Giải pt sau:\(x^3-3x^2+9x-9=0\)
Dùng hằng đẳng thức giải pt sau:
a) x^3-x^2-x=1/3
b) 5x^3+6x^2+12x+8=0
c)x^3=3x^2-9x+9
giải pt \(x^3-3x^2+2\sqrt{x+3}^3-9x=0\)
giải pt:
9x^3 -1 + (3x - 1)(x+2) =0
Sửa đề: 9x^2-1+(3x-1)(x+2)=0
=>(3x-1)(3x+1)+(3x-1)(x+2)=0
=>(3x-1)(3x+1+x+2)=0
=>(3x-1)(4x+3)=0
=>x=-3/4 hoặc x=1/3
giải pt sau
a, \(x^4=4x+1\) b,\(x^3-3x^2+9x-9=0\)
a. Dat \(x^2=t\left(t\ge0\right)\)
Suy ra PT:\(\orbr{\begin{cases}t^2=-4t+1\left(1\right)\left(x< 0\right)\\t^2=4t+1\left(2\right)\left(x\ge0\right)\end{cases}}\)
(1)\(\Leftrightarrow t^2+4t-1=0\)
\(\Leftrightarrow\left(t+2\right)^2-5=0\)
\(\Leftrightarrow\left(t+2+\sqrt{5}\right)\left(t+2-\sqrt{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=-2-\sqrt{5}\left(l\right)\\t=\sqrt{5}-2\left(n\right)\end{cases}}\)
Nghiem cua PT(1) la \(t=\sqrt{5}-2\)
(2)\(\Leftrightarrow t^2-4t-1=0\)
\(\Leftrightarrow\left(t-2\right)^2-5=0\)
\(\Leftrightarrow\left(t-2+\sqrt{5}\right)\left(t-2-\sqrt{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=2-\sqrt{5}\left(l\right)\\t=2+\sqrt{5}\left(n\right)\end{cases}}\)
Nghiem cua PT(2) la \(t=2+\sqrt{5}\)
Suy ra:\(\orbr{\begin{cases}x=\sqrt{\sqrt{5}-2}\\x=\sqrt{\sqrt{5}+2}\end{cases}}\)
b.\(x^3-3x^2+9x-9=0\)
\(\Leftrightarrow\left(x-3\right)^3=-18\)
\(\Leftrightarrow x-3=-\sqrt[3]{18}\)
\(\Leftrightarrow x=3-\sqrt[3]{18}\)
\(b,x^3-3x^2+9x-9=0\)
\(\Rightarrow x^2\left(x-3\right)+9\left(x-3\right)+18=0\)
\(\Rightarrow\left(x^2+9\right)\left(x-3\right)=-18\)
từ đây bạn xét các TH nhá !
Chú ý : Vì \(x^2+9\ge9\forall\) để xét ít Th hơn
a, x^4 - 4x - 1 = 0
<=> x^4 +2x^2 + 1 -2x^2 - 4x - 2 = 0
<=> (x^2 + 1)^2 - 2(x + 1)^2 = 0
<=> (x^2 + x*can(2) + 1 + can2))*(x^2 - x*can2) + 1 - can2)) = 0
<=> x^2 + x*can(2) + 1 + can(2) = 0 (1)
hoac x^2 - x*can(2) + 1 - can(2) = 0 (2)
PT (1) vo nghiem
PT (2) co 2 nghiem phan biet:
x1 = (can2) + can(-2 + 4can(2)))/2
x2 = (can(2) + can(-2 + 4can(2)))/2
Giải bất pt sau
\(\left(x-1\right)\left(3x^2+9x-12\right)< 0\)
Giải PT sau :
\(3x\left(2+\sqrt{9x^2+3}\right)-\left(4x+1\right)\left(1+\sqrt{1+x+x^2}\right)=0\)
GIẢI PT: 3x^3 + 3x^2 +9x +1 = 0
giải pt vô tỉ sau
\(3x\left(2+\sqrt{9x^2+3}\right)+\left(4x+2\right)\left(1+\sqrt{1+x+x^2}\right)=0\)
T sợ chỉ dám liên hợp thôi, nhường cách bình phương cho 1 ng` chăm chỉ :(
\(pt\Leftrightarrow6x+3x\sqrt{9x^2+3}+4x+2+\left(4x+2\right)\sqrt{x^2+x+1}=0\)
\(\Leftrightarrow2\left(5x+1\right)+\left(3x\sqrt{9x^2+3}+\dfrac{6\sqrt{21}}{25}\right)+\left(\left(4x+2\right)\sqrt{x^2+x+1}-\dfrac{6\sqrt{21}}{25}\right)=0\)
\(\Leftrightarrow2\left(5x+1\right)+\dfrac{\dfrac{27}{625}\left(5x-1\right)\left(5x+1\right)\left(75x^2+28\right)}{3x\sqrt{9x^2+3}-\dfrac{6\sqrt{21}}{25}}+\dfrac{\dfrac{4}{625}\left(5x+1\right)\left(5x+4\right)\left(100x^2+100x+109\right)}{\left(4x+2\right)\sqrt{x^2+x+1}+\dfrac{6\sqrt{21}}{25}}=0\)
\(\Leftrightarrow\left(5x+1\right)\left(2+\dfrac{\dfrac{27}{625}\left(5x-1\right)\left(75x^2+28\right)}{3x\sqrt{9x^2+3}-\dfrac{6\sqrt{21}}{25}}+\dfrac{\dfrac{4}{625}\left(5x+4\right)\left(100x^2+100x+109\right)}{\left(4x+2\right)\sqrt{x^2+x+1}+\dfrac{6\sqrt{21}}{25}}\right)=0\)
\(\Rightarrow5x+1=0\Rightarrow x=-\dfrac{1}{5}\)
GIẢI PT: 3x^3 + 3x^2 + 9x + 1 = 0