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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
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 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
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 \(x=3\)
1/ Rút gọn: \(a)3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\left(a\ge0\right)=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}=3\sqrt{2a}\left(1-a\right)\)b)\(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-1-2}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3+2+1+2\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3}{1+\sqrt{2}}\)c)\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)\sqrt{2}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{6-2\sqrt{5}}}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{9-5}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\)
Làm nốt nè :3
\(2.a.P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}=\dfrac{x-1}{x}\left(x>0;x\ne1\right)\)\(b.P>\dfrac{1}{2}\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x-2}{2x}>0\)
\(\Leftrightarrow x-2>0\left(do:x>0\right)\)
\(\Leftrightarrow x>2\)
\(3.a.A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}=\dfrac{\sqrt{a}-1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\sqrt{a}-1\left(a>0;a\ne1\right)\)
\(b.Để:A< 0\Leftrightarrow\sqrt{a}-1< 0\Leftrightarrow a< 1\)
Kết hợp với DKXĐ : \(0< a< 1\)
Bài 1:
A.\(\left(\sqrt{x}+2\right)\) = -1 (ĐK: \(x\ge0\)
\(\Leftrightarrow\dfrac{1}{x-4}\left(\sqrt{x}+2\right)=-1\)
\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=-1\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}-2}=-1\)
\(\Leftrightarrow\sqrt{x}-2=-1\)
\(\Leftrightarrow\sqrt{x}=1\\ \Leftrightarrow x=1\left(TM\right)\)
Vậy x = 1
Bài 2: ĐK: \(x\ge0\)
Để \(B\in Z\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\in Z\Leftrightarrow\sqrt{x}+2\inƯ\left(3\right)\)\(\Leftrightarrow\sqrt{x}+2\in\left\{\pm1,\pm3\right\}\)\(\Leftrightarrow x\in\left\{1\right\}\)
Bài 3:
a, Ta có: \(x+\sqrt{x}+1=x+2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}+1\\ =\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
Ta có: 2 > 0 và \(x+\sqrt{x}+1>0\Rightarrow C>0\) và \(x\ne1\)
b, ĐK: \(x\ge0,x\ne1\)
\(C=\dfrac{2}{x+\sqrt{x}+1}\)
Ta có: \(x+\sqrt{x}+1=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có: \(\sqrt{x}\ge0\forall x\Rightarrow\sqrt{x}+\dfrac{1}{2}\ge\dfrac{1}{2}\forall x\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2\ge\dfrac{1}{4}\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge1\Leftrightarrow\dfrac{2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le2\)
Dấu bằng xảy ra \(\Leftrightarrow\sqrt{x}+\dfrac{1}{2}=\dfrac{1}{2}\\ \Leftrightarrow x=0\left(TM\right)\)
Vậy MaxC = 2 khi x = 0
Còn cái GTNN chưa tính ra được, để sau nha
Bài 4: ĐK: \(x\ge0,x\ne1\)
\(D=\left(\dfrac{2x+1}{\sqrt{x^3-1}}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-2\sqrt{x}+1\right)\)
\(=\left(\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\sqrt{x}-1\right)^2\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}\)
\(=\sqrt{x}-1\)
\(D=3\Leftrightarrow\sqrt{x}-1=3\Leftrightarrow x=2\left(TM\right)\)
\(D=x-3\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}-1=\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(1-\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3-\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(L\right)\\x=9\left(TM\right)\end{matrix}\right.\)
Bài 5: \(E< -1\Leftrightarrow\dfrac{-3x}{2x+4\sqrt{x}}< -1\)\(\Leftrightarrow\dfrac{-3x}{2x+4\sqrt{x}}+1< 0\Leftrightarrow\dfrac{-3x+2x+4\sqrt{x}}{2x+4\sqrt{x}}< 0\)
\(\Leftrightarrow\dfrac{4\sqrt{x}-x}{2x+4\sqrt{x}}< 0\Leftrightarrow\dfrac{\sqrt{x}\left(4-\sqrt{x}\right)}{2x+4\sqrt{x}}< 0\)
Ta có: \(\sqrt{x}>0\Leftrightarrow x>0\Leftrightarrow2x+4\sqrt{x}>0\) mà \(\dfrac{\sqrt{x}\left(4-\sqrt{x}\right)}{2x+4\sqrt{x}}< 0\)\(\Rightarrow\sqrt{x}\left(4-\sqrt{x}\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}< 0\left(L\right)\\4-\sqrt{x}>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}>0\\4-\sqrt{x}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x< 16,x\ne0\\\left\{{}\begin{matrix}x>0\\x< 16\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 16,x\ne0\\0< x< 16\end{matrix}\right.\)
a/ \(\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}=\dfrac{x+3+2\left(\sqrt{x}-3\right)-\left(\sqrt{x}+3\right)}{x-9}=\dfrac{x+3+2\sqrt{x}-6-\sqrt{x}-3}{x-9}=\dfrac{x-\sqrt{x}-6}{x-9}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-3\left(\sqrt{x}+2\right)}{x-9}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
b/ \(\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+6+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{3\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-3\right)^2\left(\sqrt{x}+3\right)}\)
P/s: câu b đề sai phải không bạn, mk nghĩ ngoài dấu ngoặc là phép chia thì đúng hơn
có phải/....
1) \(A=\dfrac{x+3}{\sqrt{x}-2}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-2}{x-4}\) hay \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\left(\sqrt{x}-2\right)}{x-4}\)
2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
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) \(C=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{4\sqrt{x}}{x-1}=\dfrac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(C=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{x}+1=3\sqrt{x}-3\Leftrightarrow2\sqrt{x}=4\)
\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)