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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) Ta thấy x=-2 thỏa mãn ĐKXĐ của B.
Thay x=-2 và B ta có :
\(B=\frac{2\cdot\left(-2\right)+1}{\left(-2\right)^2-1}=\frac{-3}{3}=-1\)
b) Rút gọn :
\(A=\frac{3x+1}{x^2-1}-\frac{x}{x-1}\)
\(=\frac{3x+1-x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{-x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)
Xấu nhỉ ??
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
\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\) ( Chữa đề nhé.)
a) \(ĐKXĐ:x\ne-3;x\ne2\)
\(\text{Với }x\ne-3;x\ne2,\text{ ta có: }A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\\ =\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\\ =\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x-2\right)\left(x+3\right)}\\ =\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\\ =\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\\ =\dfrac{\left(x+3\right)\left(x-4\right)}{\left(x-2\right)\left(x+3\right)}\\ =\dfrac{x-4}{x-2}\\ \text{Vậy }A=\dfrac{x-4}{x-2}\text{ với }x\ne-3;x\ne2\)
b) Lập bảng xét dấu:
x x-4 x-2 x-4 2 4 0 0 x-2 _ _ + _ + + 0 + _ +
\(\Rightarrow\left[{}\begin{matrix}x< 2\\x>4\end{matrix}\right.\)
Vậy để \(A>0\) thì \(x< 2\) hoặc \(x>4\)
c) \(\text{Với }x\ne-3;x\ne2\)
\(\text{Ta có : }A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}\\ =\dfrac{x-2}{x-2}-\dfrac{2}{x-2}=1-\dfrac{2}{x-2}\)
\(\Rightarrow\) Để A nhận giá trị nguyên
thì \(\Rightarrow\dfrac{2}{x-2}\in Z\)
\(\Rightarrow2⋮x-2\\ \Rightarrow x-2\inƯ_{\left(2\right)}\)
Mà \(Ư_{\left(2\right)}=\left\{\pm1;\pm2\right\}\)
Lập bảng giá trị:
| \(x-2\) | \(-2\) | \(-1\) | \(1\) | \(2\) |
| \(x\) | \(0\left(TM\right)\) | \(1\left(TM\right)\) | \(3\left(TM\right)\) | \(4\left(TM\right)\) |
\(\Rightarrow x\in\left\{-2;-1;1;2\right\}\)
Vậy với \(x\in\left\{-2;-1;1;2\right\}\)
thì \(A\in Z\)
Câu 2:
a) \(ĐKXĐ:x\ne\dfrac{3}{2};x\ne1\)
\(\text{Với }x\ne\dfrac{3}{2};x\ne1,\text{ ta có : }B=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3+\dfrac{2}{1-x}\right)\\ =\left[\dfrac{2x}{\left(2x-3\right)\left(x-1\right)}-\dfrac{5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}\right]:\left(\dfrac{3\left(1-x\right)}{1-x}+\dfrac{2}{1-x}\right)\\ =\dfrac{2x-5x+5}{\left(2x-3\right)\left(x-1\right)}:\dfrac{3-3x+2}{\left(1-x\right)}\\ =\dfrac{\left(-3x+5\right)\cdot\left(1-x\right)}{\left(2x-3\right)\left(x-1\right)\cdot\left(-3x+5\right)}\\ =-\dfrac{1}{2x-3}\)
Vậy \(B=-\dfrac{1}{2x-3}\) với \(x\ne\dfrac{3}{2};x\ne1\)
b) \(\text{Với }x\ne\dfrac{3}{2};x\ne1\)
Để \(B=\dfrac{1}{x^2}\)
\(\text{thì }\Rightarrow\dfrac{-1}{2x-3}=\dfrac{1}{x^2}\\ \Rightarrow2x-3=-x^2\\ \Leftrightarrow2x-3+x^2=0\\ \Leftrightarrow x^2-3x+x-3=0\\ \Leftrightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\left(TM\right)\)
Vậy với \(x=-1;x=3\) thì \(B=\dfrac{1}{x^2}\)
a) \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\) MTC: \(2x\left(x+3\right)\)
\(=\dfrac{3x}{2x\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-\left(x-6\right)}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
b) \(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5x+6}{4-x^2}\)
\(=\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x+6}{x^2-4}\)
\(=\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x+6}{\left(x-2\right)\left(x+2\right)}\) MTC: \(\left(x-2\right)\left(x+2\right)\)
\(=\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5x+6}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)+2\left(x+2\right)-\left(5x+6\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x-8+2x+4-5x-6}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x-10}{\left(x-2\right)\left(x+2\right)}\)
c) \(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)
\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{4x^2-2x}\)
\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{2x\left(2x-1\right)}\) MTC: \(2x\left(2x-1\right)\)
\(=\dfrac{\left(1-3x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{2x\left(3x-2\right)}{2x\left(2x-1\right)}-\dfrac{3x-2}{2x\left(2x-1\right)}\)
\(=\dfrac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)-\left(3x-2\right)}{2x\left(2x-1\right)}\)
\(=\dfrac{2x-1-6x^2+3x+6x^2-4x-3x+2}{2x\left(2x-1\right)}\)
\(=\dfrac{-2x+1}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(2x-1\right)}{2x\left(2x-1\right)}\)
\(=\dfrac{-1}{2x}\)
d) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(=\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\) MTC: \(\left(x-1\right)\left(x^2+x+1\right)\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2\left(x-2\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x^2+2\right)+2\left(x-2\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+2+2x-4-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-3+x}{\left(x-1\right)\left(x^2+x+1\right)}\)
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 \(x=-2\)