6x⁴+7x³-36x²-7x+6=0

<=> 6x⁴-2x³+9x³-3x²-33x²+11x-18x+6=0

<=> 2x³(3x-1)+3x²(3x-1)-11x(3x-1)-6(3x-1)=0

<=> (3x-1)(2x³+3x²-11x-6)=0

<=>(3x-1)(2x³-4x²+7x²-14x+3x-6)=0

<=>(3x-1)[2x²(x-2)+7x(x-2)+3(x-2)]=0

<=>(3x-1)(x-2)(2x²+7x+3)=0

<=>(3x-1)(x-2)(2x²+6x+x+3)=0

<=>(3x-1)(x-2)[2x(x+3)+(x+3)]=0

<=>(3x-1)(x-2)(x+3)(2x+1)=0

th1: 3x+1=0 <=> x=\(-\frac{1}{3}\)

th2: x-2=0 <=> x=2

th3: x+3=0 <=> x=-3

th4: 2x+1=0 <=> x=-\(\frac{1}{2}\)

Cái này t dùng máy tính

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

Đến đây thì pt có 4 nghiệm:\(x=2;-3;-\frac{1}{2};\frac{1}{3}\)

Vậy....

Yêu cầu giải không dùng máy tính.

sao mà đòi hỏi zậy còn ko cho dùng máy tính

(6x⁴-12x³)+(19³-38x²)+(2x²-4x)-(3x-6)=0

6x^3(x-2)+19x^2(x-2)+2x(x-2)-3(x-2)=0

(x-2)(6x^3+19x^2+2x-3)=0

(x-2)[(6x^3+18x^2)+(x^2+3x)-(x+3)]=0

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

(x-2)(x+3)[(6x^2+3x)-(2x+1)]=0

(x-2)(x+3)(2x+1)(3x-1)=0

⇒ x=2

x=-3

x=-1/2

x=1/3

Lạnh xun loz

phải không mày >

haizz

\(6x^4+7x^3-36x^2-7x+6=0\)

\(\Leftrightarrow\left(6x^4-11x^3-3x^2+2x\right)+\left(18x^3-33x^2-9x+6\right)=0\)

\(\Leftrightarrow x\left(6x^3-11x^2-3x+2\right)+3\left(6x^3-11x^2-3x+2\right)=0\)

\(\Leftrightarrow\left(6x^3-11x^2-3x+2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[\left(6x^3-14x+4x\right)+\left(3x^2-7x+2\right)\right]\left(x+3\right)=0\)

\(\Leftrightarrow\left[2x\left(3x^2-7x+2\right)+\left(3x^2-7x+2\right)\right]\left(x+3\right)=0\)

\(\Leftrightarrow\left(3x^2-7x+2\right)\left(2x+1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(3x^2-6x-x+2\right)\left(2x+1\right)\left(x+3\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\3x-1=0\end{cases}}\)hoặc \(\orbr{\begin{cases}2x+1=0\\x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{3}\end{cases}}\)hoặc\(\orbr{\begin{cases}x=\frac{-1}{2}\\x=-3\end{cases}}\)

Vậy tập hợp nghiệm \(S=\left\{2;-3;\frac{1}{3};\frac{-1}{2}\right\}\)

Xét thấy x = 0 không thỏa mãn pt

Ta có : \(6x^4+7x^3-36x^2+7x+6=0\)

\(\Leftrightarrow x^2\left(6x^2+7x-36+\frac{7}{x}+\frac{6}{x^2}\right)=0\)

\(\Leftrightarrow6x^2+7x-36+\frac{7}{x}+\frac{6}{x^2}=0\)

\(\Leftrightarrow6\left(x^2+\frac{1}{x^2}\right)+7\left(x+\frac{1}{x}\right)-36=0\)

\(\Leftrightarrow6\left(x+\frac{1}{x}\right)^2-7\left(x+\frac{1}{x}\right)-36-12=0\)

\(\Leftrightarrow6\left(x+\frac{1}{x}\right)^2-7\left(x+\frac{1}{x}\right)-48=0\)

Đặt \(x+\frac{1}{x}=a\)

\(pt\Leftrightarrow6a^2-7a-48=0\)

\(\Leftrightarrow6\left(a^2-\frac{7}{6}a-8\right)=0\)

\(\Leftrightarrow a^2-\frac{7}{6}a-8=0\)

\(\Leftrightarrow a^2-2\cdot a\cdot\frac{7}{12}+\frac{49}{144}-\frac{1201}{144}=0\)

\(\Leftrightarrow\left(a-\frac{7}{12}\right)^2=\left(\frac{\pm\sqrt{1201}}{12}\right)^2\)

\(\Leftrightarrow a=\frac{\pm\sqrt{1201}+7}{12}\)

\(\Leftrightarrow x+\frac{1}{x}=\frac{\pm\sqrt{1201}+7}{12}\)

Giải nốt nha bạn. Nghiệm hơi xấu