ĐK: \(x,y\ne0\)

\(\left\{{}\begin{matrix}x-\dfrac{1}{x^3}=y-\dfrac{1}{y^3}\\\left(x-4y\right)\left(2x-y+4\right)=-36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y-\left(\dfrac{1}{x^3}-\dfrac{1}{y^3}\right)=0\\\left(x-4y\right)\left(2x-y+4\right)=-36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y+\dfrac{\left(x-y\right)\left(x^2+y^2+xy\right)}{x^3y^3}=0\\\left(x-4y\right)\left(2x-y+4\right)=-36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\left(x-y\right)\left(x^3y^3+x^2+y^2+xy\right)}{x^3y^3}=0\\\left(x-4y\right)\left(2x-y+4\right)=-36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\\left(x-3x\right)\left(2x-x+4\right)=-36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\-2x^2-8x=-36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x^2+4x-18=0\end{matrix}\right.\)

\(\Leftrightarrow x=y=-2\pm\sqrt{22}\left(tm\right)\)

1.

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-2m\left(x+\dfrac{1}{x}\right)-1+2m=0\)

Đặt \(x+\dfrac{1}{x}=t\Rightarrow\left|t\right|\ge2\)

\(\Rightarrow t^2-1-2mt+2m=0\)

\(\Leftrightarrow\left(t-1\right)\left(t+1\right)-2m\left(t-1\right)=0\)

\(\Leftrightarrow\left(t-1\right)\left(t+1-2m\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\left(loại\right)\\t=2m-1\end{matrix}\right.\)

Pt có nghiệm \(\Leftrightarrow\left[{}\begin{matrix}2m-1\ge2\\2m-1\le-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m\ge\dfrac{3}{2}\\m\le-\dfrac{1}{2}\end{matrix}\right.\)

2.

Cộng vế với vế: \(3\left|x\right|=3\Rightarrow\left|x\right|=1\)

\(\Rightarrow\left|y\right|=-1< 0\) (không thỏa mãn)

Vậy hệ pt vô nghiệm

Giải:

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

ĐKXĐ: \(x\ne\left\{1;2;3;4\right\}\)

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

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

\(\Leftrightarrow\left(x-4\right)\left[\left(x-3\right)+\left(x-1\right)\right]=\left(x-2\right)\left[\left(x-1\right)+\left(x-3\right)\right]\)

\(\Leftrightarrow x-4=x-2\)

\(\Leftrightarrow0x=2\)

Vậy ...

\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}+2\right)+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}+2\right)=\left(x-5\right)^2-5\)

\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}\right)+20+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)^2-10\left(x^2+\dfrac{1}{x^2}\right)=\left(x-5\right)^2-5\)

\(\Leftrightarrow\left(x-5\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)

\(\Leftrightarrow x-1-4=5\left(x-5\right)\)

=>x-5=5(x-5)

=>x-5-5x+25=0

=>-4x+20=0

hay x=5(loại)

`1/[x-5]-4/[(x-5)(x-1)]=5/[x-1]`       `ĐK: x \ne 5,x \ne 1`

`<=>[x-1-4]/[(x-5)(x-1)]=[5(x-5)]/[(x-5)(x-1)]`

   `=>x-5=5x-25`

`<=>4x=20`

`<=>x=5` (ko t/m)

Vậy ptr vô nghiệm

a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)

Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

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

Suy ra: \(9-3x+10x-2=4\)

\(\Leftrightarrow7x+7=4\)

\(\Leftrightarrow7x=-3\)

hay \(x=-\dfrac{3}{7}\)

Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)

a) \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}=1-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)

ĐKXĐ \(x-1\ne0\) hoặc \(x+3\ne0\)

\(\Rightarrow x\ne1\) và \(x\ne-3\)

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

\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)

\(\Leftrightarrow3x^2+9x-x-3-\left(2x^2-2x+5x-5\right)=x^2+3x-x-3-4\)

\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5=x^2+3x-x-3-4\)

\(\Leftrightarrow9x-x+2x-5x-3x+x=3-5-3-4\)

\(\Leftrightarrow3x=-9\)

\(\Leftrightarrow x=-3\) (không thỏa ĐK)

Vậy PTVN

b) \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)

ĐKXĐ: \(x-3\ne0\Rightarrow x\ne3\)

\(x+3\ne0\Rightarrow x\ne-3\)

\(2x+7\ne0\Rightarrow2x\ne-7\Rightarrow x\ne\dfrac{-7}{2}\)

\(\dfrac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}=\dfrac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)

\(\Leftrightarrow13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow13x+39+x^2+3x-3x-9=12x+42\)

\(\Leftrightarrow x^2+x-12=0\)

\(\Leftrightarrow x^2-3x+4x-12=0\)

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

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

\(\left\{{}\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\left(KTĐK\right)\\x=-4\left(TĐK\right)\end{matrix}\right.\)

Vậy S={-4}

a) \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}=1-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\) ( đk: x ≠ 1 ; x ≠ -3 )

\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)

\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5=x^2+3x-x-3-4\)

\(\Leftrightarrow3x=-9\)

\(\Rightarrow x=-3\left(KTM\right)\)

S = ∅

b) \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)

( đk: x ≠ ± 3 ; x ≠ \(\dfrac{-7}{2}\) )

\(\Leftrightarrow13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow13x+39+x^2-9=12x+42\)

\(\Leftrightarrow x^2-x-12=0\)

\(\Leftrightarrow x^2+3x-4x-12=0\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}x-4=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\)

S = \(\left\{4\right\}\)

Bài 1:
\(\frac{(x+1)^4}{(x^2+1)^2}+\frac{4x}{x^2+1}=6\)

\(\Leftrightarrow \frac{(x+1)^4+4x(x^2+1)}{(x^2+1)^2}=6\)

\(\Leftrightarrow \frac{x^4+8x^3+6x^2+8x+1}{(x^2+1)^2}=6\Rightarrow x^4+8x^3+6x^2+8x+1=6(x^2+1)^2\)

\(\Leftrightarrow x^4+8x^3+6x^2+8x+1=6(x^4+2x^2+1)\)

\(\Leftrightarrow 5x^4-8x^3+6x^2-8x+5=0\)

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

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

\(\Leftrightarrow (x-1)[5(x-1)(x^2+x+1)-3x(x-1)]=0\)

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

Dễ thấy \(5x^2+2x+5>0\), do đó \((x-1)^2=0\Leftrightarrow x=1\)

Bài 2: ĐK: \(x\geq 0\)

\(A=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\)

\(A=\frac{\sqrt{x}(\sqrt{x^3}-1)}{x+\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x^3}+1)}{x-\sqrt{x}+1}+x+1\)

\(A=\frac{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}+1)(x-\sqrt{x}+1)}{x-\sqrt{x}+1}+x+1\)

\(A=\sqrt{x}(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}+1)+x+1\)

\(A=x-2\sqrt{x}+1=(\sqrt{x}-1)^2\)

Bài 3:

Ta có:

\(\Delta'_1=a^2-(2b-1)=a^2-2b+1\)

\(\Delta'_2=b^2-(2c-1)=b^2-2c+1\)

\(\Delta'_3=c^2-(2a-1)=c^2-2a+1\)

Do đó:

\(\Delta'_1+\Delta'_2+\Delta'_3=a^2-2b+1+b^2-2c+1+c^2-2a+1\)

\(=(a-1)^2+(b-1)^2+(c-1)^2\geq 0,\forall a,b,c\in\mathbb{R}\)

Suy ra ít nhất một trong ba số \(\Delta'_1; \Delta'_2; \Delta'_3\geq 0\) vì nếu tất cả đều âm thì tổng của chúng âm( mâu thuẫn)

Điều đó đồng nghĩa với việc ít nhất một trong 3 phương trình đã cho có nghiệm.