Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
B=\(\frac{x\sqrt{x}-1}{x-1}\)(x>0,x≠1)
=\(\frac{\sqrt{x^3}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
a) Ta có:
\(P=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right)\div\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\)
\(P=\frac{\left(\sqrt{x}+1\right)\sqrt{x}-x-2}{\sqrt{x}+1}\div\frac{\left(\sqrt{x}-1\right)\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x-\sqrt{x}+\sqrt{x}-4}\)
\(P=\frac{\sqrt{x}-2}{\sqrt{x}+1}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
b) Đề đánh kia ai hiểu được đây, lm đại 3 TH ra nè:
Nếu \(P=\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{1}{2}\)
\(\Leftrightarrow\sqrt{x}+2=2\sqrt{x}-2\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Rightarrow x=16\)
Nếu \(P>\frac{1}{2}\) mà \(\sqrt{x}+2>0\left(\forall x\right)\)
\(\Rightarrow\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Rightarrow x>1\)
Nếu \(P< \frac{1}{2}\) mà \(\sqrt{x}+2>0\left(\forall x\right)\)
\(\Rightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\Rightarrow x< 1\)
\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)
\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)
\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(=\frac{\sqrt{x}^3+\sqrt{y}^3}{\sqrt{x}+\sqrt{y}}-\left(x+2\sqrt{xy}+y\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)
\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)
\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé
\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
a) đk : \(x\ge0\) ; \(x\ne1\)
A=\(\left(\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}+1\right)}-\frac{x+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(=\left(\frac{-\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\) \(=\frac{1-\sqrt{x}}{x+1}\)
b) đk : \(x\ne0;x\ne1\)
B=\(\left(\frac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{x-1}\right):\left(\frac{1-x}{2\sqrt{x}}\right)^2\) \(=\left(\frac{-2\sqrt{x}}{x-1}\right):\left(\frac{1-x}{2\sqrt{x}}\right)^2\) \(=\frac{-4x}{\left(x-1\right)^3}\)