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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: \(A=\dfrac{1}{\sqrt{x}-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1-x-2-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{-2x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
Để A>0 thì \(\dfrac{-2x+\sqrt{x}}{\sqrt{x}-1}>0\)
=>\(\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}< 0\)
=>1/2<căn x<1
=>1/4<x<1
b: \(B=\dfrac{2}{A}+\sqrt{x}\)
\(=\dfrac{2\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{-2x+\sqrt{x}}+\sqrt{x}\)
\(=\dfrac{2\left(x\sqrt{x}-1\right)-2x\sqrt{x}+x}{-2x+\sqrt{x}}=\dfrac{x-2}{-2x+\sqrt{x}}=\dfrac{-\left(x-2\right)}{2x-\sqrt{x}}< =0\)
Dấu '=' xảy ra khi x=2
a , thu gọn
\(A=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3x+3}{x-9}\right]:\left[\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}\right]\)
\(A=\left(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(A=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(A=\dfrac{-3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(A=-\dfrac{3}{\sqrt{x}+3}\)
b , tự làm
\(a\text{) Để biểu thức xác định }\\ \text{thì }\Rightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-3\ne0\\x-9\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
\(\text{b) }A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\\ =\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}\right)\\ =\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ =\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ =\dfrac{-3}{\sqrt{x}+3}\)
\(c\text{) Để }A\le-\dfrac{1}{3}\\ \text{thì }\Rightarrow\dfrac{-3}{\sqrt{x}+3}\le-\dfrac{1}{3}\\ \Rightarrow\dfrac{3}{\sqrt{x}+3}\ge\dfrac{1}{3}\\ \Rightarrow\dfrac{3}{\sqrt{x}+3}-\dfrac{1}{3}\ge0\\ \Rightarrow\dfrac{9}{3\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}\ge0\\ \Rightarrow\dfrac{9-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}\ge0\\ \Rightarrow\dfrac{\sqrt{x}-6}{\sqrt{x}+3}\le0\\ \Leftrightarrow\sqrt{x}-6\ge0\left(\text{Vì }\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}\ge6\\ \Leftrightarrow x\ge36\)
\(d\text{) Do }\sqrt{x}\ge0\\ \Rightarrow\sqrt{x}+3\ge3\\ \Rightarrow\dfrac{-3}{\sqrt{x}+3}\ge-1\\ \text{Dấu }"="\text{ }xảy\text{ }ra\text{ }khi:\text{ }x=0\)
Vậy..............
a, ĐK: \(x\ge0,x\ne9\)
b, \(A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{2}{\sqrt{x}+3}\)
c, ĐK: \(x\ge0,x\ne9\)
\(A>\dfrac{1}{3}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}>\dfrac{1}{3}\Leftrightarrow\sqrt{x}+3>6\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
Vậy \(A>\dfrac{1}{3}\Leftrightarrow x>9\)
d, ĐK: \(x\ge0,x\ne9\)
Ta có: \(x\ge0\forall x\Leftrightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+3\ge3\)\(\Leftrightarrow\dfrac{1}{\sqrt{x}+3}\le\dfrac{1}{3}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}\le\dfrac{2}{3}\)
Dấu bằng xảy ra \(\Leftrightarrow\sqrt{x}+3=3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\)
Vậy MaxA = \(\dfrac{2}{3}\Leftrightarrow x=0\)
Kết quả rút gọn: \(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(M=\frac{x+12}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{x+12}{\sqrt{x}+2}\)
\(M=\frac{x-4+16}{\sqrt{x}+2}=\sqrt{x}-2+\frac{16}{\sqrt{x}+2}=\left(\sqrt{x}+2+\frac{16}{\sqrt{x}+2}\right)-4\)
Âp dụng BĐT AM-GM cho 2 số không âm ta có:
\(M\ge2\sqrt{\left(\sqrt{x}+2\right).\frac{16}{\sqrt{x}+2}}-4=2.4-4=4\)
Vậy min M =4. Dấu bằng xảy ra \(\Leftrightarrow\left(\sqrt{x}+2\right)^2=16\Leftrightarrow\sqrt{x}+2=4\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
\(P=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\) \(ĐKXĐ:x\ne1\)
\(P=\left(\frac{3}{x-1}+\frac{\sqrt{x}-1}{x-1}\right):\frac{1}{\sqrt{x}+1}\)
\(P=\frac{\sqrt{x}+2}{x-1}.\left(\sqrt{x}+1\right)\)
\(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
b) theo câu a) \(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}\) với \(ĐKXĐ:x\ne1\)
theo bài ra \(P=\frac{5}{4}\)thì \(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{5}{4}\)
\(\Leftrightarrow\left(\sqrt{x}+2\right).4=\left(\sqrt{x}-1\right).5\)
\(\Leftrightarrow4\sqrt{x}+8=5\sqrt{x}-5\)
\(\Leftrightarrow-\sqrt{x}+13=0\)
\(\Leftrightarrow-\sqrt{x}=-13\)
\(\Leftrightarrow\sqrt{x}=13\)
\(\Leftrightarrow x=169\)
vậy \(x=169\)khi \(P=\frac{5}{4}\)
\(a,đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(b,\)\(A=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right).\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\left(1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right).\left(1-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
\(c,A_{max}\Leftrightarrow1-x\)lớn nhất \(\Rightarrow x\)nhỏ nhất
Mà \(x\ge0\)\(\Rightarrow x\)nhỏ nhất \(\Leftrightarrow x=0\)
\(\Rightarrow A_{max}=1\Leftrightarrow x=0\)
Câu a : ĐK : \(x>1\)
\(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Câu b : \(A=\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\Leftrightarrow2\sqrt{x}=3\Leftrightarrow x=\dfrac{9}{4}\)
Câu c : \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}}-9\sqrt{x}=1-\left(9\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)\le1-2\sqrt{9}=-5\)
Vậy GTLN là -5 . Khi \(x=\dfrac{1}{9}\)
$a) ĐK:$ \(x>1\)
\(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(A=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}-1}\right].\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\\ A=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\\ A=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
$b)$ \(A=\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\Leftrightarrow2\sqrt{x}=3\Leftrightarrow x=\dfrac{9}{4}\)
$c)$ \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}}-9\sqrt{x}=1-\left(9\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)\le1-2\sqrt{9}=-5\)
Vậy $GTLN$ là $-5$ . Khi \(x=\dfrac{1}{9}\)