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1 tháng 2 2019
\(PT< =>x^4+5x^3-6x^2-6x+5x^2-6x-6=0\)
\(< =>x^4+5x^3-x^2-12x-6=0\)
\(< =>\left(x^2-x-1\right)\left(x^2+6x+6\right)=0\)
<=>\(\orbr{\begin{cases}x=\frac{1+\sqrt{5}}{2}\\x=\frac{1-\sqrt{5}}{2}\end{cases}}\)hay \(\orbr{\begin{cases}x=-3+\sqrt{3}\\x=-3-\sqrt{3}\end{cases}}\)
Vậy \(S=\left\{\frac{1+\sqrt{5}}{2};\frac{1-\sqrt{5}}{2};-3+\sqrt{3};-3-\sqrt{3}\right\}\)
1 tháng 3 2019
1) \(x^4-6x^3-x^2+54x-72=0\)
\(\Leftrightarrow x^3\left(x-2\right)-4x^2\left(x-2\right)-9x\left(x-2\right)+36\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-4x^2-9x+36\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-4\right)-9\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x-3\right)\left(x+3\right)=0\)
Tự làm nốt...
2) \(x^4-5x^2+4=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)-4\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
Tự làm nốt...
1 tháng 3 2019
\(x^4-2x^3-6x^2+8x+8=0\)
\(\Leftrightarrow x^3\left(x-2\right)-6x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-6x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+2\right)-2x\left(x+2\right)-2\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2-2x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left[\left(x-1\right)^2-\left(\sqrt{3}\right)^2\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{3}\right)=0\)
...
\(2x^4-13x^3+20x^2-3x-2=0\)
\(\Leftrightarrow2x^3\left(x-2\right)-9x^2\left(x-2\right)+2x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-9x^2+2x+1\right)=0\)
Bí
TD
0
LC
1
4 tháng 2 2018
1 ) \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
Đặt \(t=x^2+x\), ta được :
\(t^2+4t-12=0\)
\(\Leftrightarrow t^2-2t+6t-12=0\)
\(\Leftrightarrow\left(t-2\right)\left(t+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-6\end{matrix}\right.\)
+ ) Khi \(t=2,\) thì :
\(x^2+x=2\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
+ ) Khi \(t=-6,\) thì :
\(x^2+x=-6\)
\(\Leftrightarrow x^2+x+6=0\)
\(\Leftrightarrow x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{23}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}=0\) ( vô lí )
Vậy .........
2 ) \(6x^4-5x^3-38x^2-5x+6=0\)
\(\Leftrightarrow6x^4-18x^3+13x^3-39x^2+x^2-3x-2x+6=0\)
\(\Leftrightarrow6x^3\left(x-3\right)+13x^2\left(x-3\right)+x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(6x^3+13x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(6x^3+12x^2+x^2+2x-x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[6x^2\left(x+2\right)+x\left(x+2\right)-\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(6x^2+x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(6x^2+3x-2x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left[3x\left(2x+1\right)-\left(2x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(3x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=\dfrac{1}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
TN
1
4 tháng 4 2020
a) x^4 - 5x^2 + 4 = 0
<=> (x^2 - 1)(x^2 - 4) = 0
<=> x^2 - 1 = 0 hoặc x^2 - 4 = 0
<=> x = +-1 hoặc x = +-2
b) x^4 - 10x^2 + 9 = 0
<=> (x^2 - 1)(x^2 - 9) = 0
<=> x^2 - 1 = 0 hoặc x^2 - 9 = 0
<=> x = +-1 hoặc x = +-3
c) x^3 + 6x^2 + 11x + 6 = 0
<=> (x^2 + 5x + 6)(x + 1) = 0
<=> (x + 2)(x + 3)(x + 1) = 0
<=> x + 2 = 0 hoặc x + 3 = 0 hoặc x + 1 = 0
<=> x = -2 hoặc x = -3 hoặc x = -1
d) x^3 + 9x^2 + 26x + 24 = 0
<=> (x^2 + 7x + 12)(x + 2) = 0
<=> (x + 3)(x + 4)(x + 2) = 0
<=> x + 3 = 0 hoặc x + 4 = 0 hoặc x + 2 = 0
<=> x = -3 hoặc x = -4 hoặc x = -2
5 tháng 7 2016
1.
Đặt \(x^2-5x=a\Rightarrow a^2=\left(x^2-5x\right)^2\)
Thay vào pt:
\(\Rightarrow a^2+10a+24=0\)
\(\Leftrightarrow a^2+6a+4a+24=0\)
\(\Leftrightarrow a\left(a+6\right)+4\left(a+6\right)=0\)
\(\Leftrightarrow\left(a+6\right)\left(a+4\right)=0\)
\(\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-3x-2x+6\right)\left(x^2-4x-x+4\right)=0\)
\(\Leftrightarrow\left[x\left(x-3\right)-2\left(x-3\right)\right]\left[x\left(x-4\right)-\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x-4\right)\left(x-1\right)=0\)
\(\Rightarrow x-3=0,x-2=0,x-4=0,x-1=0\)
\(\Rightarrow x=3,x=2,x=4,x=1\)
T I C K mình sẽ giải típ cho cảm ơn
\(x^4+\left(x+1\right)\left(5x^2-6x-6\right)=0\)
\(\Leftrightarrow x^4+5x^3-x^2-12x-6=0\)
\(\Leftrightarrow x^4-x^3+6x^3-x^2-6x^2+6x^2\)
\(-6x-6x-6=0\)
\(\Leftrightarrow\left(x^4-x^3-x^2\right)+\left(6x^3-6x^2-6x\right)+\)
\(\left(6x^2-6x-6\right)=0\)
\(\Leftrightarrow x^2\left(x^2-x-1\right)+6x\left(x^2-x-1\right)+\)
\(6\left(x^2-x-1\right)=0\)
\(\Leftrightarrow\left(x^2+6x+6\right)\left(x^2-x-1\right)=0\)
\(TH1:x^2+6x+6=0\)
Ta có: \(\Delta=6^2-4.6=12\sqrt{\Delta}=\sqrt{12}\)
pt có 2 nghiệm:
\(x_1=\frac{-6+\sqrt{12}}{2}=-3+\sqrt{3}\)
\(x_2=\frac{-6-\sqrt{12}}{2}=-3-\sqrt{3}\)
\(TH2:x^2-x-1=0\)
Ta có: \(\Delta=1^2+4.1=5,\sqrt{\Delta}=\sqrt{5}\)
pt có 2 nghiệm:
\(x_1=\frac{1+\sqrt{5}}{2}\)và \(x_2=\frac{1-\sqrt{5}}{2}\)
Vậy pt có 4 nghiệm \(x_1=\frac{-6+\sqrt{12}}{2}=-3+\sqrt{3}\);\(x_2=\frac{-6-\sqrt{12}}{2}=-3-\sqrt{3}\);
\(x_3=\frac{1+\sqrt{5}}{2}\);\(x_4=\frac{1-\sqrt{5}}{2}\)
Làm tốt rồi nhưng mà lớp 8 chưa học cách giải pt bậc 2 \(\Delta\). Thì chúng ta có thể:
VD TH1: \(x^2+6x+6=0\)
<=> \(x^2+6x+9-9+6=0\)
<=> \(\left(x+3\right)^2=3\)
<=> \(\orbr{\begin{cases}x+3=\sqrt{3}\\x+3=-\sqrt{3}\end{cases}}\)<=> \(\orbr{\begin{cases}x=-3+\sqrt{3}\\x=-3-\sqrt{3}\end{cases}}\)
tương tự Th2.