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a: \(Q=\dfrac{x^2-4x+3+2x+6-x^2-3}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x-1-2x-1}{2x+1}\)
\(=\dfrac{-2x+6}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{2x+1}{-2}\)
\(=\dfrac{2x+1}{x+3}\)
b: ta có: |x+1|=1/2
=>x+1=1/2 hoặc x+1=-1/2
=>x=-3/2
Thay x=-3/2 vào A, ta được:
\(A=\left(2\cdot\dfrac{-3}{2}+1\right):\left(\dfrac{-3}{2}+3\right)=-2:\dfrac{3}{2}=-\dfrac{4}{3}\)
c: Để Q=2 thì 2x+1=2x+6
=>\(x\in\varnothing\)
a) điều kiện xát định : \(x\ne\pm1\)
ta có : \(P=\left(\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\right).\left(\dfrac{1-x^2}{2}\right)^2\)
\(P=\left(\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\right).\dfrac{\left(1-x\right)^2\left(1+x\right)^2}{4}\)
\(P=\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)^2\left(x+1\right)^2}{4}-\dfrac{x+2}{\left(x+1\right)^2}.\dfrac{\left(x-1\right)^2\left(x+1\right)^2}{4}\)
\(P=\dfrac{\left(x-2\right)\left(x-1\right)\left(x+1\right)}{4}-\dfrac{\left(x+2\right)\left(x-1\right)^2}{4}\)
\(P=\dfrac{\left(x-2\right)\left(x-1\right)\left(x+1\right)-\left(x+2\right)\left(x-1\right)^2}{4}\)
\(P=\dfrac{\left(x-1\right)\left(\left(x-2\right)\left(x+1\right)-\left(x+2\right)\left(x-1\right)\right)}{4}\)
\(P=\dfrac{\left(x-1\right)\left(x^2-x-2-\left(x^2+x-2\right)\right)}{4}\)
\(P=\dfrac{\left(x-1\right)\left(x^2-x-2-x^2-x+2\right)}{4}=\dfrac{\left(x-1\right)\left(-2x\right)}{4}\)
\(P=\dfrac{-2x^2+2x}{4}\)
b) ta có : \(P-4=5x\Leftrightarrow\dfrac{-2x^2+2x}{4}-4=5x\)
\(\Leftrightarrow\dfrac{-2x^2+2x-16}{4}=5x\Leftrightarrow-2x^2+2x-16=20x\)
\(\Leftrightarrow20x-\left(-2x^2+2x-16\right)=0\Leftrightarrow2x^2+18x+16=0\)
\(\Leftrightarrow2x^2+2x+16x+16=0\Leftrightarrow2x\left(x+1\right)+16\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x+16\right)\left(x+1\right)\Leftrightarrow\left[{}\begin{matrix}2x+16=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(tmđk\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
vậy \(x=-8\) thỏa mãng điều kiện bài toán
\(A=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\\ A=\left(x^2-5x+4\right)\left(x^2-5x+6\right)\\ A=\left(x^2-5x+5-1\right)\left(x^2-5x+5+1\right)\\ A=\left(x^2-5x+5\right)^2-1\ge-1\)
đẳng thức xảy ra khi :
\(x^2-5x+5=0\\ x^2-2.\dfrac{5}{2}x+\dfrac{25}{4}=\dfrac{25}{4}-5\\ \left(x-\dfrac{5}{2}\right)^2=\dfrac{5}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\sqrt{\dfrac{5}{4}}\\x-\dfrac{5}{2}=-\sqrt{\dfrac{5}{4}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{\sqrt{5}+5}{2}\\x=-\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{5-\sqrt{5}}{2}\end{matrix}\right.\)
vậy GTNN của A =-1 tại \(\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{\sqrt{5}+5}{2}\\x=-\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{5-\sqrt{5}}{2}\end{matrix}\right.\)
a) ĐKXĐ : \(x\ne0,x\ne\frac{3}{2},x\ne-\frac{3}{2}\)
Ta có : \(M=\frac{\left(2x^3+3x^2\right)\left(2x+1\right)}{4x^3-9x}\)
\(=\frac{x^2\left(2x+3\right)\left(2x+1\right)}{x\left(4x^2-9\right)}\)
\(=\frac{x^2\left(2x+3\right)\left(2x+1\right)}{x\left(2x+3\right)\left(2x-3\right)}\)
\(=\frac{x\left(2x+1\right)}{2x-3}\)
Vậy : \(M=\frac{x\left(2x+1\right)}{2x-3}\) với \(x\ne0,x\ne\frac{3}{2},x\ne-\frac{3}{2}\)
b) Để \(M=0\Leftrightarrow\frac{x\left(2x+1\right)}{2x-3}=0\)
\(\Rightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(loại\right)\\x=-\frac{1}{2}\left(tm\right)\end{cases}}\)
Vậy : \(x=-\frac{1}{2}\) để M=0.
\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm\frac{3}{2}\end{cases}}\)
a) \(M=\frac{\left(2x^3+3x^2\right)\left(2x+1\right)}{4x^3-9x}\)
\(\Leftrightarrow M=\frac{x^2\left(2x+3\right)\left(2x+1\right)}{x\left(4x^2-9\right)}\)
\(\Leftrightarrow M=\frac{x\left(2x+3\right)\left(2x+1\right)}{\left(2x+3\right)\left(2x-3\right)}\)
\(\Leftrightarrow M=\frac{x\left(2x+1\right)}{2x-3}\)
b) Để M =0
\(\Leftrightarrow\frac{x\left(2x+1\right)}{2x-3}=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(KTM\right)\\x=\frac{-1}{2}\left(TM\right)\end{cases}}}\)
Vậy ..........
c) Ta có :
\(M=\frac{x\left(2x+1\right)}{2x-3}=x+2+\frac{6}{2x-3}\)
Để M có giá trị nguyên
\(\Leftrightarrow2x-3\inƯ\left(6\right)=\left\{1;2;3;6\right\}\)( Không lấy âm vì n thuộc N )
Ta có bảng sau :
| 2x-3 | 1 | 2 | 3 | 6 |
| x | 2 | 5/2(L) | 3 | 9/2(L) |
Vậy..........
a) M xác định \(\Leftrightarrow4x^3-9x\ne0\)
\(\Leftrightarrow x\left(4x^2-9\right)\ne0\\ \Leftrightarrow\left[{}\begin{matrix}x\ne0\\x\ne\pm\dfrac{3}{2}\end{matrix}\right.\)
b)
\(M=\dfrac{\left(2x^3+3x^2\right)\left(2x+1\right)}{4x^3-9x}=\dfrac{4x^4+2x^3+6x^3+3x^2}{4x^3-9x}\\ =\dfrac{4x^4+8x^3+3x^2}{4x^3-9x}\\ =\dfrac{x\left(4x^3+8x^2+3x\right)}{x\left(x^2-9\right)}\\ =\dfrac{4x^3+8x^2+3x}{x^2-9}\)
c)
\(M=0\\ \Leftrightarrow\left(2x^3+3x^2\right)\left(2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x^3+3x^2=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2\left(2x+3\right)=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)