\(\frac{2x-1}{3x^2+7x+2}+\frac{3}{9x^2+15x+4}-\frac{2x+7}{3x^2-5x-12}=\frac{5}{x+2}\)

\(\Leftrightarrow\frac{2x-1}{\left(3x+1\right)\left(x+2\right)}+\frac{3}{\left(3x+1\right)\left(3x+4\right)}-\frac{2x+7}{\left(4x+3\right)\left(x-3\right)}=\frac{5}{\left(x+2\right)}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{3x+1}+\frac{1}{3x+1}-\frac{1}{3x+4}+\frac{1}{3x+4}-\frac{1}{x-3}=\frac{5}{x+2}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x-3}=\frac{5}{x+2}\)

\(\Leftrightarrow\frac{x-3-x-2}{\left(x+2\right)\left(x-3\right)}=\frac{5\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}\)

\(\Leftrightarrow5x-3=-5\)

\(\Leftrightarrow x=-\frac{2}{5}\)

Chúc bạn học tốt !!!

Bài 1:

Đặt \(t=2x^2+3x-1\) ta có:

\(t^2-5\left(t+4\right)+24=0\)

\(\Rightarrow t^2-5t-20+24=0\)

\(\Rightarrow t^2-5t+4=0\)

\(\Rightarrow\left(t-4\right)\left(t-1\right)=0\)\(\Rightarrow\left[\begin{matrix}t=4\\t=1\end{matrix}\right.\)

*)Xét \(2x^2+3x-1=4\)

\(\Rightarrow\left(x-1\right)\left(2x+5\right)=0\)\(\Rightarrow\left[\begin{matrix}x=1\\x=-\frac{5}{2}\end{matrix}\right.\)

*)Xét \(2x^2+3x-1=1\)

\(\Rightarrow\left(x+2\right)\left(2x-1\right)=0\)\(\Rightarrow\left[\begin{matrix}x=-2\\x=\frac{1}{2}\end{matrix}\right.\)

Bài 2:

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

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

\(\Rightarrow\left(x^2-4\right)\left[x+3-\left(x-1\right)\right]=0\)

\(\Rightarrow4\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x+2\right)=0\)\(\Rightarrow\left[\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

$\dfrac{4}{-25x^2+30x-3}=\dfrac{3}{5x-1}-\dfrac{2}{5x-1}$

Điều kiện: $5x-1\ne0$

$\Leftrightarrow x\ne\dfrac{1}{5}$

Ta có:

$\dfrac{3}{5x-1}-\dfrac{2}{5x-1}=\dfrac{1}{5x-1}$

Do đó:

$\dfrac{4}{-25x^2+30x-3}=\dfrac{1}{5x-1}$

$\Leftrightarrow 4(5x-1)=-25x^2+30x-3$

$\Leftrightarrow 20x-4=-25x^2+30x-3$

$\Leftrightarrow 25x^2-10x-1=0$

$\Leftrightarrow x=\dfrac{10\pm\sqrt{(-10)^2-4\cdot25\cdot(-1)}}{2\cdot25}$

$\Leftrightarrow x=\dfrac{10\pm\sqrt{200}}{50}$

$\Leftrightarrow x=\dfrac{10\pm10\sqrt2}{50}$

$\Leftrightarrow x=\dfrac{1\pm\sqrt2}{5}$

Hai nghiệm đều thỏa mãn điều kiện.

Vậy: $x\in\left{\dfrac{1-\sqrt2}{5};\dfrac{1+\sqrt2}{5}\right}$

$\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+4}-\dfrac{2}{x^2-4x+3}=0$

Điều kiện:

$x^2-3x+2=(x-1)(x-2)\ne0$

$x^2-5x+4=(x-1)(x-4)\ne0$

$x^2-4x+3=(x-1)(x-3)\ne0$

$\Rightarrow x\ne1,2,3,4$

Ta có:

$\dfrac{1}{(x-1)(x-2)}+\dfrac{1}{(x-1)(x-4)}-\dfrac{2}{(x-1)(x-3)}=0$

$\Leftrightarrow \dfrac{1}{x-2}+\dfrac{1}{x-4}-\dfrac{2}{x-3}=0$

Quy đồng:

$\Leftrightarrow \dfrac{(x-4)(x-3)+(x-2)(x-3)-2(x-2)(x-4)}{(x-2)(x-3)(x-4)}=0$

Khai triển tử:

$(x-4)(x-3)+(x-2)(x-3)-2(x-2)(x-4)$

$=x^2-7x+12+x^2-5x+6-2(x^2-6x+8)$

$=x^2-7x+12+x^2-5x+6-2x^2+12x-16$

$=2$

Do đó:

$\dfrac{2}{(x-2)(x-3)(x-4)}=0$

Điều này không thể xảy ra.

Vậy: phương trình vô nghiệm.

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

\(\Leftrightarrow x^4+x^3+2x^3+2x^2+2x^2+2x+x+1=0\)

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}(x+1)^2=0\\x^2+x+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x+1=0\\\varnothing\end{cases}}\Rightarrow x=-1\)

mk cảm ơn ah!!!

Bây giờ a giải đc hệ này chưa ạ? Nếu giải đc r cho e xin lời giải đc ko ạ

a/ \(=x^4+2x^3+2x^2+\left(x^3+2x^2+2x\right)-\left(5x^2+10x+10\right)\)

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

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

b/ \(=3x^4+x^3-x^2+\left(9x^3+3x^2-3x\right)-\left(18x^2+6x-6\right)\)

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

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

c/ Bạn xem lại đề, câu này ko phân tích được