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Câu 1:
\(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{\left(x-7\right)\left(x-3\right)}{\left(x-7\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)
\(\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}=\dfrac{2x^2-6x+5x-15}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{\left(2x+5\right)\left(x-3\right)}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)
Do đó: \(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}\)
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 \(x=-2\)
Câu 1:
1:
a) Ta có: \(P=\dfrac{x^3-3}{x^2-2x-3}-\dfrac{2x-6}{x+1}+\dfrac{x+3}{3-x}\)
\(=\dfrac{x^3-3}{\left(x-3\right)\left(x+1\right)}-\dfrac{\left(2x-6\right)\left(x-3\right)}{\left(x+1\right)\left(x-3\right)}-\dfrac{\left(x+3\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}\)
\(=\dfrac{x^3-3-2x^2+6x+6x-18-x^2-4x-3}{\left(x+1\right)\left(x-3\right)}\)
\(=\dfrac{x^3-3x^2+8x-24}{\left(x+1\right)\left(x-3\right)}\)
\(=\dfrac{x^2\left(x-3\right)+8\left(x-3\right)}{\left(x+1\right)\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)\left(x^2+8\right)}{\left(x+1\right)\left(x-3\right)}\)
\(=\dfrac{x^2+8}{x+1}\)
Câu 1:
a)
\(P=\frac{x^3-3}{(x+1)(x-3)}-\frac{2(x-3)^2}{(x+1)(x-3)}-\frac{(x+3)(x+1)}{(x-3)(x+1)}\)
\(=\frac{x^3-3-2(x-3)^2-(x+3)^2}{(x+1)(x-3)}\)
\(=\frac{x^3-3x^2+8x-24}{(x+1)(x-3)}=\frac{(x-3)(x^2+8)}{(x+1)(x-3)}=\frac{x^2+8}{x+1}\)
b) Với $x$ nguyên, để $P$ nguyên thì $\frac{x^2+8}{x+1}$ nguyên
Điều này xảy ra khi $x^2+8\vdots x+1$
$\Leftrightarrow x^2-1+9\vdots x+1$
$\Leftrightarrow (x-1)(x+1)+9\vdots x+1$
$\Leftrightarrow 9\vdots x+1$
$\Rightarrow x+1\in\left\{\pm 1;\pm 3;\pm 9\right\}$
$\Rightarrow x\in\left\{-2;0; -4; 2; -10; 8\right\}$ (đều thỏa mãn ĐKXĐ)
c) Cách 1:
x^4+3x^3-x^2+ax+b x^2+2x-3 x^2+x x^4+2x^3-3x^2 - x^3+2x^2+ax+b x^3+2x^2-3x - (a+3)x+b
Để \(P\left(x\right)⋮Q\left(x\right)\)
\(\Leftrightarrow\left(a+3\right)x+b=0\)
\(\Leftrightarrow\hept{\begin{cases}a+3=0\\b=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=-3\\b=0\end{cases}}\)
Vậy a=-3 và b=0 để \(P\left(x\right)⋮Q\left(x\right)\)
a)
2n^2-n+2 2n+1 n-1 2x^2+n - -2n+2 -2n-1 - 3
Để \(2n^2-n+2⋮2n+1\)
\(\Leftrightarrow3⋮2n+1\)
\(\Leftrightarrow2n+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow n\in\left\{0;1;-2;-1\right\}\)
Vậy \(n\in\left\{0;1;-2;-1\right\}\)để \(2n^2-n+2⋮2n+1\)
Câu 2:
\(A=3\left(2x+9\right)^2-1>=-1\)
Dấu '=' xảy ra khi x=-9/2
Câu 9:
=>(x-30)^2=0
=>x-30=0
=>x=30
Câu 10:
\(=2x^2+6x-4x-12-2x^2-2x=-12\)
a) \(ĐKXĐ:x\ne\pm3;x\ne-6\)
Với \(x\ne\pm3;x\ne-6\), ta có:
\(P=\left(\dfrac{x}{x-3}-\dfrac{2}{x+3}+\dfrac{x^2}{9-x^2}\right):\dfrac{x+6}{3x+9}\\ =\left(\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2}{\left(x+3\right)\left(x-3\right)}\right)\cdot\dfrac{3\left(x+3\right)}{x+6}\\ =\dfrac{x^2+3x-2x+6-x^2}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{3\left(x+3\right)}{x+6}\\ =\dfrac{x+6}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{3\left(x+3\right)}{x+6}\\ =\dfrac{3}{x-3}\)
Vậy \(P=\dfrac{3}{x-3}\) với \(x\ne\pm3;x\ne-6\)
b) Ta có: \(2x-\left|4-x\right|=5\)
+) Nếu \(x\le4\Leftrightarrow2x-\left(4-x\right)=5\)
\(\Leftrightarrow2x-4+x=5\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\left(Tm\right)\)
+) Nếu \(x>4\Leftrightarrow2x-\left(x-4\right)=5\)
\(\Leftrightarrow2x-x+4=5\\ \Leftrightarrow x=1\left(Ktm\right)\)
Với \(x\ne\pm3;x\ne-6\)
Khi \(x=3\left(Ktm\right)\rightarrow\text{loại}\)
Vậy khi \(2x-\left|4-x\right|=5\) không có giá trị.
c) Với \(x\ne\pm3;x\ne-6\)
Để P nhận giá trị nguyên
thì \(\Rightarrow\dfrac{3}{x-3}\in Z\)
\(\Rightarrow3⋮x-3\\ \Rightarrow x-3\inƯ_{\left(3\right)}\)
Mà \(Ư_{\left(3\right)}=\left\{\pm1;\pm3\right\}\)
Lập bảng giá trị:
| \(x-3\) | \(-3\) | \(-1\) | \(1\) | \(3\) |
| \(x\) | \(0\left(TM\right)\) | \(2\left(TM\right)\) | \(4\left(TM\right)\) | \(6\left(KTM\right)\) |
Vậy để P nhận giá trị nguyên
thì \(x\in\left\{0;2;4\right\}\)
d) Với \(x\ne\pm3;x\ne-6\)
Ta có : \(P^2-P+1=\dfrac{9}{\left(x-3\right)^2}-\dfrac{3}{x-3}+1\)
Đặt \(\dfrac{3}{x-3}=y\)
\(\Rightarrow P^2-P+1=y^2-y+1\\ =y^2-y+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(y^2-y+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Do \(\left(y-\dfrac{1}{2}\right)^2\ge0\forall y\)
\(\Rightarrow P^2-P+1=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall y\)
Dấu "=" xảy ra khi:
\(\left(y-\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow y-\dfrac{1}{2}=0\\ \Leftrightarrow y=\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{x-3}=\dfrac{1}{2}\\ \Leftrightarrow x-3=6\\ \Leftrightarrow x=9\left(TM\right)\)
Vậy \(GTNN\) của biểu thức là \(\dfrac{3}{4}\) khi \(x=9\)
