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\(a,M=1:\left(\frac{x^2+2}{x^3-1}+\frac{x+1}{x^2+x+1}-\frac{1}{x-1}\right)\)
\(=1:\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+1}{x^2+x+1}+\frac{-1}{x-1}\right]\)
\(=1:\left[\frac{\left(x^2+2\right)+\left(x+1\right)\left(x-1\right)+\left(-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\left[\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\left[\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left[\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\frac{x}{x^2+x+1}=\frac{x^2+x+1}{x}\)
a: \(Q=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |x|=1/3 thì x=1/3 hoặc x=-1/3
Khi x=1/3 thì \(Q=\left(\dfrac{1}{3}\right)^2:\left(\dfrac{1}{3}-1\right)=-\dfrac{1}{6}\)
Khi x=-1/3 thì \(Q=\left(-\dfrac{1}{3}\right)^2:\left(-\dfrac{1}{3}-1\right)=-\dfrac{1}{12}\)
c: Để Q là số nguyên thì \(x^2-1+1⋮x-1\)
=>\(x-1\in\left\{1;-1\right\}\)
=>x=2
d: Để Q=4 thì x^2=4x-4
=>x=2
a) Rút gọn :
P = \(\left(\dfrac{2x}{x+3}+\dfrac{10}{x-3}-\dfrac{2x^2+14}{x^2-9}\right):\dfrac{4}{x+3}\)
\(ĐKXĐ:\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne3\end{matrix}\right.\)
Ta có : \(P=\left[\dfrac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{10\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2x^2+14}{\left(x+3\right)\left(x-3\right)}\right].\dfrac{x+3}{4}\)
\(P=\dfrac{2x^2-6x+10x+30-2x^2-14}{\left(x+3\right)\left(x-3\right)}.\dfrac{x+3}{4}\)
\(P=\dfrac{4x+16}{4x-13}=\dfrac{x+4}{x-3}\)
b) |x| = 3 => \(\left\{{}\begin{matrix}\left|x\right|=3khix\ge0\\\left|x\right|=-3khix< 0\end{matrix}\right.\)
* TH1 : x \(\ge0\)
\(P=\dfrac{x+4}{x-3}=\dfrac{3+4}{3-3}\left(koTMvìmẫu\ne0\right)\)
* TH2 : x < 0
\(P=\dfrac{x+4}{x-3}=\dfrac{-3+4}{-3-3}=\dfrac{-1}{6}\left(Tm\right)\)
c) Để P = \(\dfrac{-1}{2}\) thì :
\(\dfrac{x+4}{x-3}=\dfrac{-1}{2}\)
\(\Leftrightarrow2x+8=3-x\)
\(\Leftrightarrow2x+x=-8+3\)
\(\Leftrightarrow3x=-5\Rightarrow x=\dfrac{-5}{3}\)
d) P \(\le\) 2
<=> \(\dfrac{x+4}{x-3}\le2\)
\(\Leftrightarrow\dfrac{x+4}{x-3}-\dfrac{2x-6}{x-3}\le0\)
\(\Leftrightarrow\dfrac{10-x}{x-3}\le0\)
Lập bang xét dấu và tìm x nhé!!
B3;a,ĐKXĐ:\(x\ne\pm4\)
A=\(\left(\dfrac{4}{x-4}-\dfrac{4}{x+4}\right)\dfrac{x^2+8x+16}{32}=\left(\dfrac{4x+16}{x^2-16}-\dfrac{4x-16}{x^2-16}\right)\dfrac{x^2+2.4x+4^2}{32}=\left(\dfrac{4x+16-4x+16}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\left(\dfrac{32}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\dfrac{32\left(x+4\right)^2}{32.\left(x-4\right)\left(x+4\right)}=\dfrac{x+4}{x-4}\\ \\ \\ \\ \\ \\ b,Tacó\dfrac{x+4}{x-4}=\dfrac{1}{3}\Leftrightarrow3x+12=x-4\Leftrightarrow x=-8\left(TM\right)c,TAcó\dfrac{x+4}{x-4}=3\Leftrightarrow x+4=3x-12\Leftrightarrow x=8\left(TM\right)\)
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 \(x=9\)
a: \(P=\dfrac{x\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}-\dfrac{x\left(2x+1\right)}{x}+\dfrac{2\left(x-1\right)\left(x+1\right)}{x-1}\)
\(=x^2-x-2x-1+2x+2\)
\(=x^2-x+1\)
b: \(P=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\)
Dấu '=' xảy ra khi x=1/2
a/ đkxđ: x \(\ne\pm\)2; x≠3
\(P=\left(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}-\dfrac{4x^2}{x^2-4}\right):\dfrac{x^2-6x+9}{\left(2-x\right)\left(x-3\right)}\)
\(=\left(\dfrac{\left(2+x\right)^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}+\dfrac{4x^2}{x^2-4}\right):\dfrac{\left(x-3\right)^2}{\left(2-x\right)\left(x-3\right)}\)
\(=\dfrac{x^2+4x+4-x^2+4x-4+4x^2}{\left(2-x\right)\left(2+x\right)}\cdot\dfrac{2-x}{x-3}\)
\(=\dfrac{8x+4x^2}{2+x}\cdot\dfrac{1}{x-3}=\dfrac{4x\left(2+x\right)}{2+x}\cdot\dfrac{1}{x-3}=\dfrac{4x}{x-3}\)
b/ x = \(\dfrac{1}{3}\Leftrightarrow P=\dfrac{4\cdot\dfrac{1}{3}}{\dfrac{1}{3}-3}=\dfrac{4}{3}:\left(-\dfrac{8}{3}\right)=\dfrac{4}{3}\cdot\left(-\dfrac{3}{8}\right)=-\dfrac{4}{8}=-\dfrac{1}{2}\)
c/ \(P\in Z\Rightarrow\dfrac{4x}{x-3}\in Z\)
Ta có: \(\dfrac{4x}{x-3}=\dfrac{4x-12+12}{x-3}=\dfrac{4\left(x-3\right)}{x-3}+\dfrac{12}{x-3}=4+\dfrac{12}{x-3}\)
=> \(x-3\inƯ\left(12\right)\) thì P ∈ Z
=> \(x-3=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
\(\Leftrightarrow x=\left\{-9;-3;-1;0;1;2;4;5;6;7;9;15\right\}\)
mà x>4
=> x = {5;6;7;9;15}
a, Ta có:
\(P=\left(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}-\dfrac{4x^2}{x^2-4}\right):\dfrac{x^2-6x+9}{\left(2-x\right)\left(x-3\right)}\)
\(=\left(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}+\dfrac{4x^2}{4-x^2}\right):\left[\dfrac{\left(x-3\right)^2}{\left(2-x\right)\left(x-3\right)}\right]\)
\(=\left(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}+\dfrac{4x^2}{\left(2-x\right)\left(2+x\right)}\right):\dfrac{x-3}{2-x}\)
\(=\dfrac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)
\(=\dfrac{4+4x+x^2-\left(4-4x+x^2\right)+4x^2}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)
\(=\dfrac{4+4x+x^2-4+4x-x^2+4x^2}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)
\(=\dfrac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)
\(=\dfrac{4x\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)
\(=\dfrac{4x}{x-3}\)
a: \(C=\left(\dfrac{2x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\dfrac{x^2+x+1-x^2+2}{x^2+x+1}\)
\(=\dfrac{2x^2+1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x+3}\)
\(=\dfrac{x^2-x}{\left(x-1\right)}\cdot\dfrac{1}{x+3}=\dfrac{x}{x+3}\)
b: Để C là số nguyên dương thì \(\left\{{}\begin{matrix}x⋮x+3\\\dfrac{x}{x+3}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3\in\left\{1;-1;3;-3\right\}\\x\in\left(-\infty;-3\right)\cup\left(0;+\infty\right)\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{-4;-6\right\}\)
a: \(M=1:\dfrac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=1:\dfrac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x\left(x-1\right)}=\dfrac{x^2+x+1}{x}\)
b: \(M-3=\dfrac{x^2-2x+1}{x}=\dfrac{\left(x-1\right)^2}{x}>0\)
=>M>3
e: Khi x=1/4 thì \(M=\dfrac{\dfrac{1}{16}+\dfrac{1}{4}+1}{\dfrac{1}{4}}=\dfrac{21}{4}\)