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a: ĐKXĐ: \(x\in\left\{-5;3;-3\right\}\)
\(A=\dfrac{-3\left(x+5\right)}{\left(x+5\right)^2}:\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-3}{x+5}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{-3\left(x+3\right)}\)
\(=\dfrac{x-3}{x+5}\)
b: Để A<1 thì A-1<0
=>\(\dfrac{x-3-x-5}{x+5}< 0\)
=>x+5>0
=>x>-5
c: Để A=(2x-3)/(x+1) thì \(\dfrac{2x-3}{x+1}=\dfrac{x-3}{x+5}\)
=>2x^2+10x-3x-15=x^2-2x-3
=>2x^2+7x-15-x^2+2x+3=0
=>x^2+9x-12=0
hay \(x=\dfrac{-9\pm\sqrt{129}}{2}\)
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!!
b: ĐKXĐ: x<>0; x<>-5
a: \(A=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(X+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
Bài 1:
a) x≠2x≠2
Bài 2:
a) x≠0;x≠5x≠0;x≠5
b) x2−10x+25x2−5x=(x−5)2x(x−5)=x−5xx2−10x+25x2−5x=(x−5)2x(x−5)=x−5x
c) Để phân thức có giá trị nguyên thì x−5xx−5x phải có giá trị nguyên.
=> x=−5x=−5
Bài 3:
a) (x+12x−2+3x2−1−x+32x+2)⋅(4x2−45)(x+12x−2+3x2−1−x+32x+2)⋅(4x2−45)
=(x+12(x−1)+3(x−1)(x+1)−x+32(x+1))⋅2(2x2−2)5=(x+12(x−1)+3(x−1)(x+1)−x+32(x+1))⋅2(2x2−2)5
=(x+1)2+6−(x−1)(x+3)2(x−1)(x+1)⋅2⋅2(x2−1)5=(x+1)2+6−(x−1)(x+3)2(x−1)(x+1)⋅2⋅2(x2−1)5
=(x+1)2+6−(x2+3x−x−3)(x−1)(x+1)⋅2(x−1)(x+1)5=(x+1)2+6−(x2+3x−x−3)(x−1)(x+1)⋅2(x−1)(x+1)5
=[(x+1)2+6−(x2+2x−3)]⋅25=[(x+1)2+6−(x2+2x−3)]⋅25
=[(x+1)2+6−x2−2x+3]⋅25=[(x+1)2+6−x2−2x+3]⋅25
=[(x+1)2+9−x2−2x]⋅25=[(x+1)2+9−x2−2x]⋅25
=2(x+1)25+185−25x2−45x=2(x+1)25+185−25x2−45x
=2(x2+2x+1)5+185−25x2−45x=2(x2+2x+1)5+185−25x2−45x
=2x2+4x+25+185−25x2−45x=2x2+4x+25+185−25x2−45x
=2x2+4x+2+185−25x2−45x=2x2+4x+2+185−25x2−45x
=2x2+4x+205−25x2−45x=2x2+4x+205−25x2−45x
c) tự làm, đkxđ: x≠1;x≠−1
Bài 1:
a) \(x\ne2\)
Bài 2:
a) \(x\ne0;x\ne5\)
b) \(\dfrac{x^2-10x+25}{x^2-5x}=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
c) Để phân thức có giá trị nguyên thì \(\dfrac{x-5}{x}\) phải có giá trị nguyên.
=> \(x=-5\)
Bài 3:
a) \(\left(\dfrac{x+1}{2x-2}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2x+2}\right)\cdot\left(\dfrac{4x^2-4}{5}\right)\)
\(=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{2\left(2x^2-2\right)}{5}\)
\(=\dfrac{\left(x+1\right)^2+6-\left(x-1\right)\left(x+3\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2\cdot2\left(x^2-1\right)}{5}\)
\(=\dfrac{\left(x+1\right)^2+6-\left(x^2+3x-x-3\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2\left(x-1\right)\left(x+1\right)}{5}\)
\(=\left[\left(x+1\right)^2+6-\left(x^2+2x-3\right)\right]\cdot\dfrac{2}{5}\)
\(=\left[\left(x+1\right)^2+6-x^2-2x+3\right]\cdot\dfrac{2}{5}\)
\(=\left[\left(x+1\right)^2+9-x^2-2x\right]\cdot\dfrac{2}{5}\)
\(=\dfrac{2\left(x+1\right)^2}{5}+\dfrac{18}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
\(=\dfrac{2\left(x^2+2x+1\right)}{5}+\dfrac{18}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
\(=\dfrac{2x^2+4x+2}{5}+\dfrac{18}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
\(=\dfrac{2x^2+4x+2+18}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
\(=\dfrac{2x^2+4x+20}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
c) tự làm, đkxđ: \(x\ne1;x\ne-1\)
ô hô ngộ quá nhìu người bt toán lớp 8 trong khi lớp 7 với lại óc nguyow trở lại r kaka
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 \(x=9\)
Bài làm
\(P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)
\(=\frac{x+2}{x+3}-\frac{5}{x^2+3x-2x-6}-\frac{1}{x-2}\)
\(=\frac{x+2}{x+3}-\frac{5}{x\left(x+3\right)-2\left(x+3\right)}-\frac{1}{x-2}\)
\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x^2-4x+3x-12}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x\left(x-4\right)+3\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)
b) x2 - 9 = 0 <=> ( x - 3 )( x + 3 ) = 0
<=> \(\orbr{\begin{cases}x=3\left(nhan\right)\\x=-3\left(loai\right)\end{cases}}\)
x = 3 => \(P=\frac{3-4}{3-2}=-1\)
c) \(P=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\)
Để P đạt giá trị nguyên => \(\frac{2}{x-2}\)nguyên
=> \(2⋮x-2\)
=> \(x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| x-2 | 1 | -1 | 2 | -2 |
| x | 3 | 1 | 4 | 0 |
Vậy ...
\(a,ĐK:x\ne\pm3\\ Sửa:M=\dfrac{x}{x+3}+\dfrac{2x}{x-3}+\dfrac{9-3x^2}{x^2-9}\\ M=\dfrac{x^2-3x+2x^2+6x+9-3x^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x-3}\\ b,x=2\Leftrightarrow M=\dfrac{3}{2-3}=-3\\ c,M\in Z\Leftrightarrow x-3\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{0;2;4;6\right\}\left(tm\right)\)