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a)
\(A=\dfrac{1+x^2+\dfrac{1}{x}}{2+\dfrac{1}{x}}=1\Leftrightarrow\left\{{}\begin{matrix}x\ne0;x\ne-\dfrac{1}{2}\\1+x^2+\dfrac{1}{x}=2+\dfrac{1}{x}\end{matrix}\right.\)
\(\Rightarrow x^2=1\Rightarrow x=\pm1\)
Ta có :
\(P=\left(\dfrac{8}{x^2-16}+\dfrac{1}{x+4}\right):\dfrac{1}{x^2-2x-8}\)
\(P=\left(\dfrac{8+x-4}{\left(x+4\right)\left(x-4\right)}\right):\dfrac{1}{\left(x+2\right)\left(x-4\right)}\)
\(P=\dfrac{x+4}{\left(x+4\right)\left(x-4\right)}:\dfrac{1}{\left(x+2\right)\left(x-4\right)}\)
\(P=\dfrac{1}{x-4}.\left(x+2\right)\left(x-4\right)\)
\(P=\dfrac{\left(x+2\right)\left(x-4\right)}{\left(x-4\right)}\)
\(P=x+2\)
2 . Ta có :
\(x^2-9x+20=0\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
Thay \(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\) vào biểu thức \(P=x+2\) ta được :
\(\left[{}\begin{matrix}4+2=6\\5+2=7\end{matrix}\right.\)
Kết luận __________________________________
ĐKXĐ của phân thức là : \(\left\{{}\begin{matrix}x^2-16\ne0\\x+4\ne0\\x^2-2x-8\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-4\right)\left(x+4\right)\ne0\\x\ne-4\\\left(x-4\right)\left(x+2\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne4\\x\ne-4\\x\ne-2\end{matrix}\right.\)
\(P=\left(\dfrac{8}{x^2-16}+\dfrac{1}{x+4}\right):\dfrac{1}{x^2-2x-8}\) \(=\left(\dfrac{8}{\left(x-4\right)\left(x+4\right)}+\dfrac{1}{x+4}\right).\left(x^2-2x-8\right)\) \(=\dfrac{8+x-4}{\left(x-4\right)\left(x+4\right)}.\left(x^2-4x+2x-8\right)\) \(=\dfrac{x+4}{\left(x-4\right)\left(x+4\right)}.\left(x-4\right)\left(x+2\right)\) \(=x+2\) + Tính giá trị của P tại x2 - 9x + 20 = 0 \(x^2-9x+20=0\) \(\Rightarrow x^2-4x-5x+20=0\) \(\Rightarrow\left(x^2-4x\right)-\left(5x-20\right)=0\) \(\Rightarrow x\left(x-4\right)-5\left(x-4\right)=0\) \(\Rightarrow\left(x-4\right)\left(x-5\right)=0\) \(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\left(kot/m\right)\\x=5\left(t/m\right)\end{matrix}\right.\) Thay x = 5 vào biểu thức P ,có : \(5+2=7\) Vậy tại x= 5 giá trị của P là 7
Câu trả lời sai là:
(C) Giá trị của Q tại \(x=3\) là \(\dfrac{3-3}{3+3}=0\)
Do ĐKXĐ của phương trình
\(Q=\dfrac{x^2-6x+9}{x^2-9}\) là \(x\ne\pm3\)
B3;a,ĐKXĐ:\(x\ne\pm4\)
A=\(\left(\dfrac{4}{x-4}-\dfrac{4}{x+4}\right)\dfrac{x^2+8x+16}{32}=\left(\dfrac{4x+16}{x^2-16}-\dfrac{4x-16}{x^2-16}\right)\dfrac{x^2+2.4x+4^2}{32}=\left(\dfrac{4x+16-4x+16}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\left(\dfrac{32}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\dfrac{32\left(x+4\right)^2}{32.\left(x-4\right)\left(x+4\right)}=\dfrac{x+4}{x-4}\\ \\ \\ \\ \\ \\ b,Tacó\dfrac{x+4}{x-4}=\dfrac{1}{3}\Leftrightarrow3x+12=x-4\Leftrightarrow x=-8\left(TM\right)c,TAcó\dfrac{x+4}{x-4}=3\Leftrightarrow x+4=3x-12\Leftrightarrow x=8\left(TM\right)\)
Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
\(=\left(\dfrac{1}{1-x}+\dfrac{1}{1+x}\right)+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
mk nghỉ bài này đề sai
a) điều kiện : \(x\ne0;x\ne-1;x\ne2\)
ta có : \(A=1+\left(\dfrac{x+1}{x^3+1}-\dfrac{1}{x-x^2-1}+\dfrac{2}{x+1}\right):\dfrac{x^3-2x^2}{x^3-x^2+x}\)
\(\Leftrightarrow A=1+\left(\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{1}{x^2-x+1}+\dfrac{2}{x+1}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\) \(\Leftrightarrow A=1+\left(\dfrac{x+1+x+1+2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\) \(\Leftrightarrow A=1+\left(\dfrac{2x^2+4}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x^2-x+1}{x\left(x-2\right)}\) \(\Leftrightarrow A=1+\dfrac{2x^2+4}{x\left(x+1\right)\left(x-2\right)}=\dfrac{2x^2+4+x\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}\)\(\Leftrightarrow A=\dfrac{x^3+x^2-2x+4}{x\left(x+1\right)\left(x-2\right)}\)
b) ta có : \(\left|x-\dfrac{3}{4}\right|=\dfrac{5}{4}\) \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{5}{4}\\x-\dfrac{3}{4}=\dfrac{-5}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\left(L\right)\\x=\dfrac{-1}{2}\end{matrix}\right.\)
thế vào \(A\) ta có : \(A=\dfrac{41}{5}\)
vậy ...............................................................................................................
\(=\dfrac{1}{x+1}-\dfrac{8}{\left(x+1\right)\left(x-4\right)}=\dfrac{x-4-8}{\left(x+1\right)\left(x-4\right)}=\dfrac{x-12}{\left(x+1\right)\left(x-4\right)}=\dfrac{-11}{2\cdot\left(-3\right)}=\dfrac{11}{6}\)