\right)\left[\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}\right]=0\)
giải các pt sau:
a, \(\left(x^2+4x+8\right)^2+3x.\left(x^2+4x+8\right)+2x^2=0\) 0
b, \(\frac{x-5}{2017}+\frac{x-2}{2020}=\frac{x-6}{2016}+\frac{x-68}{1954}\)
giải các pt sau
\(\frac{x+1}{3}+\frac{3\left(2x+1\right)}{4}=\frac{2x+3\left(x+1\right)}{6}+\frac{7+12x}{12}\)
\(\frac{2\left(3x+1\right)+1}{4}-5=\frac{2\left(3x-1\right)}{5}-\frac{3x+2}{10}\)
\(\frac{3\left(x-3\right)}{4}+\frac{4x-10.5}{10}=\frac{3\left(x+1\right)}{5}+6\)
\(\frac{x+1}{58}+\frac{x+2}{57}=\frac{x+3}{56}+\frac{x+4}{55}\)
B1 :Giải phương trình
a,\(\frac{3\left(x-3\right)}{4}-1=\frac{2x+3\left(x+1\right)}{6}-\frac{7+12x}{12}\)
b,\(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
c,\(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
d,I7-xI-5x=1
B2:Giải bất phương trình
a,\(\left(x-2\right)\left(x+2\right)\ge x\left(x-4\right)\)
b,\(\frac{x-1}{4}-1\ge\frac{x+1}{3}+8\)
Giải các pt sau:
a, \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)
b,\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
Giúp mình với ạ
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)2\)
giải pt