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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 \(A=\dfrac{2+2}{2+3}=\dfrac{4}{5}\)
ĐKXĐ :x\(\ge\)0
a) với x=64 thỏa mãn đk; khi đó: A=\(\dfrac{2+\sqrt{64}}{\sqrt{64}}=\dfrac{2+8}{8}=\dfrac{5}{4}\)
b)với đk của x thì B xác định ; ta có
B\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+\left(2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)\(=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
c)Xét M=A:B =\(\dfrac{2+\sqrt{x}}{\sqrt{2}}:\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
Để \(M>\dfrac{3}{2}hay\dfrac{\sqrt{x}+2}{\sqrt{x}+1}>\dfrac{3}{2}\Leftrightarrow2\sqrt{x}+4>3\sqrt{x}+3\left(do:\sqrt{x}+1>0\right)\Leftrightarrow\sqrt{x}< 1\Rightarrow x< 1\)
Kết hợp đk x\(\ge\)0. Vậy 0\(\le\)x<1 thì M=A:B>3/2
Bài 2:
a: \(A=\dfrac{2x+6\sqrt{x}-x-9\sqrt{x}}{x-9}=\dfrac{x-3\sqrt{x}}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+5\right)}{x-25}=\dfrac{\sqrt{x}}{\sqrt{x}-5}\)
b: \(P=A:B=\dfrac{\sqrt{x}}{\sqrt{x}+3}:\dfrac{\sqrt{x}}{\sqrt{x}-5}=\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\)
\(P-1=\dfrac{\sqrt{x}-5-\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{-8}{\sqrt{x}+3}< 0\)
=>P<1
Bạn nào làm được bài này thì giúp mình với ạ ! mình đang cần gấp
Bài 4:
\(AH=\sqrt{9\cdot16}=12\left(cm\right)\)
\(AB=\sqrt{9\cdot25}=15\left(cm\right)\)
AC=căn(25^2-15^2)=20(cm)
Xét ΔABC vuông tại A có sin ABC=AC/BC=4/5
nên góc ABC=53 độ
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 \(x=3\)
Bài 1:
a. ta có \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}-y\)
= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
=\(\sqrt{xy}\)
b.ĐK: x ≠ 1
Ta có: A= \(\sqrt{\dfrac{x+2\sqrt{x}+1}{x-2\sqrt{x}+1}}\)=\(\sqrt{\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)^2}}\)=\(\dfrac{\sqrt{x}+1}{\left|\sqrt{x}-1\right|}\)
*Nếu \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge1\)
⇒ A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
*Nếu \(\sqrt{x}-1< 0\Rightarrow\sqrt{x}< 1\)
⇒ A=\(\dfrac{\sqrt{x}+1}{-\sqrt{x}+1}\)
c.Ta có:
Bài 2: a) Ta có: Q=\(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) -\(\left(\dfrac{x+2}{\left(\sqrt{x}\right)^3-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\right)\) =\(\dfrac{1}{\sqrt{x}-1}\) -\(\left(\dfrac{x+2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\left(\dfrac{x+2+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\dfrac{2x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) =
a) Rut gon H
\(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}+\dfrac{1}{2-\sqrt{a}}\)
\(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}-\dfrac{1}{\sqrt{a}-2}\)
DKXD : \(\left\{{}\begin{matrix}\sqrt{a}+3\ne0\\\sqrt{a}-2\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a\ne9\\a\ne4\end{matrix}\right.\)
Ta co : \(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{\sqrt{a}+3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(H=\dfrac{a-\sqrt{a}-6}{a+\sqrt{a}-6}\)
a: \(P=\dfrac{2x+4\sqrt{x}-x-6\sqrt{x}}{x-4}=\dfrac{x-2\sqrt{x}}{x-4}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
b: Thay x=1 vào P, ta được:
\(P=\dfrac{1}{1+2}=\dfrac{1}{3}\)