Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 1:
Thay x=0 vào y=x+1, ta được:
y=0+1=1
Thay y=0 vào y=x+1, ta được:
x+1=0
hay x=-1
vậy: A(-1;0); B(0;1)
\(AB=\sqrt{\left(-1-0\right)^2+\left(0-1\right)^2}=\sqrt{2}\)
\(C_{OAB}=OA+OB+AB=2+\sqrt{2}\)
\(S_{OAB}=\dfrac{OA\cdot OB}{2}=\dfrac{1}{2}\)
a: \(x=2\sqrt{2}+2-2\sqrt{2}+2=4\)
Thay x=4 vào B, ta được:
\(B=\dfrac{2-3}{2+1}=\dfrac{-1}{3}\)
\(K=\dfrac{2\sqrt{3a+1}+2\sqrt{3b+1}+2\sqrt{3c+1}}{2}\)\(\le\)\(\dfrac{3a+1+4+3b+1+4+3c+1+4}{4}=\dfrac{24}{4}=6\)
Vậy \(K_{max}=6\)
Dấu bằng xảy ra khi a=b=c=1
1, Với x > 0 ; x khác 4
\(A=\left(\dfrac{\sqrt{x}-1}{x-4}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}\right):\dfrac{x\sqrt{x}}{\left(x-4\right)^2}\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right):\dfrac{x\sqrt{x}}{\left(x-4\right)^2}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(x-4\right)\left(\sqrt{x}+2\right)}:\dfrac{x\sqrt{x}}{\left(x-4\right)^2}\)
\(=\dfrac{2\sqrt{x}\left(x-4\right)^2}{x\sqrt{x}\left(x-4\right)\left(\sqrt{x}+2\right)}=\dfrac{2\left(\sqrt{x}-2\right)}{x}\)
2, Ta có \(x=4+2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
Thay vào ta được \(A=\dfrac{2\left(\sqrt{3}-1\right)}{4+2\sqrt{3}}=-5+3\sqrt{3}\)
3, Ta có \(\dfrac{2\left(\sqrt{x}-2\right)}{x}-\dfrac{1}{4}\ge0\Leftrightarrow\dfrac{8\left(\sqrt{x}-2\right)-x}{4x}\ge0\)
\(\Rightarrow-x+8\sqrt{x}-16\ge0\Leftrightarrow-\left(\sqrt{x}-4\right)^2\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-4\right)^2\le0\Leftrightarrow\sqrt{x}-4\le0\Leftrightarrow x\le16\)
Kết hợp đk vậy 0 < x =< 16 ; x khác 4
\(x^3=8+3\sqrt[3]{\left(4-2\sqrt[]{2}\right)\left(4+2\sqrt[]{2}\right)}\left(\sqrt[3]{4-2\sqrt[]{2}}+\sqrt[]{4+2\sqrt[]{2}}\right)\)
\(\Rightarrow x^3=8+6x\)
\(\Rightarrow x^3-6x=8\)
Do đó:
\(P=x\left(x^3-6x\right)-8x+24=8x-8x+24=24\)
1) \(\dfrac{4\sqrt{x}-x-4}{x-4}\)
\(=\dfrac{-x+4\sqrt{x}-4}{x-4}\)
\(=\dfrac{-\left(x-4\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+2}\)
\(=\dfrac{-\sqrt{x}+2}{\sqrt{x}+2}\)
2) \(\dfrac{x+y-2\sqrt{xy}}{x\sqrt{y}-y\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
3) \(\dfrac{x-9}{x\sqrt{x}-27}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}\right)^3-3^3}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)}\)
\(=\dfrac{\sqrt{x}+3}{x+3\sqrt{x}+9}\)
\(1,\dfrac{4\sqrt{x}-x-4}{x-4}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{-\left(x-4\sqrt{x}+4\right)}{\sqrt{x^2}-2^2}=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+2}=\dfrac{2-\sqrt{x}}{\sqrt{x}+2}\)
\(2,\dfrac{x+y-2\sqrt{xy}}{x\sqrt{y}-y\sqrt{x}}\left(dk:x,y\ge0\right)\\ =\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
\(3,\dfrac{x-9}{x\sqrt{x}-27}\left(dk:x\ge0\right)\\ =\dfrac{\sqrt{x^2}-3^2}{\sqrt{x^3}-3^3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)}=\dfrac{\sqrt{x}+3}{x+3\sqrt{x}+9}\)
giúp e với aaa, e cảm ơn
giúp e với a, e cảm ơn
Mn giúp e với 
giúp e với mng ơi. e cảm ơn 
giúp e gấp với ạ, e cảm ơn