\(P=\dfrac{x^4+x^3-x^2-2x-2}{x^4+2x^3-x^2-4x-2}\)
a/ Rút gọn P
b/ Tìm x để P đạt GTNN
Cho phân thức: \(P=\dfrac{x^4+x^3-x^2-2x-2}{x^4+2x^3-x^2-4x-2}\)
a, Rút gọn phân thức P
b, Với x > 0. Tìm giá trị của x để phân thức P có GTNN.
A=\(\dfrac{4x^2+\left(2x+3\right)\left(x+1\right)-9}{9x^2-4}\)
a) Rút gọn A
b) Tìm các số nguyên x để A đạt giá trị nguyên
a, \(A=\dfrac{4x^2+2x^2+5x+3-9}{9x^2-4}=\dfrac{6x^2+5x-6}{9x^2-4}=\dfrac{\left(3x-2\right)\left(2x+3\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{2x+3}{3x+2}\)
b, Ta có \(6x+9⋮3x+2\Leftrightarrow2\left(3x+2\right)+5⋮3x+2\Rightarrow3x+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
3x+2 | 1 | -1 | 5 | -5 |
x | loại | -1 | 1 | loại |
Bài 2: Cho biểu thức B=(\(\dfrac{3X}{2X+3}\)+\(\dfrac{4}{3-2x}\)-\(\dfrac{4x^2-23x-12}{4x^2-9}\)):(\(\dfrac{x+3}{2x+3}\) )với x khác 3/2;-3/2;-3
a) Rút gọn B
b) Tính giá trị của B biết 2x^2+7x+3=0
c) Tìm x thuộc Z để B thuộc Z
d) Tìm x để |B|<1
CỨU MÌNH CÂU d NHA MÌNH CẢM ƠN!
a: \(B=\dfrac{3x\left(2x-3\right)-4\left(2x+3\right)-4x^2+23x+12}{\left(2x-3\right)\left(2x+3\right)}\cdot\dfrac{2x+3}{x+3}\)
\(=\dfrac{6x^2-9x-8x-12-4x^2+23x+12}{2x-3}\cdot\dfrac{1}{x+3}\)
\(=\dfrac{2x^2+6x}{\left(2x-3\right)}\cdot\dfrac{1}{x+3}=\dfrac{2x}{2x-3}\)
b: 2x^2+7x+3=0
=>(2x+3)(x+2)=0
=>x=-3/2(loại) hoặc x=-2(nhận)
Khi x=-2 thì \(A=\dfrac{2\cdot\left(-2\right)}{-2-3}=\dfrac{-4}{-7}=\dfrac{4}{7}\)
d: |B|<1
=>B>-1 và B<1
=>B+1>0 và B-1<0
=>\(\left\{{}\begin{matrix}\dfrac{2x+2x-3}{2x-3}>0\\\dfrac{2x-2x+3}{2x-3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3< 0\\\dfrac{4x-3}{2x-3}>0\end{matrix}\right.\Leftrightarrow x< \dfrac{3}{4}\)
Cho P=\(\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}\)
a/ Tìm điều kiện của x để giá trị P xác định. b/ Rút gọn P
c/ Tính giá trị của P với 2(x-1)=6 d/ Tìm x để giá trị của x để P < 0
a: ĐKXĐ: x<>2; x<>-2; x<>0; x<>3
b: \(P=\left(\dfrac{-\left(x+2\right)}{x-2}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(=\dfrac{-x^2-4x-4+4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)
\(=\dfrac{4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{\left(x-3\right)}=\dfrac{-4x^2\left(x-2\right)}{\left(x+2\right)\left(x-3\right)}\)
c: 2(x-1)=6
=>x-1=3
=>x=4
Thay x=4 vào P, ta đc:
\(P=\dfrac{-4\cdot4^2\cdot\left(4-2\right)}{\left(4+2\right)\left(4-3\right)}=\dfrac{-64\cdot2}{6}=\dfrac{-128}{6}=-\dfrac{64}{3}\)
Cho biểu thức A =(\(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\)):\(\dfrac{x^2-3x}{2x^2-x^3}\)
a) Rút gọn A
b) Tìm giá trị của A biết |x-5|=2
c) Tìm x∈Z để A⋮ 4
a) đk: x khác 0;2;-2;3
A = \(\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}\)
= \(\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{\left(2-x\right)\left(2+x\right)}-\dfrac{2-x}{2+x}\right):\dfrac{x-3}{2x-x^2}\)
= \(\left(\dfrac{\left(x+2\right)^2+4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\right):\dfrac{x-3}{x\left(2-x\right)}\)
= \(\dfrac{x^2+4x+4+4x^2-x^2+4x-4}{\left(2-x\right)\left(2+x\right)}.\dfrac{x\left(2-x\right)}{x-3}\)
= \(\dfrac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}.\dfrac{x\left(2-x\right)}{x-3}\)
= \(\dfrac{4x\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}.\dfrac{x\left(2-x\right)}{x-3}=\dfrac{4x^2}{x-3}\)
b) Có \(\left|x-5\right|=2\)
<=> \(\left[{}\begin{matrix}x-5=2< =>x=7\left(Tm\right)\\x-5=-2< =>x=3\left(L\right)\end{matrix}\right.\)
Thay x = 7 vào A, ta có:
\(A=\dfrac{4.7^2}{7-3}=49\)
c) A = \(\dfrac{4x^2}{x-3}⋮4\left(\forall x\right)\)
Cho \(P=\left(\dfrac{x}{x+2}-\dfrac{x^3-8}{x^3+8}.\dfrac{x^2-2x+4}{x^2-4}\right):\dfrac{4}{x+2}\)
a ) Rút gọn P
b ) Tìm x để P<0
c ) Tìm x để \(P=\dfrac{1}{x}+1\)
d ) Tính P khi \(\left|2x-1\right|=3\)
e ) Tính giá trị nhỏ nhất của P
`a)P=(x/(x+2)-(x^3-8)/(x^3+8)*(x^2-2x+4)/(x^2-4)):4/(x+2)`
`đk:x ne 0,x ne -2`
`P=(x/(x+2)-((x-2)(x^2+2x+4))/((x+2)(x^2-2x+4))*(x^2-2x+4)/((x-2)(x+2)))*(x+2)/4`
`=(x/(x+2)-(x^2+2x+4)/(x+2)^2)*(x+2)/4`
`=(x^2+2x-x^2-2x-4)/(x+2)^2*(x+2)/4`
`=-4/(x+2)^2*(x+2)/4`
`=-1/(x+2)`
`b)P<0`
`<=>-1/(x+2)<0`
Vì `-1<0`
`<=>x+2>0`
`<=>x> -2`
`c)P=1/x+1(x ne 0)`
`<=>-1/(x+2)=1/x+1`
`<=>1/x+1+1/(x+2)=0``
`<=>x+2+x(x+2)+x=0`
`<=>x^2+4x+2=0`
`<=>` \(\left[ \begin{array}{l}x=\sqrt2-2\\x=-\sqrt2-2\end{array} \right.\)
`d)|2x-1|=3`
`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2(l)\\x=-1(tm)\end{array} \right.\)
`x=-1=>P=-1/(-1+2)=-1`
`e)P=-1/(x+2)` thì nhỏ nhất cái gì nhỉ?
a) đk: \(x\ne-2;2\)
\(P=\left[\dfrac{x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}.\dfrac{x^2-2x+4}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{4}{x+2}\)
= \(\left[\dfrac{x}{x+2}-\dfrac{x^2+2x+4}{\left(x+2\right)^2}\right].\dfrac{x+2}{4}\)
= \(\dfrac{x^2+2x-x^2-2x-4}{\left(x+2\right)^2}.\dfrac{x+2}{4}\) = \(\dfrac{-4}{4\left(x+2\right)}=\dfrac{-1}{x+2}\)
b) Để P < 0
<=> \(\dfrac{-1}{x+2}< 0\)
<=> x +2 > 0
<=> x > -2 ( x khác 2)
c) Để P= \(\dfrac{1}{x}+1\)
<=> \(\dfrac{-1}{x+2}=\dfrac{1}{x}+1\)
<=> \(\dfrac{1}{x}+\dfrac{1}{x+2}+1=0\)
<=> \(\dfrac{x+2+x+x\left(x+2\right)}{x\left(x+2\right)}=0\)
<=> x2 + 4x + 2 = 0
<=> (x+2)2 = 2
<=> \(\left[{}\begin{matrix}x=\sqrt{2}-2\left(c\right)\\x=-\sqrt{2}-2\left(c\right)\end{matrix}\right.\)
d) Để \(\left|2x-1\right|=3\)
<=> \(\left[{}\begin{matrix}2x-1=3< =>x=2\left(l\right)\\2x-1=-3< =>x=-1\left(c\right)\end{matrix}\right.\)
Thay x = -1, ta có:
P = \(\dfrac{-1}{-1+2}=-1\)
a) ĐKXĐ: \(x\ne2;-2\)
\(P=\left(\dfrac{x}{x+2}-\dfrac{x^3-8}{x^3+8}.\dfrac{x^2-2x+4}{x^2-4}\right):\dfrac{4}{x+2}\)
\(=\left(\dfrac{x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}.\dfrac{x^2-2x+4}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{4}{x+2}\)
\(=\left(\dfrac{x}{x+2}-\dfrac{x^2+2x+4}{x+2}.\dfrac{1}{x+2}\right):\dfrac{4}{x+2}\)
\(=\left(\dfrac{x}{x+2}-\dfrac{x^2+2x+4}{\left(x+2\right)^2}\right):\dfrac{4}{x+2}\)
\(=\dfrac{x\left(x+2\right)-x^2-2x-4}{\left(x+2\right)^2}.\dfrac{x+2}{4}=-\dfrac{4}{\left(x+2\right)^2}.\dfrac{x+2}{4}=-\dfrac{1}{x+2}\)
b) \(P< 0\Rightarrow-\dfrac{1}{x+2}< 0\Rightarrow x+2>0\Rightarrow x>-2\)
\(\Rightarrow x>-2;x\ne2\)
c) \(P=\dfrac{1}{x}+1\Rightarrow\dfrac{-1}{x+2}=\dfrac{x+1}{x}\Rightarrow-x=\left(x+2\right)\left(x+1\right)\)
\(\Rightarrow-x=x^2+3x+2\Rightarrow x^2+4x+2=0\)
\(\Delta=4^2-2.4=8\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-4-2\sqrt{2}}{2}=-2-\sqrt{2}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-4+2\sqrt{2}}{2}=-2+\sqrt{2}\end{matrix}\right.\)
d) \(\left|2x-1\right|=3\Rightarrow\left[{}\begin{matrix}2x-1=3\\1-2x=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}P=-\dfrac{1}{2+2}=-\dfrac{1}{4}\\P=-\dfrac{1}{-1+2}=-1\end{matrix}\right.\)
Cho 2 biểu thức:
A=\(\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}\) B=\(\dfrac{2x+1}{x^2-4}\)
a) Tính giá trị của biểu thức B khi x thỏa mãn \(|4x-2|=6\)
b)Rút gọn biểu thức A
c)Tìm x để P=\(\dfrac{2A}{B}>1\)
a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}
Thay x = 2, ta có B không tồn tại
Thay x = -1, ta có B = \(\dfrac{1}{3}\)
b)ĐKXĐ:x ≠ 2,-2
Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x
Do đó không tồn tại x thỏa mãn đề bài
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}\)-\(\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)+\(\dfrac{2\left(x+\sqrt{x}-2\right)}{\sqrt{x}-1}\) (x>0, x khác 1)
a) Rút gọn P
b) Tìm x để \(\dfrac{P}{2012\sqrt{x}}\) đạt GTNN
a) \(P=\dfrac{x^2-\sqrt[]{x}}{x+\sqrt[]{x}+1}-\dfrac{2x+\sqrt[]{x}}{\sqrt[]{x}}+\dfrac{2\left(x+\sqrt[]{x}-2\right)}{\sqrt[]{x}-1}\)
Điều kiện xác định \(\Leftrightarrow\left\{{}\begin{matrix}x>0\\\sqrt[]{x}-1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{\sqrt[]{x}\left[\left(\sqrt[]{x}\right)^3-1\right]}{x+\sqrt[]{x}+1}-\dfrac{\sqrt[]{x}\left(2\sqrt[]{x}+1\right)}{\sqrt[]{x}}+\dfrac{2\left(\sqrt[]{x}-1\right)\left(\sqrt[]{x}+2\right)}{\sqrt[]{x}-1}\)
\(\Rightarrow P=\dfrac{\sqrt[]{x}\left(\sqrt[]{x}-1\right)\left(x+\sqrt[]{x}+1\right)}{x+\sqrt[]{x}+1}-\left(2\sqrt[]{x}+1\right)+2\left(\sqrt[]{x}+2\right)\)
\(\Rightarrow P=\sqrt[]{x}\left(\sqrt[]{x}-1\right)-\left(2\sqrt[]{x}+1\right)+2\left(\sqrt[]{x}+2\right)\)
\(\Rightarrow P=x-\sqrt[]{x}-2\sqrt[]{x}-1+2\sqrt[]{x}+4\)
\(\Rightarrow P=x-\sqrt[]{x}+3\)
b) \(A=\dfrac{P}{2012\sqrt[]{x}}=\dfrac{x-\sqrt[]{x}+3}{2012\sqrt[]{x}}\)\(\)
\(=\dfrac{x-\sqrt[]{x}+\dfrac{1}{4}-\dfrac{1}{4}+3}{2012\sqrt[]{x}}\)
\(=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2+\dfrac{11}{4}}{2012\sqrt[]{x}}\)
\(\Rightarrow A=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}+\dfrac{\dfrac{11}{4}}{2012\sqrt[]{x}}=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}+\dfrac{11}{4.2012\sqrt[]{x}}\)
Ta lại có \(\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}\ge0,\forall x\ne0\)
\(\dfrac{1}{\sqrt[]{x}}>0\Rightarrow\dfrac{11}{4.2012\sqrt[]{x}}\ge\dfrac{11}{4.2012}=\dfrac{11}{8048}\)
\(\Rightarrow A=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}+\dfrac{11}{4.2012\sqrt[]{x}}\ge\dfrac{11}{8048}\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt[]{x}=1\Leftrightarrow x=1\)
Vậy \(GTNN\left(A\right)=\dfrac{11}{8048}\left(tạix=1\right)\)
\(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x+\sqrt{x}-2\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right).\left(\sqrt{x}+2\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2.\left(\sqrt{x}+2\right)\)
\(=x-\sqrt{x}+3\)
b) \(\dfrac{P}{2012\sqrt{x}}=\dfrac{x-\sqrt{x}+3}{2012\sqrt{x}}=\dfrac{\sqrt{x}}{2012}-\dfrac{1}{2012}+\dfrac{3}{2012\sqrt{x}}\)
\(=\left(\dfrac{\sqrt{x}}{2012}+\dfrac{3}{2012\sqrt{x}}\right)-\dfrac{1}{2012}\)
\(\ge2\sqrt{\dfrac{\sqrt{x}.3}{2012^2\sqrt{x}}}-\dfrac{1}{2012}\) (BĐT Cauchy)
\(=\dfrac{2\sqrt{3}}{2012}-\dfrac{1}{2012}=\dfrac{2\sqrt{3}-1}{2012}\)
Dấu "=" xảy ra khi \(\dfrac{\sqrt{x}}{2012}=\dfrac{3}{2012\sqrt{x}}\Leftrightarrow x=3\)(tm)
Rút gọn biểu thức rồi tìm giá trị x để biểu thức rút gọn âm:
\(\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}\)
\(\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}\)
Để biểu thức trên nhận giá trị âm khi \(\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}< 0\)
\(\Rightarrow x^3-2x^2-4x+8< 0\)do \(\left(x-2\right)^2\ge0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)-2x\left(x+2\right)< 0\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)^2< 0\Leftrightarrow x< -2\)