P+(√x/x-4+1/√x+2):2/√x-2
1 rút gọn P
2 tìm x để P=4/5
3 so sánh P với 1
4 tìm x để P nhỏ hơn 2/√x+1
5 tìm GTNN của P
K
Khách
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14 tháng 10 2021
\(a,P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\left(x\ge0;x\ne1\right)\\ P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,P=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\\ \Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\\ c,P\in Z\Leftrightarrow\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{1;2\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}=0\left(x\ne1\right)\\ \Leftrightarrow x=0\)
\(d,P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\left(\dfrac{2}{\sqrt{x}+1}>0\right)\\ e,P=1-\dfrac{2}{\sqrt{x}+1}\\ \sqrt{x}+1\ge1\Leftrightarrow-\dfrac{2}{\sqrt{x}+1}\ge-\dfrac{2}{1}=-2\\ \Leftrightarrow P=1-\dfrac{2}{\sqrt{x}+1}\ge1-\left(-2\right)=3\)
Dấu \("="\Leftrightarrow x=0\)
14 tháng 10 2021
a) ĐKXĐ: \(x\ge0,x\ne1\)
\(P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=-1\)
\(\Leftrightarrow-\sqrt{x}-1=\sqrt{x}-1\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)
c) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\)
\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Kết hợp đk:
\(\Leftrightarrow x\in\left\{0\right\}\)
d) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\)
e) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
Do \(\sqrt{x}+1\ge1\Leftrightarrow-\dfrac{2}{\sqrt{x}+1}\ge-2\)
\(\Leftrightarrow P=1-\dfrac{2}{\sqrt{x}+1}\ge1-2=-1\)
\(minP=-1\Leftrightarrow x=0\)
NV
11 tháng 6 2016
ĐKXĐ: \(\hept{\begin{cases}x\ne1\\x^2+x+1\ne0\end{cases}}\)
a/ \(R=1:\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+1}{x^2+x+1}-\frac{1}{x-1}\right]\)
\(=1:\left[\frac{x^2+2+\left(x+1\right)\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left(\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\)
\(=1:\left[\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left[\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left(\frac{x}{x^2+x+1}\right)\)
\(=\frac{x^2+x+1}{x}\)
b/ Ta có: \(R=\frac{x^2+x+1}{x}=3+\frac{\left(x-1\right)^2}{x}>3\)
Vậy R > 3
28 tháng 6 2017
a.ĐKXĐ \(x\ne0,x\ne1\),\(x\ne-1\)
B=\(\frac{4}{\left(x-1\right)^2}-\frac{x^2-1}{x^3-x}.\frac{x^3+x}{\left(x-1\right)^2}\)=\(\frac{4}{\left(x-1\right)^2}-\frac{x.\left(x^2+1\right)\left(x^2-1\right)}{x\left(x^2-1\right)\left(x-1\right)^2}\)=\(\frac{4}{\left(x-1\right)^2}-\frac{x^2+1}{\left(x-1\right)^2}\)
=\(\frac{3-x^2}{\left(x-1\right)^2}\)
b.TH1 x=3\(\Rightarrow\)B=\(\frac{3-3^2}{2^2}=\frac{-3}{2}\)
TH2 x=-1\(\Rightarrow\)B=\(\frac{3-\left(-1\right)^2}{4}=\frac{1}{2}\)
c.B=-1\(\Leftrightarrow\frac{3-x^2}{\left(x-1\right)^2}=-1\)\(\Leftrightarrow x^2-3=x^2-2x+1\)\(\Leftrightarrow2x=4\Leftrightarrow x=2\)
d.B+2=\(\frac{3-x^2}{\left(x-1\right)^2}+2=\frac{x^2-4x+5}{\left(x-1\right)^2}=\frac{\left(x-2\right)^2+1}{\left(x-1\right)^2}\ge0\)với mọi x\(\Rightarrow B\)>-2
1 tháng 11 2016
a/ ĐKXĐ : \(0\le x\ne4\)
\(B=\frac{x\sqrt{x}+15\sqrt{x}-35}{x-\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{\sqrt{x}-1}{\sqrt{x}-2}\)
\(=\frac{x\sqrt{x}+15\sqrt{x}-35-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x\sqrt{x}+15\sqrt{x}-35-x+4-x+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x\sqrt{x}-2x+15\sqrt{x}-30}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}-2\right)\left(x+15\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{x+15}{\sqrt{x}+1}\)
c/ \(x=21-4\sqrt{5}=\left(2\sqrt{5}-1\right)^2\) thay vào B được
\(B=\frac{21-4\sqrt{5}+15}{2\sqrt{5}-1+1}=\frac{36-4\sqrt{5}}{2\sqrt{5}}=\frac{-10+18\sqrt{5}}{5}\)
d/ Đặt \(t=\sqrt{x},t\ge0\) thì \(B=\frac{t^2+15}{t+1}=6\Leftrightarrow t^2+15=6\left(t+1\right)\Leftrightarrow t^2-6t+9=0\Leftrightarrow t=3\)
=> x = 9
e/ \(B=\frac{t^2+15}{t+1}=\frac{6\left(t+1\right)+\left(t^2-6t+9\right)}{t+1}=\frac{\left(t-3\right)^2}{t+1}+6\ge6\)
Đẳng thức xảy ra khi t = 3 <=> x = 9
Vậy B đạt giá trị nhỏ nhất bằng 6 khi x = 9
OM
7 tháng 8 2018
a/ ĐKXĐ : 0≤x≠4
B=x√x+15√x−35x−√x−2 −√x+2√x+1 −√x−1√x−2
=x√x+15√x−35−(√x+2)(√x−2)−(√x+1)(√x−1)(√x+1)(√x−2)
=x√x+15√x−35−x+4−x+1(√x+1)(√x−2)
=x√x−2x+15√x−30(√x+1)(√x−2) =(√x−2)(x+15)(√x+1)(√x−2) =x+15√x+1
c/ x=21−4√5=(2√5−1)2 thay vào B được
B=21−4√5+152√5−1+1 =36−4√52√5 =−10+18√55
d/ Đặt t=√x,t≥0 thì B=t2+15t+1 =6⇔t2+15=6(t+1)⇔t2−6t+9=0⇔t=3
=> x = 9
e/ B=t2+15t+1 =6(t+1)+(t2−6t+9)t+1 =(t−3)2t+1 +6≥6
Đẳng thức xảy ra khi t = 3 <=> x = 9
Vậy B đạt giá trị nhỏ nhất bằng 6 khi x = 9
31 tháng 7 2023
a: Sửa đề: \(P=\left(\dfrac{x}{2x-2}+\dfrac{3-x}{2x^2-2}\right):\left(\dfrac{x+1}{x^2+x+1}+\dfrac{x+2}{x^3-1}\right)\)\(P=\left(\dfrac{x}{2\left(x-1\right)}+\dfrac{3-x}{2\left(x-1\right)\left(x+1\right)}\right):\dfrac{\left(x+1\right)\left(x-1\right)+x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3-x}{2\left(x-1\right)\left(x+1\right)}:\dfrac{x^2-1+x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}\)
\(=\dfrac{x^2+3}{2\left(x+1\right)}\)
b: P=3
=>x^2+3=6(x+1)=6x+6
=>x^2-6x-3=0
=>\(x=3\pm2\sqrt{3}\)
c: P>4
=>P-4>0
=>\(\dfrac{x^2+3-8\left(x+1\right)}{2\left(x+1\right)}>0\)
=>\(\dfrac{x^2-8x-5}{x+1}>0\)
TH1: x^2-8x-5>0 và x+1>0
=>x>-1 và (x<4-căn 21 hoặc x>4+căn 21)
=>-1<x<4-căn 21 hoặc x>4+căn 21
Th2: x^2-8x-5<0 và x+1<0
=>x<-1 và (4-căn 21<x<4+căn 21)
=>Vô lý
13 tháng 1 2019
\(a,M=1:\left(\frac{x^2+2}{x^3-1}+\frac{x+1}{x^2+x+1}-\frac{1}{x-1}\right)\)
\(=1:\left[\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+1}{x^2+x+1}+\frac{-1}{x-1}\right]\)
\(=1:\left[\frac{\left(x^2+2\right)+\left(x+1\right)\left(x-1\right)+\left(-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\left[\frac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\left[\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\right]=1:\left[\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right]\)
\(=1:\frac{x}{x^2+x+1}=\frac{x^2+x+1}{x}\)
1:
ĐKXĐ: x>=0; x<>4
\(P=\dfrac{\sqrt{x}+\sqrt{x}-2}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}\)
\(=\dfrac{2\sqrt{x}-2}{2}\cdot\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
3: \(P-1=\dfrac{\sqrt{x}-1-\sqrt{x}-2}{\sqrt{x}+2}=\dfrac{-3}{\sqrt{x}+2}< 0\)
=>P<1
giúp mình đặc biệt là câu 3 ạ