=> (x-5)x+1-(x-5)x+1+20=0
=> (x-5)x+1.1 - (x-5)x+1.(x-5)20=0
=> (x-5)x+1.[1-(x-5)20 ]=0
=> (x-5)x+1 = 0 hoặc 1-(x-5)20=0
=> x-5 = 0 hoặc (x-5)20 = 1
=> x=5 hoặc x-5=1
=> x=5 hoặc x=6.
tim x biet :
x-128 = \(\left(4\frac{20}{21}-5\right):\left(\frac{4141}{4242}-1\right):\left(\frac{636363}{646464}-1\right)\)
tim x biet
\(\left(x-\frac{1}{3}\right).\left(y-\frac{1}{2}\right).\left(z-5\right)=0\)
và x+2=y+1=z+3
Tim x,y,z biet:
\(x+1=y+2=z+3và\left(x-\frac{1}{5}\right)\left(y+\frac{1}{3}\right)\left(z-6\right)=0\)
Tim x biet
\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(\right)x+10}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
1. Tim x biet : \(2015-\left|x-2015\right|=x;\left|x\right|=-x-5\)
Tim x biet neu \(\frac{x-2}{\left(a+3\right)\left(a-5\right)}=\frac{1}{2\left(a+3\right)}+\frac{1}{2\left(a+5\right)}\)
va x khac -3;5
tim x,y,z biet \(\sqrt{\left(x-\sqrt{5}\right)^2}+\sqrt{\left(y+\sqrt{3}\right)^2}+\left|x-y-z\right|\)
tim x
a)\(2x\left(x-\frac{1}{7}\right)=0\)
b)\(\left(x-9\right).\left(x+\frac{3}{5}\right)=0\)
c)\(\left(\frac{-4}{7}-2x\right)\left(x-\frac{5}{4}\right)=0\)
tim x
a, \(\left(x+1\right)\left(2-x\right)\left(3+x^2\right)\ge0\)
b, \(\left(2x+1\right)\left(x+3\right)\left(x-5\right)>0\)
c,\(\frac{\left(3x-6\right)\left(x-4\right)}{x+3}< 0\)
