a, Rút gọn: B=\(\dfrac{4x^3+8x^2-x-2}{4x^2+4x+1}\)
b, tìm x \(\in\) Z để B \(\in\) Z
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KN
18 tháng 8 2021
a. ĐKXĐ : \(x\ne\frac{1}{2};\frac{5}{2};4;-\frac{3}{2};\frac{1\pm\sqrt{43}}{2}\)
\(A=\left(\frac{2x-3}{4x^2-12x+5}+\frac{3x-8}{13x-2x^2-20}-\frac{3}{2x-1}\right):\frac{21+2x-2x^2}{4x^2+4x-3}+\)
\(=\left(\frac{2x-3}{\left(2x-1\right)\left(2x-5\right)}-\frac{3x-8}{\left(2x-5\right)\left(x-4\right)}-\frac{3}{2x-1}\right).\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)
\(=\frac{\left(2x-3\right)\left(x-4\right)-\left(3x-8\right)\left(2x-1\right)-3\left(2x-5\right)\left(x-4\right)}{\left(2x-1\right)\left(2x-5\right)\left(x-4\right)}.\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)
\(=\frac{-10x^2+47x-56}{\left(2x-5\right)\left(x-4\right)}.\frac{2x+3}{-2x^2+2x+21}+1\) số to wa
30 tháng 4 2023
a: \(B=\dfrac{3x\left(2x-3\right)-4\left(2x+3\right)-4x^2+23x+12}{\left(2x-3\right)\left(2x+3\right)}\cdot\dfrac{2x+3}{x+3}\)
\(=\dfrac{6x^2-9x-8x-12-4x^2+23x+12}{2x-3}\cdot\dfrac{1}{x+3}\)
\(=\dfrac{2x^2+6x}{\left(2x-3\right)}\cdot\dfrac{1}{x+3}=\dfrac{2x}{2x-3}\)
b: 2x^2+7x+3=0
=>(2x+3)(x+2)=0
=>x=-3/2(loại) hoặc x=-2(nhận)
Khi x=-2 thì \(A=\dfrac{2\cdot\left(-2\right)}{-2-3}=\dfrac{-4}{-7}=\dfrac{4}{7}\)
d: |B|<1
=>B>-1 và B<1
=>B+1>0 và B-1<0
=>\(\left\{{}\begin{matrix}\dfrac{2x+2x-3}{2x-3}>0\\\dfrac{2x-2x+3}{2x-3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3< 0\\\dfrac{4x-3}{2x-3}>0\end{matrix}\right.\Leftrightarrow x< \dfrac{3}{4}\)
a, \(B=\dfrac{4x^3+8x^2-x-2}{4x^2+4x+1}\)
\(=\dfrac{4x^3+2x^2+6x^2+3x-4x-2}{\left(2x+1\right)^2}\)
\(=\dfrac{2x^2\left(2x+1\right)+3x\left(2x+1\right)-2\left(2x+1\right)}{\left(2x+1\right)^2}\)
\(=\dfrac{\left(2x^2+3x-2\right)\left(2x+1\right)}{\left(2x+1\right)}\)
\(=\dfrac{2x^2+3x-2}{2x+1}\)
b, Để \(B\in Z\Leftrightarrow2x^2+3x-2⋮2x+1\)
\(\Leftrightarrow2x^2+x+2x+1-3⋮2x+1\)
\(\Leftrightarrow x\left(2x+1\right)+\left(2x+1\right)-3⋮2x+1\)
\(\Leftrightarrow\left(x+1\right)\left(2x+1\right)-3⋮2x+1\)
\(\Leftrightarrow3⋮2x+1\)
\(\Leftrightarrow2x+1\in\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow x\in\left\{0;-1;1;-2\right\}\)
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