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a/ \(P=12\)
b/ \(Q=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c/ Ta có:
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Dấu = xảy ra khi x = 3 (thỏa tất cả các điều kiện )
a. Thay x = 3 vào biểu thức P ta được :
\(p=\frac{x+3}{\sqrt{x}-2}=\frac{9+3}{\sqrt{9}-2}=12\)
b, \(Q=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c, Ta có :
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Vậy GTNN \(\frac{P}{Q}=2\sqrt{3}\) khi và chỉ khi \(x=3\)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)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 1:
a: \(=\dfrac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{x-9}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{x-9}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
b: Để A=3/4 thì căn x+2=3
=>x=1
c: Khi x=4 thì \(A=\dfrac{2+2}{2+3}=\dfrac{4}{5}\)
\(Q=\frac{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\cdot\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(Q=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(Q=x+1\)
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: \(Q=\dfrac{9-x+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-\left(\sqrt{x}+2\right)}{\sqrt{x}+3}\)
\(P=\dfrac{x-9-x+3\sqrt{x}}{x-9}=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}=\dfrac{3}{\sqrt{x}+3}\)
\(M=\dfrac{3}{\sqrt{x}+3}:\dfrac{-\left(\sqrt{x}+2\right)}{\sqrt{x}+3}=\dfrac{-3}{\sqrt{x}+2}\)
Để M>0 thì căn x+2<0(vô lý)
b: \(M\left(x+1\right)=\dfrac{-3\left(x+1\right)}{\sqrt{x}+2}\)
\(=\dfrac{-3x-3}{\sqrt{x}+2}=\dfrac{-3x+12-15}{\sqrt{x}+2}=-\dfrac{3\left(x-4\right)}{\sqrt{x}+2}-\dfrac{15}{\sqrt{x}+2}\)
\(=-3\left(\sqrt{x}-2\right)-\dfrac{15}{\sqrt{x}+2}\)
\(=-3\sqrt{x}+6-\dfrac{15}{\sqrt{x}+2}\)
\(=-3\sqrt{x}-6-\dfrac{15}{\sqrt{x}+2}+12\)
\(=-\left(3\sqrt{x}+6+\dfrac{15}{\sqrt{x}+2}\right)+12\)
\(3\left(\sqrt{x}+2\right)+\dfrac{15}{\sqrt{x}+2}>=2\cdot\sqrt{3\left(\sqrt{x}+2\right)\cdot\dfrac{15}{\sqrt{x}+2}}=6\sqrt{5}\)
=>M(x+1)<=-6căn 5+12
Dấu = xảy ra khi \(3\left(\sqrt{x}+2\right)^2=15\)
=>(căn x+2)^2=5
=>căn x+2=căn 5
=>căn x=căn 5-2
=>x=9-4căn 5
Bài 2: a) Ta có: Q=\(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) -\(\left(\dfrac{x+2}{\left(\sqrt{x}\right)^3-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\right)\) =\(\dfrac{1}{\sqrt{x}-1}\) -\(\left(\dfrac{x+2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\left(\dfrac{x+2+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\dfrac{2x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) =
1/ \(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x}{x-4}+\dfrac{\sqrt{x}+2}{x-4}+\dfrac{\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
2/ Khi x = 25 thì:
\(A=\dfrac{\sqrt{25}}{\sqrt{25}-2}=\dfrac{5}{5-2}=\dfrac{5}{3}\)
3/ A = \(-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=-\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{x}=-\dfrac{1}{3}\sqrt{x}+\dfrac{2}{3}\)
\(\Leftrightarrow\sqrt{x}+\dfrac{1}{3}\sqrt{x}=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{4}{3}\sqrt{x}=\dfrac{2}{3}\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)
vậy............
CHÚC EM HỌC TỐT NHÁ
giải phương trình
a)\(\sqrt{x^8}=256\) b)\(\sqrt{x^2-2x+1}=x-1\)