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\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)ĐK x>=0 x khác -1
=\(\frac{\sqrt{x}+1}{x+1}.\frac{x+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/ x =\(\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{3+2\sqrt{3}+1}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)
\(\Rightarrow\sqrt{x}=\frac{\sqrt{3}+1}{2}\)
Em thay vào tính nhé!
c) với x>1
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}.\sqrt{x}=\frac{x+\sqrt{x}}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\)
Áp dụng bất đẳng thức Cosi
A\(\ge2\sqrt{2}+3\)
Xét dấu bằng xảy ra ....
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
\(ĐKXĐ:x>1\)
\(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right)\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(A=\dfrac{x-1}{2\sqrt{x}}.\left[\dfrac{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)+\left(x+\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(A=\dfrac{x-1}{2\sqrt{x}}.\left(\dfrac{x\sqrt{x}-x-x+\sqrt{x}+x\sqrt{x}+x+x+\sqrt{x}}{x-1}\right)\)
\(A=\dfrac{x-1}{2\sqrt{x}}.\dfrac{2x\sqrt{x}+2\sqrt{x}}{x-1}\)
\(A=\dfrac{2x\sqrt{x}+2\sqrt{x}}{2\sqrt{x}}=\dfrac{2\sqrt{x}\left(x+1\right)}{2\sqrt{x}}=x+1\)
b. Để: \(A=x+1>-6\Rightarrow x+1+6>0\)
\(\Rightarrow x+7>0\Rightarrow x>-7\)
Mà: theo ĐKXĐ: \(x>1\)
⇒ \(x\in R\&x>1\)
\(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right)\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\\ ĐKXĐ:x>0;x\ne1\\ \Rightarrow A=\left(\dfrac{x}{2}-\dfrac{1}{2}\right)\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\\ =\dfrac{x-1}{2}\cdot\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\\ =\dfrac{x-1}{2}\cdot\dfrac{x-2\sqrt{x}+1+x+2\sqrt{x}+1}{x-1}\\ =\dfrac{2x+2}{2}=x+1\)
b) Với \(x>0;x\ne1\)
Để \(A>-6\)
thì \(\Rightarrow x+1>-6\)
\(\Rightarrow x>-7\)
bài 2 ) a) đk : \(a>0;b>0\)
b) P = \(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)
P = \(\dfrac{a-2\sqrt{ab}+b+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
P = \(\dfrac{a+2\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}.\sqrt{a}-\sqrt{b}\) = \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}.\sqrt{a}-\sqrt{b}\) = \(\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\) = \(a-b\)
c) ta có P = \(a-b\) thay \(a=2\sqrt{3};b=\sqrt{3}\) vào ta có
P = \(2\sqrt{3}-\sqrt{3}=\sqrt{3}\) vậy khi \(a=2\sqrt{3};b=\sqrt{3}\) thì P = \(\sqrt{3}\)
bài 1) a) P = \(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
P = \(\dfrac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{\left(a+\sqrt{a}\right)\left(a-\sqrt{a}\right)}+\dfrac{a-1}{\sqrt{a}}.\dfrac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
P = \(\dfrac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{\left(a+\sqrt{a}\right)\left(a-\sqrt{a}\right)}+\dfrac{a-1}{\sqrt{a}}.\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{a-1}\)
P = \(\dfrac{a^2\sqrt{a}+a^2-a-\sqrt{a}-a^2\sqrt{a}+a^2-a+\sqrt{a}}{\left(a+\sqrt{a}\right)\left(a-\sqrt{a}\right)}+\dfrac{2a+2}{\sqrt{a}}\)
P = \(\dfrac{2a^2-2a}{a^2-a}+\dfrac{2a+1}{\sqrt{a}}\) = \(\dfrac{2\left(a^2-a\right)}{a^2-a}+\dfrac{2a+2}{\sqrt{a}}\)
P = \(2+\dfrac{2a+2}{\sqrt{a}}\) = \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)
b) ta có P = 7 \(\Leftrightarrow\) \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}=7\) \(\Leftrightarrow\) \(2a+2\sqrt{a}+2=7\sqrt{a}\)
\(\Leftrightarrow\) \(2a-5\sqrt{a}+2=0\) (1)
đặc \(\sqrt{a}=u\) \(\left(u\ge0\right)\) (1) \(\Leftrightarrow\) \(2u^2-5u+2\)
\(\Delta=\left(-5\right)^2-4.2.2\) = \(25-16=9>0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(u_1=\dfrac{5+3}{4}=\dfrac{8}{4}=2\left(tmđk\right)\)
\(u_2=\dfrac{5-3}{4}=\dfrac{2}{4}=\dfrac{1}{2}\left(tmđk\right)\)
ta có : \(u=\sqrt{a}=2\Leftrightarrow x=4\)
\(u=\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\)
vậy \(a=4;a=\dfrac{1}{4}\) thì P = 7
a/ đkxđ: x > 0; x≠1
b/ \(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\dfrac{x-1}{2\sqrt{x}}\cdot\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{x-1}{2\sqrt{x}}\cdot\dfrac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)
\(=\dfrac{-4x}{2\sqrt{x}}=-2\sqrt{x}\)
1. \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)\)
\(=\left(1+\sqrt{2}\right)^2-\sqrt{3}^2\)
\(=1+2\sqrt{2}+2-3\)
\(=2\sqrt{2}\)
3. \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\cdot\left(1+\dfrac{1}{\sqrt{x}}\right)\)(1)
ĐKXĐ \(x>0,x\ne1\)
pt (1) <=> \(\left(\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\right)\cdot\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+1+\sqrt{x}-1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}\)
\(\Leftrightarrow\dfrac{\sqrt{x}\cdot2}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\)
b) Để \(\sqrt{A}>A\Leftrightarrow\sqrt{\dfrac{2}{\sqrt{x}-1}}>\dfrac{2}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}>\dfrac{4}{x-2\sqrt{x}+1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}-\dfrac{4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\cdot\left(\sqrt{x}-1\right)-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-2-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-6}{x-2\sqrt{x}+1}>0\)
Vì \(2\sqrt{2}-6< 0\Rightarrow x-2\sqrt{x}+1< 0\)
mà \(x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\ge0\forall x\)
Vậy không có giá trị nào của x thỏa mãn \(\sqrt{A}>A\)
(P/s Đề câu b bị sai hay sao vậy, chả có số nào mà \(\sqrt{A}>A\) cả, check lại đề giùm với nhé)
Câu 1:
Sửa đề: \(P=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
a: \(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b: Để P>0 thì căn a-2>0
=>a>4
Bài 2:
\(=\sqrt{8-4\sqrt{3}}\cdot\sqrt{\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}}\)
\(=\sqrt{8-4\sqrt{3}}\cdot\sqrt{\dfrac{\left(\sqrt{6}+\sqrt{2}\right)^2}{6-2}}\)
\(=\left(\sqrt{6}-\sqrt{2}\right)\cdot\dfrac{\left(\sqrt{6}+\sqrt{2}\right)}{2}\)
\(=\dfrac{6-2}{2}=\dfrac{4}{2}=2\)
ĐKXĐ : \(x\ne1;x\ne0;x>0\)
a) \(A=\left(\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\cdot\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\dfrac{x-1}{\sqrt{x}}\right)\)
\(=\dfrac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)+4x\sqrt{x}-4\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{4x\sqrt{x}}{\sqrt{x}}=4x\)
b) \(x=\dfrac{\sqrt{6}}{2+\sqrt{6}}=\dfrac{\sqrt{6}\cdot\left(2-\sqrt{6}\right)}{-2}=\dfrac{2\sqrt{6}-6}{-2}=3-\sqrt{6}\)
Suy ra : \(A=4\cdot\left(3-\sqrt{6}\right)=12-4\sqrt{6}\)
c) \(\sqrt{A}>A\Leftrightarrow\sqrt{4x}>4x\)
\(\Leftrightarrow2\sqrt{x}>4x\)
\(\Leftrightarrow2\sqrt{x}-4x>0\)
\(\Leftrightarrow2\sqrt{x}\cdot\left(1-2\sqrt{x}\right)>0\)
\(\Rightarrow1-2\sqrt{x}>0\) (do x > 0 nên \(2\sqrt{x}>0\))
\(\Leftrightarrow1>2\sqrt{x}\)
\(\Leftrightarrow\dfrac{1}{2}>\sqrt{x}\Rightarrow x< \dfrac{1}{4}\)
Theo điều kiện suy ra giá trị của x để \(\sqrt{A}>A\) là \(0< x< \dfrac{1}{4}\)
a) \(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right)\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right)\cdot\dfrac{x-1}{\sqrt{x}}\)
\(=\left(\dfrac{2\cdot2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\left(\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+4\sqrt{x}\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=4\sqrt{x}\cdot\left[1+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\right]\cdot\dfrac{1}{\sqrt{x}}\)
\(=4\left(1+x-1\right)\)
\(=4x\)
b) Thay \(x=\dfrac{\sqrt{6}}{2+\sqrt{6}}\) vào biểu thức A.
Ta có:
\(4\cdot\dfrac{\sqrt{6}}{2+\sqrt{6}}=\dfrac{4\sqrt{6}}{2+\sqrt{6}}=\dfrac{4\sqrt{6}\cdot\left(2-\sqrt{6}\right)}{-2}\\ =-2\sqrt{6}\cdot\left(2-\sqrt{6}\right)=-4\sqrt{6}+12\)
Vậy giá trị biểu thức A tại \(x=\dfrac{\sqrt{6}}{2+\sqrt{6}}\) là \(-4\sqrt{6}+12\)
c) Để \(\sqrt{A}>A\)
\(\Rightarrow\sqrt{4x}>4x\)
\(\Leftrightarrow\sqrt{4x}>4x\left(đk:x\ge0\right)\)