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\(Q=\frac{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\cdot\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(Q=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(Q=x+1\)
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
a) Ta có:
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{2\sqrt{x}\left(\sqrt{x-3}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x-3}\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3x+3}{x-9}\right):\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\left(\frac{2x-6}{x-9}+\frac{x+3\sqrt{x}}{x-9}-\frac{3x+3}{x-9}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2x-6+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\)
b) \(P< \frac{-1}{2}\Rightarrow\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}< \frac{-1}{2}\)
.....Chưa nghĩ ra....
c) Ta có: \(\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-3=0\Rightarrow x=9\)
Vậy Min P = 0 khi x =9.
k - kb với tớ nhia mn!
\(P=B:A\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
P nguyên
=>căn x+3 thuộc Ư(-3)
=>căn x+3 thuộc {1;-1;3;-3}
=>căn x+3=3
=>x=0
Lời giải:
ĐK: $x\geq 0; x\neq 4; x\neq 9$
a)
\(P=\frac{2\sqrt{x}-9}{(\sqrt{x}-3)(\sqrt{x}-2)}+\frac{(2\sqrt{x}+1)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}-\frac{(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}-2)}\)
\(=\frac{2\sqrt{x}-9+(2\sqrt{x}+1)(\sqrt{x}-2)-(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}-2)}=\frac{x-\sqrt{x}-2}{(\sqrt{x}-3)(\sqrt{x}-2)}\)
\(=\frac{(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-3)(\sqrt{x}-2)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b) \(P=\frac{\sqrt{x}+1}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Với $x$ nguyên, để $P$ nguyên thì $\sqrt{x}-3$ phải là ước nguyên của $4$
Mà $\sqrt{x}-3\geq -3$ nên:
$\Rightarrow \sqrt{x}-3\in\left\{\pm 1;\pm 2;4\right\}$
$\Rightarrow x\in \left\{4;16;1;25;49\right\}$ (đều thỏa mãn.
\(P=B:A\)
\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}:\dfrac{\sqrt{x}+3}{\sqrt{x}-1}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{\sqrt{x}+3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\)
\(P=\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{1}{3}\Leftrightarrow3\sqrt{x}-6=\sqrt{x}+3\)
\(\Leftrightarrow2\sqrt{x}=9\Leftrightarrow\sqrt{x}=4,5\Leftrightarrow x=\dfrac{81}{4}\)
b. \(P=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-5}{\sqrt{x}+3}=1-\dfrac{5}{\sqrt{x}+3}\)
Ta có: \(-\dfrac{5}{\sqrt{x}+3}\ge-\dfrac{5}{\sqrt{0}+3}=-\dfrac{5}{3}\)
\(\Rightarrow1-\dfrac{5}{\sqrt{x}+3}\ge1-\dfrac{5}{3}=-\dfrac{2}{3}\)
Suy ra: \(P\ge-\dfrac{2}{3}\) khi \(x=0\)
\(a,P=B:A\)
\(=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\right):\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\left(ĐKXĐ:x\ge0;x\ne9\right)\)
\(=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\right):\left[\dfrac{2\left(\sqrt{x}+3\right)+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\right):\left[\dfrac{3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{3\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+3}{3}\)
\(b,\) Để \(P=\dfrac{\sqrt{x}+3}{3}\) có giá trị nguyên
thì \(\sqrt{x}+3⋮3\)
\(\Leftrightarrow\sqrt{x}+3\in B\left(3\right)\)
\(\Leftrightarrow\sqrt{x}\in B\left(3\right)\)
Kết hợp với điều kiện, ta được:
\(P\) nguyên khi \(x=m^2\left(m\in Z;m⋮3;m\ne3\right)\)
#Toru
a:
ĐKXĐ: x>=0; x<>9
\(A=\dfrac{2\sqrt{x}+6+\sqrt{x}-3}{\left(x-9\right)}=\dfrac{3\sqrt{x}+3}{x-9}\)
\(P=B:A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\cdot\dfrac{x-9}{3\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+3}{3}\)
b: P nguyên khi \(\sqrt{x}+3⋮3\)
=>\(\sqrt{x}\in B\left(3\right)\)
=>\(x=k^2\left(k\in Z;k⋮3\right)\)