ĐKXĐ : \(x\inℝ\)

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

\(\Leftrightarrow1+\dfrac{5x}{x^2-x+5}-\dfrac{3x}{x^2-3x+5}=1\)

\(\Leftrightarrow x.\left(\dfrac{5}{x^2-x+5}-\dfrac{3}{x^2-3x+5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{5}{x^2-x+5}=\dfrac{3}{x^2-3x+5}\left(1\right)\end{matrix}\right.\)

Phương trình (1) <=> 5(x² - 3x + 5) = 3(x² - x + 5)

<=> 2x² - 12x + 10 = 0

<=> x² - 6x + 5 = 0

<=> (x - 1)(x - 5) = 0

<=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

Tập nghiệm \(S=\left\{0;1;5\right\}\)

\(ĐK:x^2-3x+5\ge0\)

Đặt \(\sqrt{x^2-3x+5}=a\ge0\)

\(PT\Leftrightarrow a+a^2-5=7\\ \Leftrightarrow a^2+a-12=0\\ \Leftrightarrow\left(a-3\right)\left(a+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=3\left(tm\right)\\a=-4\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow\sqrt{x^2-3x+5}=3\\ \Leftrightarrow x^2-3x+5=9\\ \Leftrightarrow x^2-3x-4=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)

đặt \(x^2-3x=y\)

\(pt\Leftrightarrow\sqrt{y+5}+y=7\\ \Leftrightarrow\sqrt{y+5}=7-y\\ \Leftrightarrow\left\{{}\begin{matrix}y+5=\left(7-y\right)^2\\7-y\ge0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y+5=49-14y+y^2\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y^2-15y+44=0\\y\le7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(y^2-11y\right)-\left(4y-44\right)=0\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(y-11\right)\left(y-4\right)=0\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y=4\\y=11\end{matrix}\right.\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=4\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x^2-3x=4\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x^2-3x-4=0\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x-4\right)\left(x+1\right)\\y\le7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\\y\le7\end{matrix}\right.\)

Vậy \(x\in\left\{4;-1\right\}\)

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

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

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

\(a,2x\left(x-5\right)+4\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\2x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{5;-2\right\}\)

\(b,3x-15=2x\left(x-5\right)\\ \Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(-2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\-2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{5;\dfrac{3}{2}\right\}\)

\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-1\\2x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{1}{2};3\right\}\)

Câu d xem lại đề

`|5x| = - 3x + 2`

Nếu `5x>=0<=> x>=0` thì phương trình trên trở thành :

`5x =-3x+2`

`<=> 5x +3x=2`

`<=> 8x=2`

`<=> x= 2/8=1/4` ( thỏa mãn )

Nếu `5x<0<=>x<0` thì phương trình trên trở thành :

`-5x = -3x+2`

`<=>-5x+3x=2`

`<=> 2x=2`

`<=>x=1` ( không thỏa mãn ) 

Vậy pt đã cho có nghiệm `x=1/4`

__

`6x-2<5x+3`

`<=> 6x-5x<3+2`

`<=>x<5`

Vậy bpt đã cho có tập nghiệm `x<5`

\(-3x^2+15x+5x-5+3x^2=4-x\)

<=> 21x=9

<=> x=3/7