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a: Thay x=-4 vào B, ta được:

\(B=\dfrac{-4+3}{-4}=\dfrac{-1}{-4}=\dfrac{1}{4}\)

b: \(P=A\cdot B=\dfrac{x^2-3x+2x-9+3x+9}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}\)

\(=\dfrac{x^2+2x}{\left(x-3\right)}\cdot\dfrac{1}{x}=\dfrac{x+2}{x-3}\)

c: Để P nguyên thì \(x-3\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{4;2;8;-2\right\}\)

6 tháng 1 2022

cảm on tiên sinh

 

a: \(D=B\cdot C\)

\(=\left(\dfrac{x}{x+3}+\dfrac{2x-9}{x^2-9}+\dfrac{3}{x-3}\right)\cdot\dfrac{x+3}{x}\)

\(=\dfrac{x^2-3x+2x-9+3x+9}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}\)

\(=\dfrac{x^2+2x}{x-3}\cdot\dfrac{1}{x}=\dfrac{x+2}{x-3}\)

b: Để D nguyên thì \(x-3\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{4;2;8;-2\right\}\)

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+3-9-3-1139
x-12(C)-6(C)-4(C)-2(C)0(C)6(C)

 

1 tháng 1 2019

dell hiểu đc đề ntn cả?

5 tháng 4 2020

a) \(A=\frac{2x}{x+3}+\frac{2}{x-3}+\frac{x^2-x+6}{9-x^2}\left(x\ne\pm3\right)\)

\(\Leftrightarrow A=\frac{2x}{x+3}+\frac{2}{x-3}-\frac{x^2-x+6}{x^2-9}\)

\(\Leftrightarrow A=\frac{2x}{x+3}+\frac{2}{x-3}-\frac{x^2-x+6}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x^2-x+6}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{2x^2-6x+2x+6-x^2+x-6}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{x}{x+3}\)

Vậy \(A=\frac{x}{x+3}\left(x\ne\pm3\right)\)

b) Ta có \(A=\frac{x}{x+3}\left(x\ne\pm3\right)\)

Để A nhạn giá trị nguyên thì \(\frac{x}{x+3}\)nhận gái trị nguyên

Ta có \(\frac{x}{x+3}=\frac{x+3-3}{x+3}=1-\frac{3}{x+3}\)

=> \(\frac{3}{x+3}\)nguyên thì \(1-\frac{3}{x+3}\)nguyên

=> 3 chia hết cho x+2.

x nguyên => x+3 nguyên => x+3\(\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

Ta có bảng

x+3-3-113
x-6-4-20

Đối chiếu điều kiện x\(\ne\pm3;x\inℤ\)

=> x={-6;-4;-2;0}

Vậy x={-6;-4;-2;0} thì A nhận giá trị nguyên

27 tháng 11 2021

bạn ktra lại đề ở chỗ 2/3/-x 

a: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{4}{x-2}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)

\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)

\(=\dfrac{6\left(x+1\right)\cdot x\left(x+2\right)}{3\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{2x}{x-2}\)

a: Khi x=5 thì A=5/(5+3)=5/8

b: \(C=A+B=\dfrac{x}{x+3}+\dfrac{2}{x-3}+\dfrac{3-5x}{x^2-9}\)

\(=\dfrac{x^2-3x+2x+6+3-5x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{x+3}\)

c: Để C nguyên thì x+3-6 chia hết cho x+3

=>\(x+3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

=>\(x\in\left\{-2;-4;-1;-5;0;-6;-9\right\}\)