\(\Leftrightarrow x^3-6x^2+12x-8=-27\)
\(\Leftrightarrow\left(x-2\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x-2=-3\)
\(\Leftrightarrow x=-1\)
x3+6x^2+12x−19=0
⇔(x^3+6x^2+12x+8)−27=0
⇔(x+2)^3=3
⇔x+2=3
⇒x=1
Vậy...
\(\Leftrightarrow x^3-6x^2+12x-8=-27\)
\(\Leftrightarrow\left(x-2\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x-2=-3\)
\(\Leftrightarrow x=-1\)
x3+6x^2+12x−19=0
⇔(x^3+6x^2+12x+8)−27=0
⇔(x+2)^3=3
⇔x+2=3
⇒x=1
Vậy...
giải phương trình
a)(x+3)^2-(x-3)^2=6x+18
b)x+3 phần x-2=5 phần (x-2)(3-x)
c)12x^2+30x-21 phần 16x^2-9
-3x-7 phần 3-4x =6x+5 phần 4x+3
d)4 phần x+1-2 phần x-2=x+3 phần x^2-x-2
ai giúp mik với mik nhiều bt lắm
a, 1/x-1 - 7/x+2 = 3/x2+x-2
b,x+3/x-4 + x-1/x-2 = 2/6x-8-x2
c,1/x+1 + 2/x3-x2x+1 = 3/1-x2
d, 3x-1/x-1 - 2x+5/x+3 = 1 - 4/x2+2x-3
e, x2+2x+1/x2+2x+2 + x2+2x+2/x2+2x+3 = 7/6
f, 1/4x2-12x+9 - 3/9-4x2=4/4x2+12x+9
câu 1: x+1/x-1-x+2/x+3+4/x^2+2x-3=0
câu 2 : 1/x-5 - 3/x^2-6x+5 = 5/x-1
câu 3 : x^2-3x-4 = 0
Giải Pt \(\dfrac{14}{20-6x-2x^2}+\dfrac{x^2+4x}{x^2+5x}-\dfrac{x+3}{2-x}+3=0\)
23) \(\dfrac{1}{x^2+4x+3}+\dfrac{1}{x^2+8x+15}+\dfrac{1}{x^2+12x+35}=\dfrac{1}{9}\)
24) \(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)
25) \(\dfrac{x^2+2x+2}{x+1}+\dfrac{x^2+8x+20}{x+4}=\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+6x+12}{x+3}\)
\(\frac{4}{x^2-3x+2}-\frac{3}{2x^2-6x+1}+1=0\)
\(\frac{1}{x-1}+\frac{2}{x-2}+\frac{3}{x-3}=\frac{6}{x-6}\)
\(\frac{12x^2+30x-21}{16x^2-9}\)-\(\frac{3x-7}{3-4x}\)=\(\frac{6x+5}{4x+3}\)
Giúp mình giải với ạ, cảm ơn nhiều ạ
\(\frac{4}{x^2-3x+2}-\frac{3}{2x^2-6x+1}+1=0\)
Em cảm ơn