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a) ĐK: \(x\ne0,x\ne\pm3\)
\(A=\left(\frac{x-3}{x^2-9}+\frac{1}{x+3}\right)\div\frac{x}{x+3}\)
\(=\left(\frac{1}{x+3}+\frac{1}{x+3}\right)\div\frac{x}{x+3}\)
\(=\frac{2}{x+3}\times\frac{x+3}{x}=\frac{2}{x}\)
b) \(\left|A\right|=\left|\frac{2}{x}\right|=3\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{x}=3\\\frac{2}{x}=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{2}{3}\end{cases}}\)(thỏa mãn)
a)Với x \(\ne\)-1
Ta có: x2 + x = 0
=> x(x + 1) = 0
=> \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-1\left(ktm\right)\end{cases}}\)
Với x = 0 => A = \(\frac{0-3}{0+1}=-3\)
b) Ta có: B = \(\frac{3}{x-3}+\frac{6x}{9-x^3}+\frac{x}{x+3}\)
B = \(\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
B = \(\frac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
B = \(\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)
B = \(\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\)
B = \(\frac{x+3}{x-3}\)
c) Với x \(\ne\)\(\pm\)3; x \(\ne\)-1
Ta có: P = AB = \(\frac{x-3}{x+1}\cdot\frac{x+3}{x-3}=\frac{x+3}{x+1}=\frac{\left(x+1\right)+2}{x+1}=1+\frac{2}{x+1}\)
Để P \(\in\)Z <=> 2 \(⋮\)x + 1
<=> x + 1 \(\in\)Ư(2) = {1; -1; 2; -2}
<=> x \(\in\){0; -2; 1; -3}
\(A=\left(\frac{-\left(x-3\right)}{\left(x+3\right)}.\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right).\left(\frac{x+3}{3x^2}\right)\)
\(=\left(-1+\frac{x}{x+3}\right)\left(\frac{x+3}{3x^2}\right)=\frac{-3}{\left(x+3\right)}.\frac{\left(x+3\right)}{3x^2}=\frac{-1}{x^2}\)
\(A< 0\Rightarrow\frac{-1}{x^2}< 0\Rightarrow-1< 0\) (luôn đúng)
Vậy \(x\ne0;x\ne\pm3\) thì \(A< 0\)
2) a) Ta có B = \(\frac{x+2}{x-2}-\frac{x-2}{x+2}-\frac{16}{4-x^2}=\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}=\frac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{8}{x-2}\)
Khi |x - 1| = 2
=> \(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Khi x = 3 (thỏa mãn) => A = \(\frac{3^2-2.3}{3+1}=\frac{3}{4}\)
Khi x = - 1 (không thỏa mãn) => Không tìm được A
b) Ta có P = \(A.B=\frac{x^2-2x}{x+1}.\frac{8}{x-2}=\frac{8x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{8x}{x+1}\)
Đẻ P < 8
=> \(\frac{8x}{x+1}< 8\Leftrightarrow\frac{x}{x+1}< 1\)
=> \(\orbr{\begin{cases}x< x+1\left(x>-1\right)\\x>x+1\left(x< -1\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}0x< 1\left(tm\right)\\0x>1\left(\text{loại}\right)\end{cases}}\)
Vậy x > - 1 thì P < 8
a)Ta có : \(4x^2=1\)
\(\Rightarrow\orbr{\begin{cases}2x=1\\2x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
mà \(x\ne-\frac{1}{2}\Rightarrow x=\frac{1}{2}\)
Thay \(x=\frac{1}{2}\)vào B , ta được:
\(B=\frac{\left(\frac{1}{2}\right)^2-\frac{1}{2}}{2.\frac{1}{2}+1}=\frac{\frac{1}{4}-\frac{1}{2}}{1+1}=\frac{-\frac{1}{4}}{2}=-\frac{1}{8}\)
Vậy \(B=-\frac{1}{8}\)khi \(4x^2=1\)
b)Ta có : \(A=\frac{1}{x-1}-\frac{x}{1-x^2}\)
\(=\frac{1}{x-1}+\frac{x}{x^2-1}\)
\(=\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow M=A.B=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x^2-x}{2x+1}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x\left(x-1\right)}{2x+1}\)
\(=\frac{x}{x+1}\)
Vậy \(M=\frac{x}{x+1}\)
c)Ta có: \(x< x+1\forall x\)
\(\Rightarrow M=\frac{x}{x+1}< \frac{x+1}{x+1}=1\forall x\ne-1\)
Vậy với mọi \(x\ne-1\)thì \(M< 1\)
đkcđ: x khác 0 và -3
\(A=\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x.\left(x-3\right)}\)
\(A=\frac{\left(x-3\right)^2}{x.\left(x-3\right)}-\frac{x^2}{x.\left(x-3\right)}+\frac{9}{x.\left(x-3\right)}\)
\(A=\frac{x^2-6x+9-x^2+9}{x.\left(x-3\right)}=\frac{-6x+18}{x.\left(x-3\right)}=\frac{-6.\left(x-3\right)}{x.\left(x-3\right)}=-\frac{6}{x}\)
để A thuộc Z => 6 chia hết cho x
=>....
\(Taco\)
\(ĐKXD:x\ne0;x\ne3\)
\(\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}=\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-6x+9}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}=\frac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\)
\(=\frac{18-6x}{x-3}\)
\(A\inℤ\Leftrightarrow18-6x⋮x-3\Leftrightarrow18-6x+6x-18⋮x-3\Leftrightarrow0⋮x-3\)
Vậy vs mọi GT của x thì A nguyên
\(a.2x^2-6x=0\)
\(2x\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(t/mđk\right)\\x=3\left(loại,kot/mđk\right)\end{cases}}\)
\(Thay:x=0\left(t/mđk\right)\Leftrightarrow A=\frac{x-3}{x+3}\Rightarrow\frac{0-3}{0+3}=-\frac{3}{3}=-1\left(t/mđk\right)\)