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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 ....
a: \(A=\left(\dfrac{\left(2x+2\sqrt{x}\right)-\left(\sqrt{x}+1\right)}{1-x}+\dfrac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{1+x\sqrt{x}}\right)\cdot\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\)
\(=\left(2\sqrt{x}-1\right)\cdot\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\cdot\left(\dfrac{\sqrt{x}+1}{1-x}+\dfrac{\sqrt{x}+1}{1+x\sqrt{x}}\right)\)
\(=\sqrt{x}\left(1-\sqrt{x}\right)\cdot\left(\dfrac{-1}{\sqrt{x}-1}+\dfrac{1}{x-\sqrt{x}+1}\right)\)
\(=\sqrt{x}\left(1-\sqrt{x}\right)\cdot\dfrac{-x+\sqrt{x}-1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\cdot\left(-x+2\sqrt{x}-2\right)}{-x+\sqrt{x}-1}=\dfrac{\sqrt{x}\left(x-2\sqrt{x}+2\right)}{x-\sqrt{x}+1}\)
c: \(x-2\sqrt{x}+2=\left(\sqrt{x}-1\right)^2+1>=1\)
\(\left(x-\sqrt{x}+1\right)=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\)
căn x>=0
=>A không có giá trị lớn nhất
Bài 1 : Rút gọn biểu thức :
\(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=\left(-10\sqrt{2}+10\right)-\left(18-30\sqrt{2}+25\right)\)
\(=\left(-10\sqrt{2}+10\right)-\left(7-30\sqrt{2}\right)\)
\(=-10\sqrt{2}+10-7+30\sqrt{2}\)
\(=20\sqrt{2}+3\)
Bài 2:
a) ĐKXĐ : x # 4 ; x # - 4
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
P =\(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{x+2\sqrt{x}+\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b ) Để P = 2 \(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}\) = 2
\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
Vậy, để P = 2 thì x = 16.
ĐK:\(x>0,x\ne1\)
a) \(A=\left(\frac{1}{x-\sqrt{x}}+\frac{x+\sqrt{x}+1}{x\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right].\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right].\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
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: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
\(Q=\frac{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\cdot\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(Q=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(Q=x+1\)
Không thể tìm được GTLN hay GTNN của Q.
b)
\(\frac{3x+3}{\sqrt{x}}=3\sqrt{x}+\frac{3}{\sqrt{x}}\)
Để \(\frac{3Q}{\sqrt{x}}\) nguyên thì \(\frac{3}{\sqrt{x}}\)nguyên hay \(\sqrt{x}\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Vì \(\sqrt{x}\)dương nên \(\sqrt{x}\in\left\{1;3\right\}\)
Vậy x=1, x=9 là các giá trị cần tìm
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)
a) \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\) (ĐK: \(x\ne1,x\ge0\))
\(A=\left[\dfrac{x+2}{\left(\sqrt{x}\right)^3-1^3}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right]\cdot\dfrac{2}{\sqrt{x}-1}\)
\(A=\left[\dfrac{\left(x+2\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]\cdot\dfrac{2}{\sqrt{x}-1}\)
\(A=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(A=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(A=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(A=\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(A=\dfrac{2}{x+\sqrt{x}+1}\)
b) Ta có:
\(A=\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{x+2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{2}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)
Mà: \(2>0\Rightarrow\dfrac{2}{\left(x+\dfrac{1}{2}\right)+\dfrac{3}{4}}\le\dfrac{2}{\dfrac{3}{4}}=\dfrac{8}{3}\)
Dấu "=" xảy ra:
\(\dfrac{2}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}=\dfrac{8}{3}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=2:\dfrac{8}{3}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\Leftrightarrow x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy: \(A_{max}=\dfrac{8}{3}\) khi \(x=-\dfrac{1}{2}\)