tôi cx ko chưa chắc chắn câu này nên chưa giải đc đâu

nha pn

\(x^6+6x^4-36x^3+6x^2+1=0\)

\(\Leftrightarrow\left(x^2-3x+1\right)\left(x^4+3x^3+14x^2+3x+1\right)=0\)

Dễ thấy \(x^4+3x^3+14x^2+3x+1>0\)

\(\Rightarrow x^2-3x+1=0\)

\(\Leftrightarrow x=\dfrac{3\pm\sqrt{5}}{2}\)

Huong dan
1) (x² - 5x + 1)(x² - 4) = 6(x - 1)²
<=> [(x² - 4) - 5(x - 1)](x² - 4) - 6(x - 1)² = 0
<=> (x² - 4)² - 5(x - 1)(x² - 4) - 6(x - 1)² = 0
Nhan thay x = 1 khong phai la nghiem => x - 1 ≠ 0 nen co the chia 2 ve cua pt cho (x - 1)² ≠ 0 va dat y = (x² - 4)/(x - 1) ta co pt bac 2 theo y
y² - 5y - 6 = 0 => y = - 1; y = 6
Ban tu giai tip

2) 3√(x³ + 8) = 2x² - 6x + 4 (x ≥ - 2 )
<=> 3√[(x + 2)(x² - 2x + 4)] = 2(x² - 2x + 4) - 2(x + 2)
<=> 2(x + 2) + 3√[(x + 2)(x² - 2x + 4)] - 2(x² - 2x + 4) = 0
Chia 2 ve pt cho √(x² - 2x + 4) = √[(x - 1)² + 3]> 0 va dat y = √[(x + 2)/(x² - 2x + 4)] ta co pt bac 2 theo y:
2y² + 3y - 2 = 0 => y = 1/2 ( loai nghiem y = - 2)
Ban tu giai tiep

a. Phương trình tương đương với \(\left(x^2-2x-2\right)\left(x^2+5x-2\right)=0\)  hay \(x^2-2x-2=0\)  hoặc \(x^2+5x-2=0\). Đến đây sử dụng Delta hoặc viết hai phương trình dưới dạng \(\left(x-1\right)^2=3,\left(2x+5\right)^2=33\) ta được bốn nghiệm là \(x=1\pm\sqrt{3},-\frac{5}{2}\pm\frac{\sqrt{33}}{2}\)

b. Phương trình tương đương với \(3\left(x+5\right)\left(x+6\right)\left(x+9\right)=8x+6\left(x+5\right)\left(x+6\right)\leftrightarrow3\left(x+5\right)\left(x+6\right)\left(x+9\right)=\left(x+9\right)\left(6x+20\right)\)

hay \(\left(x+9\right)\left(3x^2+27x+70\right)=0\leftrightarrow x=-9.\)

a) \(\left(4x^2-25\right)\left(2x^2-7x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x^2-25=0\left(1\right)\\2x^2-7x-9=0\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x^2=\frac{25}{4}\Leftrightarrow x=\pm\frac{5}{2}\)

\(\left(2\right)\Leftrightarrow2x^2-9x+2x-9=0\)

\(\Leftrightarrow2x\left(x+1\right)-9\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{9}{2}\end{matrix}\right.\)

Vậy....

b) \(\left(2x^2-3\right)^2-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(2x^2-3\right)^2-\left(2x-2\right)^2=0\)

\(\Leftrightarrow\left(2x^2-3-2x+2\right)\left(2x^2-3+2x-2\right)=0\)

\(\Leftrightarrow\left(2x^2-2x-1\right)\left(2x^2+2x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-2x-1=0\left(3\right)\\2x^2+2x-5=0\left(4\right)\end{matrix}\right.\)

\(\left(3\right)\Delta=2^2-4\cdot2\cdot\left(-1\right)=12\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2-\sqrt{12}}{4}=\frac{1-\sqrt{3}}{2}\\x=\frac{2+\sqrt{12}}{4}=\frac{1+\sqrt{3}}{2}\end{matrix}\right.\)

\(\left(4\right)\Delta=2^2-4\cdot2\cdot\left(-5\right)=44\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2-\sqrt{44}}{4}=\frac{-1-\sqrt{11}}{2}\\x=\frac{-2+\sqrt{44}}{4}=\frac{-1+\sqrt{11}}{2}\end{matrix}\right.\)

Vậy...

c) \(x^3+5x^2+7x+3=0\)

\(\Leftrightarrow x^3+3x^2+2x^2+6x+x+3=0\)

\(\Leftrightarrow x^2\left(x+3\right)+2x\left(x+3\right)+\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)

Vậy...

d) \(x^3-6x^2+11x-6=0\)

\(\Leftrightarrow x^3-2x^2-4x^2+8x+3x-6=0\)

\(\Leftrightarrow x^2\left(x-2\right)-4x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-4x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=3\end{matrix}\right.\)

Vậy...

pt bậc 4 => có 4 nghiệm.

bấm máy tính tìm nghiệm đẹp (-2 và 3). Chia sơ đồ hoocne.

2 nghiệm đẹp (-2 và 3) được rồi, còn 2 nghiệm còn lại thì giải pt bậc 2 là ra.

kq: x=-2, x=3, x=1/3 , x=-1/2

Ta có \(6x^4-5x^3-38x^2-5x+6=0\Leftrightarrow6x^4+12x^3-17x^3-34x^2-4x^2-8x+3x+6=0\Leftrightarrow6x^3\left(x+2\right)-17x^2\left(x+2\right)-4x\left(x+2\right)+3\left(x+2\right)=0\Leftrightarrow\left(x+2\right)\left(6x^3-17x^2-4x+3\right)=0\Leftrightarrow\left(x+2\right)\left(6x^3-18x^2+x^2-3x-x+3\right)=0\Leftrightarrow\left(x+2\right)\left[6x^2\left(x-3\right)+x\left(x-3\right)-\left(x-3\right)\right]=0\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(6x^2+x-1\right)=0\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(6x^2-2x+3x-1\right)=0\Leftrightarrow\left(x+2\right)\left(x-3\right)\left[2x\left(3x-1\right)+\left(3x-1\right)\right]=0\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(3x-1\right)\left(2x+1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x+2=0\\x-3=0\\3x-1=0\\2x+1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-2\\x=3\\x=\dfrac{1}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy S={\(-\dfrac{1}{2};-2;\dfrac{1}{3};3\)}

Ta có : \(6x^4+5x^3-38x^2+5x+6=0\)

\(\Leftrightarrow6x^4+20x^3+6x^2-15x^3-50x^2-15x+6x^2+20x+6=0\)

\(\Leftrightarrow2x^2\left(3x^2+10x+3\right)-5x\left(3x^2+10x+3\right)+2\left(3x^2+10x+3\right)=0\)

\(\Leftrightarrow\left(3x^2+10x+3\right)\left(2x^2-5x+2\right)=0\)

\(\Leftrightarrow\left(3x^2+x+9x+3\right)\left(2x^2-x-4x+2\right)=0\)

\(\Leftrightarrow\left[x\left(3x+1\right)+3\left(3x+1\right)\right]\left[x\left(2x-1\right)-2\left(2x-1\right)\right]=0\)

\(\Leftrightarrow\left(3x+1\right)\left(x+3\right)\left(2x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\)\(3x+1=0\)

hoặc    \(x+3=0\)

hoặc   \(2x-1=0\)

hoặc    \(x-2=0\)

\(\Leftrightarrow\)\(x=-\frac{1}{3}\)

hoặc   \(x=-3\)

hoặc   \(x=\frac{1}{2}\)

hoặc   \(x=2\)

Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{1}{3};-3;\frac{1}{2};2\right\}\)

\(\left(x^2-2x+6\right)\left(x^2-8x+4\right)+\left(5x+1\right)\left(x+1\right)-\left(x^2-3x-3\right)\left(x^2+x-3\right)=0\)

\(\Leftrightarrow x^8-5x^2+7x-2=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-3x+1\right)=0\)

Xong rồi nhé

\(\left(x^2-2x+6\right)\left(x^2-8x-4\right)+\left(5x+1\right)\)\(\left(x-1\right)-\left(x^2-3x-3\right)\left(x^2+x-3\right)=\)\(0\)

\(\Leftrightarrow x^8-5x^2+7x-2=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-3x+1\right)=0\)

~ 양 셜 김 ~