a, \(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) ĐKXĐ: t\(\ne\)2,t\(\ne\)-3
\(\Leftrightarrow\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{\left(t+3\right)\left(t-2\right)}\)
\(\Rightarrow\left(t+3\right)\left(t+3\right)+\left(t-2\right)\left(t-2\right)=5t+15\)
\(\Leftrightarrow t^2+6t+9+t^2-4t+4-5t-15=0\)
\(\Leftrightarrow-3t-2=0\)
\(\Leftrightarrow-3t=2\)
\(\Leftrightarrow t=\dfrac{-2}{3}\) (tđk)
\(\Rightarrow S=\left\{\dfrac{-2}{3}\right\}\)
b, \(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)ĐKXĐ: x\(\ne\)\(\dfrac{2}{7}\)
\(\Leftrightarrow\) \(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)=0\)
\(\Rightarrow\left(\dfrac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)
\(\Leftrightarrow\) \(\Rightarrow\left(\dfrac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x+8}{2-7x}+1=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x+8+2-7x=0\\x=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-4x+10=0\\x=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-8\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{5}{2};-8\right\}\)
ĐKXĐ: x khác 2 và x khác -3
\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\)
\(\Leftrightarrow\dfrac{\left(t+3\right)\left(t+3\right)}{\left(t+3\right)\left(t-2\right)}+\dfrac{\left(t-2\right)\left(t-2\right)}{\left(t+3\right)\left(t-2\right)}=\dfrac{5t+15}{t^2+t-6}\)
\(\Rightarrow t^2+6t+9+t^2-4=5t+15\)
\(\Leftrightarrow2t^2+t-10=0\)
\(\Leftrightarrow2t^2-4t+5t-10=0\)
\(\Leftrightarrow2t\left(t-2\right)+5\left(t-2\right)=0\)
\(\Leftrightarrow\left(2t+5\right)\left(t-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(loại\right)\\t=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy..................
ĐKXĐ: x khác \(\dfrac{2}{7}\)\(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)-\left(x+5\right)\left(\dfrac{3x+8}{2-7x}+1\right)=0\)
\(\Leftrightarrow\left(\dfrac{3x+8}{2-7x}+1\right)\left(2x+3-x-5\right)=0\)
\(\Leftrightarrow\left(\dfrac{-4x+10}{2-7x}\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{-4x+10}{2-7x}=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=2\end{matrix}\right.\)
Vậy...............
á chỗ kia mình bị sai dấu rồi
làm như Ngô Thị Anh Minh mới đúng nha
a, t+3t−2+t−2t+3=5t+15t2+t−6t+3t−2+t−2t+3=5t+15t2+t−6 ĐKXĐ: t≠≠2,t≠≠-3
⇔t+3t−2+t−2t+3=5t+15(t+3)(t−2)⇔t+3t−2+t−2t+3=5t+15(t+3)(t−2)
⇒(t+3)(t+3)+(t−2)(t−2)=5t+15⇒(t+3)(t+3)+(t−2)(t−2)=5t+15
⇔t2+6t+9+t2−4t+4−5t−15=0⇔t2+6t+9+t2−4t+4−5t−15=0
⇔−3t−2=0⇔−3t−2=0
⇔−3t=2⇔−3t=2
⇔t=−23⇔t=−23 (tđk)
⇒S={−23}⇒S={−23}
b, (2x+3)(3x+82−7x+1)=(x−5)(3x+82−7x+1)(2x+3)(3x+82−7x+1)=(x−5)(3x+82−7x+1)ĐKXĐ: x≠≠2727
⇔⇔ (2x+3)(3x+82−7x+1)−(x−5)(3x+82−7x+1)=0(2x+3)(3x+82−7x+1)−(x−5)(3x+82−7x+1)=0
⇒(3x+82−7x+1)(2x+3−x+5)=0⇒(3x+82−7x+1)(2x+3−x+5)=0
⇔⇔ ⇒(3x+82−7x+1)(x+8)=0⇒(3x+82−7x+1)(x+8)=0
⇔⎡⎣3x+82−7x+1=0x+8=0⇔[3x+8+2−7x=0x=−8⇔[3x+82−7x+1=0x+8=0⇔[3x+8+2−7x=0x=−8
⇔[−4x+10=0x=−8⇔[−4x+10=0x=−8
⇔⎡⎣x=52x=−8⇔[x=52x=−8
⇒S={52;−8}
