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\(a.2x^2-6x=0\)
\(2x\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(t/mđk\right)\\x=3\left(loại,kot/mđk\right)\end{cases}}\)
\(Thay:x=0\left(t/mđk\right)\Leftrightarrow A=\frac{x-3}{x+3}\Rightarrow\frac{0-3}{0+3}=-\frac{3}{3}=-1\left(t/mđk\right)\)
\(A=\left(\frac{3-x}{x+3}\times\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\) \(\left(ĐKXĐ:x\ne\pm3\right)\)
\(A=\left(\frac{3-x}{x+3}\times\frac{x+3}{x-3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{3-x}{x-3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left[\frac{\left(3-x\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right]:\frac{3x^2}{x+3}\)
\(A=\left(\frac{9-3x}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)
\(A=\frac{-3}{x+3}\times\frac{x+3}{3x^2}\)
\(A=\frac{-1}{x^2}\)
Ta có :\(x^2+x-6=0\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\left(L\right)\\x=2\left(tm\right)\end{cases}}\)
\(\Rightarrow A=\frac{-1}{2^2}\)
\(A=\frac{-1}{4}\)
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) 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)
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) \(ĐKXĐ:x\ne\pm3\)
b) \(A=\left(\frac{x}{x+3}+\frac{3-x}{x+3}\cdot\frac{x^2+3x+9}{x^2-9}\right):\frac{3}{x+3}\)
\(\Leftrightarrow A=\left(\frac{x}{x+3}-\frac{\left(x-3\right)\left(x^2+3x+9\right)}{\left(x+3\right)\left(x^2-9\right)}\right):\frac{3}{x+3}\)
\(\Leftrightarrow A=\left(\frac{x}{x+3}-\frac{x^2+3x+9}{\left(x+3\right)^2}\right):\frac{3}{x+3}\)
\(\Leftrightarrow A=\frac{x^2+3x-x^2-3x-9}{\left(x+3\right)^2}:\frac{3}{x+3}\)
\(\Leftrightarrow A=\frac{-9\left(x+3\right)}{3\left(x+3\right)^2}\)
\(\Leftrightarrow A=\frac{-3}{x+3}\)
c) Tại \(x=-\frac{1}{2}\)
\(\Leftrightarrow A=\frac{-3}{-\frac{1}{2}+3}\)
\(\Leftrightarrow A=\frac{-6}{5}\)
d) Để \(A>0\)
\(\Leftrightarrow\frac{-3}{x+3}>0\)
\(\Leftrightarrow x+3< 0\)(Vì -3 < 0)
\(\Leftrightarrow x< -3\)
e) +) Với \(A>\frac{-1}{2}\)
\(\Leftrightarrow\frac{-3}{x+3}>-\frac{1}{2}\)
\(\Leftrightarrow-6>-x-3\)
\(\Leftrightarrow x>3\)(tm)
+) Với \(A< -\frac{1}{2}\)
\(\Leftrightarrow\frac{-3}{x+3}< -\frac{1}{2}\)
\(\Leftrightarrow-6< -x-3\)
\(\Leftrightarrow x< 3\)(chú ý : \(x\ne-3\))
+) Với \(A=-\frac{1}{2}\)
\(\Leftrightarrow-\frac{3}{x+3}=-\frac{1}{2}\)
\(\Leftrightarrow x+3=6\)
\(\Leftrightarrow x=3\)(ktm)
Vậy \(\orbr{\begin{cases}A>-\frac{1}{2}\\A< -\frac{1}{2}\end{cases}}\)
\(A=\left(\frac{-\left(x-3\right)}{\left(x+3\right)}.\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right).\left(\frac{x+3}{3x^2}\right)\)
\(=\left(-1+\frac{x}{x+3}\right)\left(\frac{x+3}{3x^2}\right)=\frac{-3}{\left(x+3\right)}.\frac{\left(x+3\right)}{3x^2}=\frac{-1}{x^2}\)
\(A< 0\Rightarrow\frac{-1}{x^2}< 0\Rightarrow-1< 0\) (luôn đúng)
Vậy \(x\ne0;x\ne\pm3\) thì \(A< 0\)