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1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
a,\(\dfrac{3}{x-3}\) - \(\dfrac{6x}{9-x^2}\) + \(\dfrac{x}{x+3}\) (*)
đkxđ: x khác 3, x khác -3
(*) \(\dfrac{3(x+3)}{\left(x-3\right).\left(x+3\right)}\)- \(\dfrac{6x}{\left(x-3\right).\left(x+3\right)}\) + \(\dfrac{x\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}\)
=>3x+9 -6x + x2+3x
<=>x2 + 3x-6x+3x + 9
<=>x2 +9
<=>(x-3).(x+3)
a) \(A = \frac{2x^2 - 16x+43}{x^2-8x+22}\) = \(\frac{2(x^2-8x+22)-1}{x^2-8x+22}\) = \(2 - \frac{1}{x^2-8x+22}\)
Ta có : \(x^2-8x+22 \) = \(x^2-8x+16+6 = ( x-4)^2 +6 \)
Vì \((x-4)^2 \ge 0 \) với \( \forall x\in R\) Nên \(( x-4)^2 +6 \ge 6 \)
\(\Rightarrow \) \(x^2-8x+22 \) \( \ge 6\)\(\Rightarrow \) \(\frac{1}{x^2-8x+22} \) \(\le \frac{1}{6}\) \(\Rightarrow \) - \(\frac{1}{x^2-8x+22} \) \(\ge - \frac{1}{6}\)
\(\Rightarrow \) A = \(2 - \frac{1}{x^2-8x+22}\) \( \ge 2-\frac{1}{6}\) = \(\frac{11}{6}\) Dấu "=" xảy ra khi và chỉ khi x=4
Vậy GTNN của A = \(\frac{11}{6}\) khi và chỉ khi x=4
A = \(\dfrac{5}{x+3}-\dfrac{2}{3-x}-\dfrac{3x^2-2x-9}{x^2-9}\)
= \(\dfrac{5}{x+3}+\dfrac{2}{x-3}-\dfrac{3x^2-2x-9}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{-3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{-3x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{-3x}{x+3}\)
TH1: x - 2 = 1
<=> x = 3
Thay x = 3 vào A, ta có:
A = \(\dfrac{-9}{6}\)
= \(\dfrac{-3}{2}\)
TH2: - x + 2 = 1
<=> x = 1
Thay x = 1 vào A, ta có:
A = \(\dfrac{-3}{4}\)
1,
a, \(A=\dfrac{2x}{x-3}-\dfrac{3x^2+9}{x^2-9}+\dfrac{x}{x+3}\) (ĐK: \(x\ne\pm3\))
\(=\dfrac{2x\left(x+3\right)+x\left(x-3\right)-3x^2-9}{x^2-9}\)
\(=\dfrac{2x^2+6x+x^2-3x-3x^2-9}{x^2-9}\)
\(=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
b, ĐK: \(x\pm3\)
\(A=\dfrac{2}{x-1}\Leftrightarrow\dfrac{3}{x+3}=\dfrac{2}{x-1}\)\(\Leftrightarrow3x-3=2x+6\)\(\Leftrightarrow x=9\left(TM\right)\)
Vậy với \(x=9\) thì A = \(\dfrac{2}{x-1}\)
2,
a, \(A=\dfrac{x}{x-1}+\dfrac{2x^2}{x^2-1}-\dfrac{x}{x+1}\) (ĐK: \(x\pm1\))
\(=\dfrac{x\left(x+1\right)-x\left(x-1\right)+2x^2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+x-x^2+x+2x^2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2x^2+2x}{\left(x-1\right)\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{2x}{x-1}\)
b, ĐK: \(x\pm1\)
\(A=\dfrac{2x}{x-1}=\dfrac{2x-2+2}{x-1}=\dfrac{2\left(x-1\right)}{x-1}+\dfrac{2}{x-1}=2+\dfrac{2}{x-1}\)
Để \(A\in Z\) \(\Leftrightarrow2+\dfrac{2}{x-1}\in Z\Leftrightarrow\dfrac{2}{x-1}\in Z\Leftrightarrow x-1\inƯ_{\left(2\right)}\)
\(\Leftrightarrow x-1\in\left\{\pm1\right\}\) \(\Leftrightarrow x\in\left\{0;2\right\}\)
Vậy với \(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\) thì A \(\in Z\)
Bài 2:
\(A=\dfrac{5x^3+5x}{x^4-1}=\dfrac{5x\left(x^2+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)
.....= \(\dfrac{5x}{x^2-1}\)
\(B=\dfrac{x^2+5x+6}{x^2+6x+9}=\dfrac{x^2+2x+3x+6}{\left(x+3\right)^2}\)
.....= \(\dfrac{x\left(x+2\right)+3\left(x+2\right)}{\left(x+3\right)^2}=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+3\right)^2}\)
.....= \(\dfrac{x+2}{x+3}\)
Câu 1:
B = \(\dfrac{32x-8x^2+2x^3}{x^3+64}\)
....= \(\dfrac{2x\left(x^2-4x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\dfrac{2x}{x+4}\)
a: A=[(3x^2+3-x^2+2x-1-x^2-x-1)/(x-1)(x^2+x+1)]*(x-2)/2x^2-5x+5
=(x^2+x+1)/(x-1)(x^2+x+1)*(x-2)/2x^2-5x+5
=(x-2)/(2x^2-5x+5)(x-1)
a: \(A=\dfrac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x^2+9x}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x}{x+3}\)
\(ĐK:x\ne\pm3\\ a,A=\dfrac{5x-15+2x+6-3x^2+2x+9}{\left(x+3\right)\left(x-3\right)}\\ A=\dfrac{-3x^2+9x-1}{\left(x-3\right)\left(x+3\right)}\\ b,\left|x-2\right|=1\Leftrightarrow x=1\left(x\ne3\right)\\ \Leftrightarrow A=\dfrac{-3+9-1}{\left(-2\right)\cdot4}=\dfrac{5}{-8}\)