`(2x)/(3x^2-x+2)-(7x)/(3x^2+5x+2)=1(x ne -1,-2/3)`

Đặt `a=3x^2+2x+2(a>=5/3)`

`pt<=>(2x)/(a-3x)-(7x)/(a+3x)=1`

`=>2x(a+3x)-7x(a-3x)=a^2-9x^2`

`<=>2ax+6x^2-7ax+21x^2=a^2-9x^2`

`<=>-5ax+27x^2=a^2-9x^2`

`<=>a^2-36x^2+5ax=0`

`<=>a^2-4ax+9ax-36x^2=0`

`<=>a(a-4x)+9x(a-4x)=0`

`<=>(a-4x)(a+9x)=0`

`+)a=4x`

`=>3x^2+2x+2=4x`

`=>3x^2-2x+2=0`

`=>x^2-2/3x+2/3=0`

`=>(x-1/3)^2+5/9=0` vô lý

`+)a+9x=0`

`=>3x^2+2x+2+9x=0`

`=>3x^2+11x+2=0`

`=>x^2+11/3x+2/3=0`

`=>x=(-11+-\sqrt{97})/6`

ĐKXĐ: \(x\ne-1;x\ne-\dfrac{2}{3}\)

Ta có: \(\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{3x^2+5x+2}=1\)(1)

\(\Leftrightarrow\dfrac{2}{3x-1+\dfrac{2}{x}}-\dfrac{7}{3x+5+\dfrac{2}{x}}=1\)

Đặt: \(3x+\dfrac{2}{x}=a\)  (x khác 0) thì pt(1) trở thành:

\(\dfrac{2}{a-1}-\dfrac{7}{a+5}=1\)

\(\Leftrightarrow\dfrac{2\left(a+5\right)-7\left(a-1\right)}{\left(a-1\right)\left(a+5\right)}=1\)

\(\Leftrightarrow2\left(a+5\right)-7\left(a-1\right)=\left(a-1\right)\left(a+5\right)\)

\(\Leftrightarrow-5a+17=a^2+4a-5\)

\(\Leftrightarrow a^2+4a+5-5-17=0\)

\(\Leftrightarrow a^2+9a-22=0\)

\(\Leftrightarrow\left(a-2\right)\left(a+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2\\a=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{2}{x}=2\\3x+\dfrac{2}{x}=-11\end{matrix}\right.\)

Vì \(\left\{{}\begin{matrix}3x^2+2-2x\ne0\\3x^2+11x+2\ne0\end{matrix}\right.\)

=> PT vô nghiệm 

Ủa hình như sai:vvv

bài nó dàiiiiiiii , khôg hiểu chỗ nèo hỏi lại mình hen

\(\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{3x^2+5x+2}=1\)

\(\Leftrightarrow\left(\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{\left(3x+2\right)\left(x+1\right)}\right)=1\)

\(\Leftrightarrow\dfrac{2x\left(3x+2\right)\left(x+1\right)-\left(7x.\left(3x^2-x+2\right)\right)}{\left(3x^2-x+2\right).\left(3x+2\right)\left(x+1\right)}=\dfrac{-15x^3+17x^2-10x}{\left(3x^2-x+2\right)\left(3x+2\right)\left(x+1\right)}\)

\(\Leftrightarrow\dfrac{-15x^3+17^2-10x }{\left(3x^2-x+2\right)\left(3x+2\right)\left(x+1\right)}-1=0\)

rồi quy đồng tùm lum từa lưa nữa được như này:

\(\Leftrightarrow\dfrac{-9x^4-27x^3+10x^2-18x-4}{\left(3x^2-x+2\right)\left(3x+2\right)\left(x+1\right)}=0\)

\(\Leftrightarrow-9x^4-27x^3+10x^2-18x-4=0\)

\(\Leftrightarrow x^2+\dfrac{5}{3}.x+\dfrac{25}{26}=0\)

\(\Leftrightarrow x+\left(\dfrac{5}{6}\right)^2=\dfrac{1}{36}\)

Sử dụng công thức bậc 2 hen:

\(\Leftrightarrow x=\dfrac{-5\pm\sqrt{1}}{6}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{-5+\sqrt{1}}{6}\\x_2=\dfrac{-5-\sqrt{1}}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=-\dfrac{2}{3}\\x_2=-1\end{matrix}\right.\)

Câu 1: 

\(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{\left(x-7\right)\left(x-3\right)}{\left(x-7\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)

\(\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}=\dfrac{2x^2-6x+5x-15}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{\left(2x+5\right)\left(x-3\right)}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)

Do đó: \(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}\)

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

a: \(\dfrac{x}{2x^2+7x-15}=\dfrac{x}{\left(x+5\right)\left(2x-3\right)}=\dfrac{x^2-2x}{\left(x+5\right)\left(x-2\right)\left(2x-3\right)}\)

\(\dfrac{x+2}{x^2+3x-10}=\dfrac{x+2}{\left(x+5\right)\left(x-2\right)}=\dfrac{\left(x+2\right)\left(2x-3\right)}{\left(2x-3\right)\left(x+5\right)\left(x-2\right)}\)

\(\dfrac{1}{x+5}=\dfrac{\left(2x-3\right)\left(x-2\right)}{\left(2x-3\right)\left(x-2\right)\left(x+5\right)}\)

b: \(\dfrac{1}{-x^2+3x-2}=\dfrac{-1}{\left(x-1\right)\left(x-2\right)}=\dfrac{-\left(x+6\right)\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x+6\right)\left(x-3\right)}\)

\(\dfrac{1}{x^2+5x-6}=\dfrac{1}{\left(x+6\right)\left(x-1\right)}=\dfrac{\left(x-2\right)\left(x-3\right)}{\left(x+6\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)

\(\dfrac{1}{-x^2+4x-3}=\dfrac{-1}{\left(x-1\right)\left(x-3\right)}=\dfrac{-\left(x-2\right)\left(x+6\right)}{\left(x-1\right)\left(x-3\right)\left(x+6\right)\left(x-2\right)}\)

c: \(\dfrac{3}{x^3-1}=\dfrac{3}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\dfrac{2x}{x^2+x+1}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\dfrac{x}{x-1}=\dfrac{x\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\dfrac{3x^5+5x^3+1}{4x^4-7x^2+2}.\dfrac{x}{2x+3}.\dfrac{4x^4-7x^2+2}{3x^5+5x^3+1}\) ( sửa đề )

\(=\left[\dfrac{3x^5+5x^3+1}{4x^4-7x^2+2}.\dfrac{4x^4-7x^2+2}{3x^5+5x^3+1}\right].\dfrac{x}{2x+3}\)

\(=\dfrac{x}{2x+3}\)

\(=\dfrac{x}{2x+3}\)

a: ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le2\end{matrix}\right.\)

b: ĐKXĐ: \(\left[{}\begin{matrix}x>\dfrac{2\sqrt{14}}{7}\\x< -\dfrac{2\sqrt{14}}{7}\end{matrix}\right.\)

c: ĐKXĐ: \(x=\dfrac{1}{3}\)

d: ĐKXĐ: \(-\dfrac{2}{3}< x\le\sqrt{3}\)

1/

Ta có: 6x⁴ -x³-7x²+x+1=0

<=> 6x⁴-6x³+5x³-5x²-2x²+2x-x+1=0

<=> 6x³(x-1)+5x²(x-1)-2x(x-1)-(x-1)=0

<=> (x-1) ( 6x³+5x²-2x-1)=0

<=> ( x-1) ( 6x³-3x²+8x²-4x+2x-1)=0

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

<=> (x-1) ( 2x-1) ( 3x²+4x+1)=0

<=> (x-1) ( 2x-1) (3x²+3x+x+1)=0

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

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

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

vậy \(S=\left\{\pm1;\dfrac{1}{2};\dfrac{-1}{3}\right\}\)

\(6x^4-x^3-7x^2+x+1=0\)

\(\Leftrightarrow6x^4-6x^3+5x^3-5x^2-2x^2+2x-x+1=0\)

\(\Leftrightarrow6x^3\left(x-1\right)+5x^2\left(x-1\right)-2x\left(x-1\right)-\left(x-1\right)=0\)

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

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

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

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

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

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

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

a) Ta có: \(\dfrac{x^2+38x+4}{2x^2+17x+1}-\dfrac{3x^2-4x-2}{2x^2+17x+1}\)

\(=\dfrac{x^2+38x+4-3x^2+4x+2}{2x^2+17x+1}\)

\(=\dfrac{-2x^2+42x+6}{2x^2+17x+1}\)

c) Ta có: \(C=\dfrac{-x}{3x-2}+\dfrac{7x-4}{3x-2}\)

\(=\dfrac{-x+7x-4}{3x-2}\)

\(=\dfrac{6x-4}{3x-2}=2\)