Câu hỏi của nguyễn long nhật

Question

$A=\left(\dfrac{1-x^2-11}{x+1}\right).\left(\dfrac{3+x}{x-3}-\dfrac{36}{9-x^2}-\dfrac{x-3}{x+3}\right)$a)rút gọn biểu thức A b) tính giá trị của A biết $2x^2-x=0$

Nguyễn Lê Phước Thịnh · Accepted Answer

a:ĐKXĐ: $x\notin\left\{-1;3;-3\right\}$ $A=\left(\dfrac{1-x^2-11}{x+1}\right)\cdot\left(\dfrac{3+x}{x-3}-\dfrac{36}{9-x^2}-\dfrac{x-3}{x+3}\right)$$=\left(\dfrac{-x^2-10}{x+1}\right)\cdot\left(\dfrac{\left(x+3\right)^2+36-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}\right)$$=\dfrac{-x^2-10}{x+1}\cdot\dfrac{x^2+6x+9+36-x^2+6x-9}{\left(x-3\right)\left(x+3\right)}$$=\dfrac{-x^2-10}{x+1}\cdot\dfrac{12x+36}{\left(x-3\right)\left(x+3\right)}=\dfrac{12\left(-x^2-10\right)}{\left(x-3\right)\left(x+1\right)}$b: $2x^2-x=0$=>x(2x-1)=0=>$\left[{}\begin{matrix}x=0\left(nhận\right)\x=\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.$Khi x=0 thì $A=\dfrac{12\left(-0^2-10\right)}{\left(0-3\right)\left(0+1\right)}=\dfrac{12\left(-10\right)}{\left(-3\right)\cdot1}=\dfrac{120}{3}=40$Khi x=1/2 thì $A=\dfrac{12\left(-\dfrac{1}{4}-10\right)}{\left(\dfrac{1}{2}-3\right)\left(\dfrac{1}{2}+1\right)}=\dfrac{12\cdot\dfrac{-41}{4}}{\dfrac{-5}{2}\cdot\dfrac{3}{2}}=-41\cdot\dfrac{3}{-\dfrac{15}{4}}=\dfrac{164}{5}$Câu trả lời: a:ĐKXĐ: $x\notin\left\{-1;3;-3\right\}$ $A=\left(\dfrac{1-x^2-11}{x+1}\right)\cdot\left(\dfrac{3+x}{x-3}-\dfrac{36}{9-x^2}-\dfrac{x-3}{x+3}\right)$$=\left(\dfrac{-x^2-10}{x+1}\right)\cdot\left(\dfrac{\left(x+3\right)^2+36-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}\right)$$=\dfrac{-x^2-10}{x+1}\cdot\dfrac{x^2+6x+9+36-x^2+6x-9}{\left(x-3\right)\left(x+3\right)}$$=\dfrac{-x^2-10}{x+1}\cdot\dfrac{12x+36}{\left(x-3\right)\left(x+3\right)}=\dfrac{12\left(-x^2-10\right)}{\left(x-3\right)\left(x+1\right)}$b: $2x^2-x=0$=>x(2x-1)=0=>$\left[{}\begin{matrix}x=0\left(nhận\right)\x=\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.$Khi x=0 thì $A=\dfrac{12\left(-0^2-10\right)}{\left(0-3\right)\left(0+1\right)}=\dfrac{12\left(-10\right)}{\left(-3\right)\cdot1}=\dfrac{120}{3}=40$Khi x=1/2 thì $A=\dfrac{12\left(-\dfrac{1}{4}-10\right)}{\left(\dfrac{1}{2}-3\right)\left(\dfrac{1}{2}+1\right)}=\dfrac{12\cdot\dfrac{-41}{4}}{\dfrac{-5}{2}\cdot\dfrac{3}{2}}=-41\cdot\dfrac{3}{-\dfrac{15}{4}}=\dfrac{164}{5}$

HT.Phong (9A5) · Answer

a) $A=\left(\dfrac{1-x^2-11}{x+1}\right)\cdot\left(\dfrac{3+x}{x-3}-\dfrac{36}{9-x^2}-\dfrac{x-3}{x+3}\right)\left(x\notin\left\{-1;3;-3\right\}\right)\ =\dfrac{-x^2-10}{x+1}\cdot\left[\dfrac{x+3}{x-3}+\dfrac{36}{x^2-9}-\dfrac{x-3}{x+3}\right]\ =\dfrac{-x^2-10}{x+1}\cdot\left[\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{36}{\left(x+3\right)\left(x-3\right)}-\dfrac{\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\right]\ =\dfrac{-x^2-10}{x+1}\cdot\dfrac{x^2+6x+9+36-x^2+6x-9}{\left(x+3\right)\left(x-3\right)}\ =\dfrac{12x+36}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{-x^2-10}{x+1}\ =\dfrac{12\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{-x^2-10}{x+1}\ =\dfrac{12}{x-3}\cdot\dfrac{-x^2-10}{x+1}\ =\dfrac{-12x^2-120}{x^2-2x-3}$b) Ta có: $2x^2-x=0< =>x\left(2x-1\right)=0< =>\left[{}\begin{matrix}x=0\x=\dfrac{1}{2}\end{matrix}\right.\left(tm\right)$Thay x = 0 vào A ta có:$A=\dfrac{-12\cdot0^2-120}{0^2-2\cdot0-3}=\dfrac{-120}{-3}=40$Thay `x=1/2` vào A ta có:$A=\dfrac{-12\cdot\left(\dfrac{1}{2}\right)^2-120}{\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}-3}=\dfrac{164}{5}$Câu trả lời: a) $A=\left(\dfrac{1-x^2-11}{x+1}\right)\cdot\left(\dfrac{3+x}{x-3}-\dfrac{36}{9-x^2}-\dfrac{x-3}{x+3}\right)\left(x\notin\left\{-1;3;-3\right\}\right)\ =\dfrac{-x^2-10}{x+1}\cdot\left[\dfrac{x+3}{x-3}+\dfrac{36}{x^2-9}-\dfrac{x-3}{x+3}\right]\ =\dfrac{-x^2-10}{x+1}\cdot\left[\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{36}{\left(x+3\right)\left(x-3\right)}-\dfrac{\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\right]\ =\dfrac{-x^2-10}{x+1}\cdot\dfrac{x^2+6x+9+36-x^2+6x-9}{\left(x+3\right)\left(x-3\right)}\ =\dfrac{12x+36}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{-x^2-10}{x+1}\ =\dfrac{12\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{-x^2-10}{x+1}\ =\dfrac{12}{x-3}\cdot\dfrac{-x^2-10}{x+1}\ =\dfrac{-12x^2-120}{x^2-2x-3}$b) Ta có: $2x^2-x=0< =>x\left(2x-1\right)=0< =>\left[{}\begin{matrix}x=0\x=\dfrac{1}{2}\end{matrix}\right.\left(tm\right)$Thay x = 0 vào A ta có:$A=\dfrac{-12\cdot0^2-120}{0^2-2\cdot0-3}=\dfrac{-120}{-3}=40$Thay `x=1/2` vào A ta có:$A=\dfrac{-12\cdot\left(\dfrac{1}{2}\right)^2-120}{\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}-3}=\dfrac{164}{5}$

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