Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,ĐK: \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)
c, Với x = 4 thỏa mãn ĐKXĐ thì
\(A=\frac{-3}{4-3}=-3\)
d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)
Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
a) A \(=\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
\(\)\(=\left(\dfrac{9-x^2}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
\(=\dfrac{-3}{x+3}:\dfrac{3x^2}{x+3}\)
\(=\dfrac{-1}{x^2}\)
b) \(x=\dfrac{-1}{2}\) (Thỏa mãn ĐKXĐ \(x\ne3;x\ne-3\) )
Thay \(x=\dfrac{-1}{2}\) vào biểu thức A, ta có:
\(A=\dfrac{-1}{\left(\dfrac{-1}{2}\right)^2}=-4\)
Vậy với \(x=\dfrac{-1}{2}\) giá trị của biểu thức A = -4.
c) \(\dfrac{-1}{x^2}< 0\)
\(\Rightarrow x^2>0\) (Luôn đúng)
Vậy với mọi giá trị của \(x\) để A < 0
a)
A = \(\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
=\(\left(\dfrac{3-x}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{x}{x+3}\right):\dfrac{3x^3}{x+3}\) (đkxđ: x \(\ne\)\(\pm\)3)
= \(\left(\dfrac{x}{x+3}-1\right).\dfrac{x+3}{3x^2}\)
= \(\dfrac{x-x-3}{x+3}.\dfrac{x+3}{3x^2}\)
= -x2
b) Thay x = \(\dfrac{1}{2}\) vào A, ta có:
A = -\(\left(\dfrac{1}{2}\right)^2\)
= -\(\dfrac{1}{4}\)
c) Để A < 0 thì -x2 < 0
mà -x2 \(\le\) 0 \(\forall\)x
\(\Rightarrow\) Với mọi x (x\(\ne\)0) thì A < 0
\(ĐKXĐ:x\ne-3;2\)
\(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{1}{x-2}\)
\(=\frac{x^2+4x+4}{\left(x+3\right)\left(x+2\right)}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{x+3}{\left(x+2\right)\left(x+3\right)}\)
\(=\frac{x^2+4x+4-5-x-3}{\left(x+2\right)\left(x+3\right)}=\frac{x^2+3x-4}{\left(x+3\right)\left(x+2\right)}=\frac{\left(x+4\right)\left(x-1\right)}{\left(x+3\right)\left(x+2\right)}\)
\(x^2-9=0\Leftrightarrow x=3\left(vì:x\ne-3\right)\)
\(\Rightarrow P=\frac{7}{15}\)
\(P\inℤ\Leftrightarrow x^2+3x-4⋮x^2+5x+6\Leftrightarrow2x+10⋮x^2+5x+6\Leftrightarrow12⋮x^2+5xx+6\)
\(................\left(dễ\right)\)
P/s: shitbo sai rồi nha bạn!Nếu không tin thì thay x = 3 vào P ban đầu và giá trị P sau khi rút gọn sẽ thấy sự khác biệt =)
ĐK: \(x\ne-3;x\ne2\)
a) \(P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}\)
\(=\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)
b) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow x=\pm3\)
Thay vào điều kiện,tìm loại x = -3 .Tìm được x =3
Ta có: \(P=\frac{x-4}{x-2}=\frac{3-4}{3-2}=-1\)
c)Ta có: \(P=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\)
Để P có giá trị nguyên thì \(\frac{2}{x-2}\) nguyên hay \(x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Suy ra \(x=\left\{0;1;3;4\right\}\)
Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
\(ĐKXĐ:\hept{\begin{cases}x\ne\pm3\\x\ne0\end{cases}}\)
a) \(B=\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}\right):\frac{3x^2}{x+3}\)
\(\Leftrightarrow B=\left(\frac{3-x}{x+3}\cdot\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)
\(\Leftrightarrow B=\frac{\left(3-x\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{3x^2}\)
\(\Leftrightarrow B=-\frac{x+3}{3x^2}\)
b) Khi \(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=3\left(ktm\right)\end{cases}}\)
\(\Leftrightarrow x=1\)
\(\Leftrightarrow B=-\frac{1+3}{3.1^2}=-\frac{4}{3.}\)
c) Để B > 0
\(\Leftrightarrow-\frac{x+3}{3x^2}>0\)
\(\Leftrightarrow\frac{x+3}{3x^2}< 0\)
\(\Leftrightarrow x+3< 0\) (Do 3x2 > 0; loại giá trị = 0)
\(\Leftrightarrow x< -3\)
Vậy để \(B>0\Leftrightarrow x< -3\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne-3\end{cases}}\)
b) \(P=1+\frac{x+3}{x^2+5x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{\left(x-2\right)\left(x+2\right)}{6\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{x-2}{6}\)
\(\Leftrightarrow P=\frac{x+4}{6}\)
c) Để P = 0
\(\Leftrightarrow\frac{x+4}{6}=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Để P = 1
\(\Leftrightarrow\frac{x+4}{6}=1\)
\(\Leftrightarrow x+4=6\)
\(\Leftrightarrow x=2\)
d) Để P > 0
\(\Leftrightarrow\frac{x+4}{6}>0\)
\(\Leftrightarrow x+4>0\)(Vì 6>0)
\(\Leftrightarrow x>-4\)
a) A = \(\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\) ( x # 0 ; x # 3 ; x# - 3)
A = \(\left(\dfrac{-\left(x-3\right)}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A = \(\left(\dfrac{-x-3}{x+3}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A = \(\dfrac{-3}{x+3}.\dfrac{x+3}{3x^2}=\dfrac{-1}{x^2}\)
b) Với x = \(\dfrac{-1}{2}\) , ta có :
A = \(\dfrac{-1}{x^2}=\dfrac{-1}{\left(\dfrac{-1}{2}\right)^2}=-4\)
c) Để A < 0
⇔ \(\dfrac{-1}{x^2}< 0\)
⇔ x2 > 0 ( luôn đúng ∀x # 0)
KL...