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27 tháng 5 2016
a) \(P=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\left(ĐKXĐ:1\ne x\ge0\right)\)
\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+2+\left(x-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b) \(x=28-6\sqrt{3}=\left(3\sqrt{3}-1\right)^2\)thay vào P được : \(P=\frac{3\sqrt{3}-1}{28-6\sqrt{3}+3\sqrt{3}-1+1}=\frac{3\sqrt{3}-1}{28-3\sqrt{3}}\)
c) \(P=\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{3\sqrt{x}}{3\left(x+\sqrt{x}+1\right)}=\frac{\left(x+\sqrt{x}+1\right)-\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}=-\frac{\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}+\frac{1}{3}\le\frac{1}{3}\)Vì \(x\ne1\)nên dấu đẳng thức không xảy ra.
Do đó : \(P< \frac{1}{3}\)
27 tháng 5 2016
ĐKXĐ: \(x\ge0\)
a/ \(P=\frac{x+2}{\sqrt{x}^3-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\) \(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) \(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b/ Thay \(x=28-6\sqrt{3}\) vào P ta được: \(P=\frac{\sqrt{28-6\sqrt{3}}}{28-6\sqrt{3}+\sqrt{28-6\sqrt{3}}+1}\)
\(=\frac{\sqrt{\left(3\sqrt{3}-1\right)^2}}{29-6\sqrt{3}+\sqrt{\left(3\sqrt{3}-1\right)^2}}\) \(=\frac{3\sqrt{3}-1}{29-6\sqrt{3}+3\sqrt{3}-1}=\frac{3\sqrt{3}-1}{28-3\sqrt{3}}\)
c/ \(P< \frac{1}{3}\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}< \frac{1}{3}\) \(\Leftrightarrow x+\sqrt{x}+1>3\sqrt{x}\) \(\Leftrightarrow x-2\sqrt{x}+1>0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
Vậy x > 1
2 tháng 8 2017
x+√(x^2+3)=3/(y+√(y^3))=3(y-√(y^2+3)/-a(trục căn thức)
x+√(x^2+3)=-y+√(y^2+3) suy ra x+y=√(y^2+3)-√(x^2+3)(1)
Tương tự,x+y=√(x^2+3)-√(y^2+3)(2)
Cộng (1),(2) theo vế suy ra 2(x+y)=0 suy ra x+y=0
hay E=0.
Vậy E=0
2 tháng 8 2017
nhân \(-x+\sqrt{x^2+3}\) vào 2 vế ta đc : \(\left(-x^2+x^2+3\right)\left(y+\sqrt{y^2+3}\right)=\)\(3\left(-x+\sqrt{x^2+3}\right)\)
<=> \(y+\sqrt{y^2+3}=-x+\sqrt{x^2+3}\)<=> \(y+\sqrt{y^2+3}+x-\sqrt{x^2+3}=0\)__(1)___
làm tương tự ta đc \(\left(-y+\sqrt{y^2+3}\right)\left(x+\sqrt{x^2+3}\right)\)\(=3\left(-y+\sqrt{y^2+3}\right)\)
<=> \(x+\sqrt{x^2+3}=-y+\sqrt{y^2+3}\)<=> \(x+\sqrt{x^2+3}+y-\sqrt{y^2+3}=0\)__(2)__
lấy (1) + (2) => 2(x+y) =0 => x+y=0
lấy
TT
15 tháng 8 2020
Bài 2 :
b) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\) (1)
ĐKXĐ : \(x\ge1\)
Pt(1) tương đương :
\(\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\) (*)
Xét \(x\ge2\Rightarrow\sqrt{x-1}-1\ge0\)
\(\Rightarrow\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\)
Khi đó pt (*) trở thành :
\(\sqrt{x-1}+1+\sqrt{x-1}-1=2\)
\(\Leftrightarrow2\sqrt{x-1}=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\) ( Thỏa mãn )
Xét \(1\le x< 2\) thì \(x\ge2\Rightarrow\sqrt{x-1}-1< 0\)
Nên : \(\left|\sqrt{x-1}-1\right|=1-\sqrt{x-1}\). Khi đó pt (*) trở thành :
\(\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
\(\Leftrightarrow2=2\) ( Luôn đúng )
Vậy tập nghiệm của phương trình đã cho là \(S=\left\{x|1\le x\le2\right\}\)
TT
15 tháng 8 2020
Bài 1 :
a) ĐKXĐ : \(-1\le a\le1\)
Ta có : \(Q=\left(\frac{3}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\frac{3}{\sqrt{1-a^2}}\right)\)
\(=\left(\frac{3+\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}}\right)\cdot\frac{\sqrt{1-a^2}}{3}\)
\(=\frac{3+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\cdot\frac{\sqrt{\left(1-a\right)\left(1+a\right)}}{3}\)
\(=\frac{\left(3+\sqrt{1-a^2}\right).\sqrt{1-a}}{3}\)
Vậy \(Q=\frac{\left(3+\sqrt{1-a^2}\right).\sqrt{1-a}}{3}\) với \(-1\le a\le1\)
b) Với \(a=\frac{\sqrt{3}}{2}\) thỏa mãn ĐKXĐ \(-1\le a\le1\)nên ta có :
\(\hept{\begin{cases}1-a=1-\frac{\sqrt{3}}{2}=\frac{4-2\sqrt{3}}{4}=\frac{\left(\sqrt{3}-1\right)^2}{2^2}\\1-a^2=1-\frac{3}{4}=\frac{1}{4}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\sqrt{1-a}=\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2^2}}=\left|\frac{\sqrt{3}-1}{2}\right|=\frac{\sqrt{3}-1}{2}\\\sqrt{1-a^2}=\frac{1}{2}\end{cases}}\)
Do đó : \(Q=\frac{\left(3+\frac{1}{2}\right)\cdot\frac{\sqrt{3}-1}{2}}{3}=\frac{5\sqrt{3}-5}{12}\)
TN
13 tháng 8 2017
a)\(\sqrt{x+1}\left(x+4\right)=\left(x+18\right)\sqrt{6+x}-3x-40\)
\(pt\Leftrightarrow\sqrt{x+1}\left(x+4\right)-14=\left(x+18\right)\sqrt{6+x}-63-3x-9\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+4\right)^2-196}{\sqrt{x+1}\left(x+4\right)+14}=\frac{\left(x+18\right)^2\left(x+6\right)-3969}{\left(x+18\right)\sqrt{6+x}+63}-3\left(x-3\right)\)
\(\Leftrightarrow\frac{x^3+9x^2+24x-180}{\sqrt{x+1}\left(x+4\right)+14}-\frac{x^3+42x^2+540x-2025}{\left(x+18\right)\sqrt{6+x}+63}+3\left(x-3\right)=0\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x^2+12x+60\right)}{\sqrt{x+1}\left(x+4\right)+14}-\frac{\left(x-3\right)\left(x^2+45x+675\right)}{\left(x+18\right)\sqrt{6+x}+63}+3\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{x^2+12x+60}{\sqrt{x+1}\left(x+4\right)+14}-\frac{x^2+45x+675}{\left(x+18\right)\sqrt{6+x}+63}+3\right)=0\)
Pt trong ngoặc to to kia vô nghiệm
Suy ra x=3
b)\(3\left(\sqrt{x+9}-\sqrt{x+1}\right)=4-4x\)
\(pt\Leftrightarrow\sqrt{x+9}-\sqrt{x+1}=\frac{4-4x}{3}\)
\(\Leftrightarrow2x+10-2\sqrt{\left(x+1\right)\left(x+9\right)}=\frac{16x^2-32x+16}{9}\)
\(\Leftrightarrow-2\sqrt{\left(x+1\right)\left(x+9\right)}=\frac{16x^2-32x+16}{9}-\left(2x+10\right)\)
\(\Leftrightarrow4\left(x+1\right)\left(x+9\right)=\frac{256x^4-1600x^3+132x^2+7400x+5476}{81}\)
\(\Leftrightarrow\frac{-64\left(x^2-5x-5\right)\left(4x^2-5x-8\right)}{81}=0\)
mỗi lần bình phương tự rút ra điều kiện mà khử nghiệm nhé :v
HS
26 tháng 7 2018
\(A=3\sqrt{8}-\sqrt{50}-\sqrt{\sqrt{2}-1}\)
\(\Leftrightarrow6\sqrt{2}-5\sqrt{2}-\sqrt{\sqrt{2}-1}\)
\(\Leftrightarrow\sqrt{2}-\sqrt{\sqrt{2}-1}\)
\(B=2.\dfrac{2}{x-1}.\sqrt{\dfrac{x^2-2x+1}{4x^2}}\)
\(\Leftrightarrow\)\(\dfrac{2}{x-1}.\dfrac{\sqrt{x^2-2x+1}}{2x}\)
\(\Leftrightarrow\)\(\dfrac{2}{x-1}.\dfrac{\sqrt{\left(x-1\right)^2}}{x}\)
\(\Leftrightarrow\)\(\dfrac{2}{x-1}.\dfrac{x-1}{x}\)
\(\Leftrightarrow\)\(2.\dfrac{1}{x}\)
\(\Leftrightarrow\)\(\dfrac{2}{x}\)
3
26 tháng 7 2018
a) \(\sqrt{2x-1}< 3\)
\(\Leftrightarrow2x-1< 9\)
\(\Leftrightarrow2x< 10\)
\(\Leftrightarrow x< 5\)
30 tháng 8 2018
\(\sqrt{2x-1}\)có nghĩa khi \(2x-1< 0\)
\(\Leftrightarrow2x< 1\)
\(\Leftrightarrow1x\le\frac{1}{2}\)
Từ đó x<1/2
\(\Rightarrow\sqrt{2x-1}< 3\)
B tương tự
26 tháng 8 2018
a) ĐKXĐ: x≠1; x>0
\(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)^2}=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\sqrt{x}+1\right)=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}+1\right)=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)b) Ta có x=\(6+2\sqrt{5}\Rightarrow A=\dfrac{\left(\sqrt{6+2\sqrt{5}}+1\right)^2}{\sqrt{6+2\sqrt{5}}.\left(\sqrt{6+2\sqrt{5}}-1\right)}=\dfrac{\left(\sqrt{5}+1+1\right)^2}{\left(\sqrt{5}+1\right).\left(\sqrt{5}+1-1\right)}=\dfrac{5+4+4\sqrt{5}}{\left(\sqrt{5}+1\right).\sqrt{5}}=\dfrac{9+4\sqrt{5}}{5+\sqrt{5}}=\dfrac{25+11\sqrt{5}}{20}\)Ta có A<1⇒\(\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}< 1\)
TH1: Nếu \(\sqrt{x}\left(\sqrt{x}-1\right)>0\Rightarrow\left(\sqrt{x}+1\right)^2< \sqrt{x}\left(\sqrt{x}-1\right)\Rightarrow x+2\sqrt{x}+1< x-\sqrt{x}\Rightarrow3\sqrt{x}+1< 0\)
(vô lý)
TH2: Nếu \(\sqrt{x}\left(\sqrt{x}-1\right)< 0\Rightarrow\left(\sqrt{x}+1\right)^2>\sqrt{x}\left(\sqrt{x}-1\right)\Rightarrow3\sqrt{x}+1>0\Rightarrow x>0\)Ta có \(\sqrt{x}\left(\sqrt{x}-1\right)< 0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}-1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 0\\\sqrt{x}-1>0\end{matrix}\right.\end{matrix}\right.\)
Cái ngoặc nhọn thứ 2 bị loại
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}-1< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\x< 1\end{matrix}\right.\)(nhận)
Vậy 0<x<1 thì A<1
\(\text{Ta có:}\sqrt{9}=3\)
\(\Rightarrow\sqrt{x}< 3\Leftrightarrow x< 9\)
#Cừu