Cho biểu thức: \(A=\frac{4x^3-8x^2+3x-6}{2x^2-3x-2}\)
a, Rút gọn A.
b, Tìm giá trị nguyên của x để A có giá trị nguyên.
K
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29 tháng 12 2021
a: \(A=\dfrac{x^2-2x+2x^2+4x-3x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)
5 tháng 1 2023
a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{x\left(x-2\right)+2x\left(x+2\right)-3x^2-4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2}{x+2}\)
Có vài bước mình làm tắc á nha :>
CM
7 tháng 8 2018
a, Thay x = 25, ta tính được A = 10 7
b, Rút gọn được B =
2
x
-
3
c, Ta có A.B =
2
-
4
x
+
2
=>
2
+
2
∈
Ư
4
. Từ đó tìm được x = 0, x = 4
27 tháng 7 2021
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
H
12 tháng 5 2019
\(A=\left[\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right]:\left[x-2+\frac{10-x^2}{x+2}\right]\) ĐKXĐ : \(x\ne0;x\ne\pm2\)
\(A=\left[\frac{x^2}{x\left(x+2\right)\left(x-2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right]:\left[\frac{x^2-4}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\left[\frac{3x^2}{3x\left(x+2\right)\left(x-2\right)}-\frac{6x\left(x+2\right)}{3x\left(x+2\right)\left(x-2\right)}+\frac{3x\left(x+2\right)}{3x\left(x+2\right)\left(x-2\right)}\right]:\frac{6}{x+2}\)
\(A=\left[\frac{3x^2-6x^2-12x+3x^2+6x}{3x\left(x+2\right)\left(x-2\right)}\right].\frac{x+2}{6}\)
\(A=\frac{-x}{3x\left(x-2\right)}\)
\(A=\frac{-1}{3x-6}\)
5 tháng 12 2021
\(a,A=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\\ A=\dfrac{-6x+18}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-6\left(x-3\right)}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-3}{x-1}\\ b,A\in Z\Leftrightarrow x-1\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-2;0;2;4\right\}\)
a, \(A=\frac{4x^2\left(x-2\right)+3\left(x-2\right)}{2x\left(x-2\right)+x-2}\)
\(=\frac{\left(x-2\right)\left(4x^2+3\right)}{\left(x-2\right)\left(2x+1\right)}=\frac{4x^2+3}{2x-1}\left(ĐKXĐ:x\ne2;x\ne-\frac{1}{2}\right)\)
b, \(A\in Z\Leftrightarrow\frac{4x^2+3}{2x-1}\in Z\Leftrightarrow2x+1+\frac{4}{2x-1}\in Z\)
\(\Leftrightarrow\frac{4}{2x-1}\in Z\Leftrightarrow4⋮\left(2x-1\right)\)
\(\Rightarrow2x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Mà 2x - 1 là số lẻ nên \(2x-1\in\left\{-1;1\right\}\Rightarrow x\in\left\{0;1\right\}\) (thỏa mãn ĐKXĐ)