Bài 1: Cho biểu thức: B= (\(\dfrac{16x-x^2}{x^2-4}+\dfrac{3+2x}{2-x}+\dfrac{3x-2}{x+2}\)) . \(\dfrac{x^2+4x+4}{x-1}\)
a) Rút gọn B
b) Tìm x để B=0
c) Tìm giá trị nguyên của x để B có giá trị nguyên
mn giúp mk vs, mk cần nhanh gấp!!!!
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5 tháng 12 2021
\(a,A=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\\ A=\dfrac{-6x+18}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-6\left(x-3\right)}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-3}{x-1}\\ b,A\in Z\Leftrightarrow x-1\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-2;0;2;4\right\}\)
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 :>
7 tháng 10 2023
a:
ĐKXĐ: x<>-1/2
Để \(\dfrac{2x^3+x^2+2x+2}{2x+1}\in Z\) thì
\(2x^3+x^2+2x+1+1⋮2x+1\)
=>\(2x+1\inƯ\left(1\right)\)
=>2x+1 thuộc {1;-1}
=>x thuộc {0;-1}
b:
ĐKXĐ: x<>1/3
\(\dfrac{3x^3-7x^2+11x-1}{3x-1}\in Z\)
=>3x^3-x^2-6x^2+2x+9x-3+2 chia hết cho 3x-1
=>2 chia hết cho 3x-1
=>3x-1 thuộc {1;-1;2;-2}
=>x thuộc {2/3;0;1;-1/3}
mà x nguyên
nên x thuộc {0;1}
c:
ĐKXĐ: x<>2
\(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}\in Z\)
=>\(\left(x^2-4\right)\left(x^2+4\right)⋮\left(x-2\right)^2\left(x^2+4\right)\)
=>\(x+2⋮x-2\)
=>x-2+4 chia hết cho x-2
=>4 chia hết cho x-2
=>x-2 thuộc {1;-1;2;-2;4;-4}
=>x thuộc {3;1;4;0;6;-2}
4 tháng 1 2023
a) Ta có: \(A=\dfrac{x}{x+2}-\dfrac{2x}{x-2}+\dfrac{x^2+12}{x^2-4}\left(x\ne\pm2\right)\)
\(A=\dfrac{x\left(x-2\right)-2x\left(x+2\right)+x^2+12}{\left(x-2\right)\left(x+2\right)}\)
\(A=\dfrac{x^2-2x-2x^2-4x+x^2+12}{\left(x-2\right)\left(x+2\right)}\)
\(A=\dfrac{-6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(A=\dfrac{-6}{x+2}\)
b) Để A có giá trị nguyên thì \(x+2\inƯ\left(6\right)\)
Mà \(Ư\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
Từ đó, ta có:
\(x+1=1\Leftrightarrow x=0\) ( nhận )
\(x+1=-1\Leftrightarrow x=-2\) ( loại )
\(x+1=2\Rightarrow x=1\) ( nhận )
\(x+1=-2\Rightarrow x=-3\) ( nhận )
\(x+1=3\Rightarrow x=2\) ( loại )
\(x+1=-3\Rightarrow x=-4\) ( nhận )
\(x+1=6\Rightarrow x=5\) ( nhận )
\(x+1=-6\Rightarrow x=-7\) ( nhận )
Vậy để A nhận giá trị nguyên thì \(x\in\left\{-7;-4;-3;0;1;5\right\}\)
BS
3 tháng 1 2023
\(a,\dfrac{x}{x+2}-\dfrac{2x}{x-2}+\dfrac{x^2+12}{x^2-4}\)
\(=\dfrac{x}{x+2}-\dfrac{2x}{x-2}+\dfrac{x^2+12}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2+12}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x-2x^2-4x+x^2+12}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-6x+12}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-6}{x-2}\)
\(b,\) Để \(A\in Z\) thì \(\dfrac{-6}{x-2}\in Z\)
\(\Rightarrow x-2\inƯ\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
Vậy \(x\in\left\{3;1;4;0;5;-1;8;-4\right\}\)
8 tháng 12 2021
a: \(B=\dfrac{x^2-1-2x+3x+1}{x\left(x-1\right)}=\dfrac{x^2+x}{x\left(x-1\right)}=\dfrac{x+1}{x-1}\)
8 tháng 12 2021
a) B = \(\dfrac{x+1}{x}-\dfrac{2}{x-1}+\dfrac{3x+1}{x\left(x-1\right)}\) (ĐK: \(x\ne0;1\))
= \(\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}-\dfrac{2x}{x\left(x-1\right)}+\dfrac{3x+1}{x\left(x-1\right)}\)
= \(\dfrac{x^2-1-2x+3x+1}{x\left(x-1\right)}=\dfrac{x^2+x}{x\left(x-1\right)}=\dfrac{x+1}{x-1}\)
b) \(\left|x\right|=1< =>\left[{}\begin{matrix}x=1\left(L\right)\\x=-1\left(C\right)\end{matrix}\right.\)
Thay x = -1 vào B, ta có:
\(\dfrac{-1+1}{-1-1}=0\)
c) B nguyên <=> \(\dfrac{x+1}{x-1}\) nguyên <=> \(1+\dfrac{2}{x-1}\) nguyên
<=> 2\(⋮x-1\)
<=> x-1 \(\in\left\{-2;-1;1;2\right\}\)
| x-1 | -2 | -1 | 1 | 2 |
| x | -1 | 0 | 2 | 3 |
| C | L | C | C |
KL: x \(\in\left\{-1;2;3\right\}\)
1 tháng 12 2021
\(a,A=\dfrac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}\\ A=\dfrac{7x+35}{\left(x-1\right)\left(x+5\right)}=\dfrac{7\left(x+5\right)}{\left(x-1\right)\left(x+5\right)}=\dfrac{7}{x-1}\\ b,A\in Z\\ \Leftrightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;8\right\}\left(tm\right)\\ b,A< 0\Leftrightarrow x-1< 0\left(7>0\right)\\ \Leftrightarrow x< 1;x\ne-5\\ c,\left|A\right|=3\Leftrightarrow\dfrac{7}{\left|x-1\right|}=3\Leftrightarrow\left|x-1\right|=\dfrac{7}{3}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}+1=\dfrac{10}{3}\left(tm\right)\\x=-\dfrac{7}{3}+1=-\dfrac{4}{3}\left(tm\right)\end{matrix}\right.\)
8 tháng 12 2021
a)B = \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{7x+3}{9-x^2}\left(ĐK:x\ne\pm3\right)\)
= \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{7x+3}{x^2-9}\)
= \(\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)-7x-3}{\left(x+3\right)\left(x-3\right)}\)
= \(\dfrac{3x^2-9x}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x}{x+3}\)
b) \(\left|2x+1\right|=7< =>\left[{}\begin{matrix}2x+1=7< =>x=3\left(L\right)\\2x+1=-7< =>x=-4\left(C\right)\end{matrix}\right.\)
Thay x = -4 vào B, ta có:
B = \(\dfrac{-4.3}{-4+3}=12\)
c) Để B = \(\dfrac{-3}{5}\)
<=> \(\dfrac{3x}{x+3}=\dfrac{-3}{5}< =>\dfrac{3x}{x+3}+\dfrac{3}{5}=0\)
<=> \(\dfrac{15x+3x+9}{5\left(x+3\right)}=0< =>x=\dfrac{-1}{2}\left(TM\right)\)
d) Để B nguyên <=> \(\dfrac{3x}{x+3}\) nguyên
<=> \(3-\dfrac{9}{x+3}\) nguyên <=> \(9⋮x+3\)
| x+3 | -9 | -3 | -1 | 1 | 3 | 9 |
| x | -12(C) | -6(C) | -4(C) | -2(C) | 0(C) | 6(C) |
mn ơi, giúp mk nhé!!!
