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a: \(A=\dfrac{-\left(x+2\right)^2-2x\left(x-2\right)-4x^2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x-3\right)}{\left(x-3\right)^2}\)
\(=\dfrac{-x^2-4x-4-2x^2+4x-4x^2}{\left(x+2\right)}\cdot\dfrac{-1}{x-3}\)
\(=\dfrac{-7x^2-4}{\left(x+2\right)}\cdot\dfrac{-1}{x-3}=\dfrac{7x^2+4}{\left(x+2\right)\left(x-3\right)}\)
b: Khi x=1/3 thì \(A=\dfrac{7\cdot\dfrac{1}{9}+4}{\left(\dfrac{1}{3}-2\right)\left(\dfrac{1}{3}-3\right)}=\dfrac{43}{40}\)
a: \(A=\left(\dfrac{2x^2+2}{x^3-1}+\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+3}{x^3-x^2+3x-3}\right):\dfrac{1}{x-1}\)
\(=\left(\dfrac{2x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x^2-x+1}{x^4+2x^2+1-x^2}-\dfrac{x^2+3}{x^2\left(x-1\right)+3\left(x-1\right)}\right)\cdot\dfrac{x-1}{1}\)
\(=\left(\dfrac{2x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x^2-x+1\right)}{\left(x^2+1\right)^2-x^2}-\dfrac{x^2+3}{\left(x-1\right)\left(x^2+3\right)}\right)\cdot\dfrac{x-1}{1}\)
\(=\left(\dfrac{2x^2+3}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x^2-x+1}{\left(x^2+1+x\right)\left(x^2+1-x\right)}-\dfrac{1}{x-1}\right)\cdot\dfrac{x-1}{1}\)
\(=\left(\dfrac{2x^2+3}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x^2+x+1}-\dfrac{1}{x-1}\right)\cdot\dfrac{x-1}{1}\)
\(=\dfrac{2x^2+3+x-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x-1}{1}\)
\(=\dfrac{x^2+1}{x^2+x+1}\)
b: Để A là số nguyên thì \(x^2+1⋮x^2+x+1\)
=>\(x^2+x+1-x⋮x^2+x+1\)
=>\(x⋮x^2+x+1\)
=>\(x^2+x⋮x^2+x+1\)
=>\(x^2+x+1-1⋮x^2+x+1\)
=>\(-1⋮x^2+x+1\)
=>\(x^2+x+1\in\left\{1;-1\right\}\)
=>\(x^2+x+1=1\)
=>x2+x=0
=>x(x+1)=0
=>\(x\in\left\{0;-1\right\}\)
a, Với \(x=3\)\(=>A=\frac{x-1}{2}=\frac{3-1}{2}=\frac{2}{2}=1\)
Vậy A = 1 khi x = 3
b, Ta có : \(B=\frac{1}{x}-\frac{x}{2x+1}+\frac{2x^2-3x-1}{x\left(2x+1\right)}\)
\(=\frac{2x+1}{x\left(2x+1\right)}-\frac{x^2}{x\left(2x+1\right)}+\frac{2x^2-3x-1}{x\left(2x+1\right)}\)
\(=\frac{x^2-3x+2x+1-1}{x\left(2x+1\right)}=\frac{x^2-x}{x\left(2x+1\right)}=\frac{x\left(x-1\right)}{x\left(2x+1\right)}=\frac{x-1}{2x+1}\)
Ta có : \(A=\frac{x-1}{2};B=\frac{x-1}{2x+1}\)
\(=>C=A:B=\frac{x-1}{2}:\frac{x-1}{2x+1}=\frac{2x+1}{2}=x+\frac{1}{2}\)
đề sai bạn ơi
ĐKXĐ: \(x\notin\left\{-1;-\dfrac{1}{2}\right\}\)
a) Ta có: \(P=\left(\dfrac{2x}{x^3+x^2+x+1}+\dfrac{1}{x+1}\right):\left(1+\dfrac{x}{x+1}\right)\)
\(=\left(\dfrac{2x}{\left(x+1\right)\left(x^2+1\right)}+\dfrac{x^2+1}{\left(x^2+1\right)\left(x+1\right)}\right):\left(\dfrac{x+1+x}{x+1}\right)\)
\(=\dfrac{x^2+2x+1}{\left(x+1\right)\left(x^2+1\right)}:\dfrac{2x+1}{x+1}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x+1}{2x+1}\)
\(=\dfrac{x^2+2x+1}{\left(2x+1\right)\left(x^2+1\right)}\)
b) Vì \(x=\dfrac{1}{4}\) thỏa mãn ĐKXĐ
nên Thay \(x=\dfrac{1}{4}\) vào biểu thức \(P=\dfrac{x^2+2x+1}{\left(2x+1\right)\left(x^2+1\right)}\), ta được:
\(P=\left[\left(\dfrac{1}{4}\right)^2+2\cdot\dfrac{1}{4}+1\right]:\left[\left(2\cdot\dfrac{1}{4}+1\right)\left(\dfrac{1}{16}+1\right)\right]\)
\(=\left(\dfrac{1}{16}+\dfrac{1}{2}+1\right):\left[\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{16}+1\right)\right]\)
\(=\dfrac{25}{16}:\dfrac{51}{32}=\dfrac{25}{16}\cdot\dfrac{32}{51}=\dfrac{50}{51}\)
Vậy: Khi \(x=\dfrac{1}{4}\) thì \(P=\dfrac{50}{51}\)
1: Ta có: \(A=\left(\dfrac{x^2-16}{x-4}-1\right):\left(\dfrac{x-2}{x-3}+\dfrac{x+3}{x+1}+\dfrac{x+2-x^2}{x^2-2x-3}\right)\)
\(=\left(x+4-1\right):\left(\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x+1\right)\left(x-3\right)}+\dfrac{-x^2+x+2}{\left(x-3\right)\left(x+1\right)}\right)\)
\(=\left(x+3\right):\dfrac{x^2+x-2x-2+x^2-9-x^2+x+2}{\left(x-3\right)\left(x+1\right)}\)
\(=\left(x+3\right):\dfrac{x^2-9}{\left(x-3\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x-3\right)\left(x+1\right)}{x^2-9}\)
\(=x+1\)
ĐKXĐ: \(x\notin\left\{4;3;-1\right\}\)
2: Để \(\dfrac{A}{x^2+x+1}\) nhận giá trị nguyên thì \(x+1⋮x^2+x+1\)
\(\Leftrightarrow x^2+x⋮x^2+x+1\)
\(\Leftrightarrow x^2+x+1-1⋮x^2+x+1\)
mà \(x^2+x+1⋮x^2+x+1\)
nên \(-1⋮x^2+x+1\)
\(\Leftrightarrow x^2+x+1\inƯ\left(-1\right)\)
\(\Leftrightarrow x^2+x+1\in\left\{1;-1\right\}\)
\(\Leftrightarrow x^2+x\in\left\{0;-2\right\}\)
\(\Leftrightarrow x^2+x=0\)(Vì \(x^2+x>-2\forall x\))
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
Vậy: Để \(\dfrac{A}{x^2+x+1}\) nhận giá trị nguyên thì x=0
\(a,A=\dfrac{2x\left(x-3\right)+8\left(x+3\right)-2x-12}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x^2+6}\\ A=\dfrac{2x^2-6x+8x+24-2x-12}{\left(x-3\right)}\cdot\dfrac{1}{x^2+6}\\ A=\dfrac{2x^2+12}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2\left(x^2+6\right)}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2}{x-3}\)
\(b,A=5\Leftrightarrow\dfrac{2}{x-3}=5\Leftrightarrow5x-15=2\Leftrightarrow x=\dfrac{17}{5}\)
a: ĐKXĐ: x<>1; x<>-1
\(A=\dfrac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x-1}\)
\(=\dfrac{x+1}{x-1}-\dfrac{1}{x-1}=\dfrac{x}{x-1}\)
b: x^2+3x+2=0
=>x=-1(loại) hoặc x=-2(nhận)
Khi x=-2 thì A=-2/(-3)=2/3
Bạn ghi lại đề đi bạn. Khó hiểu quá!