Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:

\(\frac{4}{x-8+\frac{7}{x}}+\frac{5}{x-10+\frac{7}{x}}=-1\)

Đặt \(x-10+\frac{7}{x}=a\)

\(\frac{4}{a+2}+\frac{5}{a}=-1\)

\(\Leftrightarrow4a+5\left(a+2\right)=-a\left(a+2\right)\)

\(\Leftrightarrow a^2+11a+10=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-10+\frac{7}{x}=-1\\x-10+\frac{7}{x}=-10\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-9x+7=0\\x^2+7=0\end{matrix}\right.\)

thay k=0 vào pt ta được 

\(9x^2-25-0^2-2.0x=0\)

=>\(9x^2-25=0\)

=>\(\left(3x-5\right)\left(3x+5\right)=0\)

=>\(3x+5=0=>x=\dfrac{-5}{3}\)

hoặc \(3x-5=0=>x=\dfrac{5}{3}\)

Thay `k=0` vào pt ta có:

`9x^2-25-0-0=0`

`<=>9x^2=25`

`<=>x^2=25/9`

`<=>x=+-5/3`

`b)x=-1` làm nghiệm nên ta thay `x=-1` vào pt thì pt =0

`=>9.1-25-k^2-2k(-1)=0`

`<=>-16-k^2+2k=0`

`<=>k^2-2k+16=0`

`<=>(k-1)^2+15=0` vô lý

Vậy khong có giá trị của k thỏa mãn đề bài

\(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)=> \(\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)=>\(\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-2000}{29}+\dfrac{x^2-10x-2000}{27}\) => \(\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)\)

=> \(x^2-10x-2000=0\)

Tự giải ra nhé hi hi

Lời giải:

\((x^2-1)(x^2-25)=25x^2\)

\(\Leftrightarrow x^4-26x^2+25=25x^2\)

\(\Leftrightarrow x^4-51x^2+25=0\)

\(\Leftrightarrow a^2-51a+25=0\) (đặt \(a=x^2)\)

\(\Leftrightarrow (a-\frac{51}{2})^2=\frac{2501}{4}\Rightarrow a-\frac{51}{2}=\pm \frac{\sqrt{2501}}{2}\)

\(\Rightarrow a=\frac{51\pm \sqrt{2501}}{2}\)

\(\Rightarrow x=\pm \sqrt{\frac{51\pm \sqrt{2501}}{2}}\)

Lời giải:

\((x^2-1)(x^2-25)=25x^2\)

\(\Leftrightarrow x^4-26x^2+25=25x^2\)

\(\Leftrightarrow x^4-51x^2+25=0\)

\(\Leftrightarrow a^2-51a+25=0\) (đặt \(a=x^2)\)

\(\Leftrightarrow (a-\frac{51}{2})^2=\frac{2501}{4}\Rightarrow a-\frac{51}{2}=\pm \frac{\sqrt{2501}}{2}\)

\(\Rightarrow a=\frac{51\pm \sqrt{2501}}{2}\)

\(\Rightarrow x=\pm \sqrt{\frac{51\pm \sqrt{2501}}{2}}\)

Lời giải:

\((x^2-1)(x^2-25)=25x^2\)

\(\Leftrightarrow x^4-26x^2+25=25x^2\)

\(\Leftrightarrow x^4-51x^2+25=0\)

\(\Leftrightarrow a^2-51a+25=0\) (đặt \(a=x^2)\)

\(\Leftrightarrow (a-\frac{51}{2})^2=\frac{2501}{4}\Rightarrow a-\frac{51}{2}=\pm \frac{\sqrt{2501}}{2}\)

\(\Rightarrow a=\frac{51\pm \sqrt{2501}}{2}\)

\(\Rightarrow x=\pm \sqrt{\frac{51\pm \sqrt{2501}}{2}}\)

\(\Leftrightarrow x^4-26x^2+25=25x^2\)

\(\Leftrightarrow x^4-51x^2+25=0\)

\(\Leftrightarrow x^2=\frac{51\pm\sqrt{2501}}{2}\Rightarrow x=\pm\sqrt{\frac{51\pm\sqrt{2501}}{2}}\)