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Ta có :
\(B=\left(\frac{1}{x-4}-\frac{1}{x+4\sqrt{x}+4}\right).\frac{x+2\sqrt{x}}{\sqrt{x}}\)
\(=\left(\frac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}-\frac{1}{\left(\sqrt{x}+2\right)^2}\right).\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right).\left(\sqrt{x}+2\right)\)
\(=\frac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}.\left(\sqrt{x}+2\right)\)
\(=\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
C =\(\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right).\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
=\(\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\)
=1-x
C=\(\left(1-\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\).\(\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
=\(\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\)
=\(1-x\)
(\(\sqrt{1+x}+\sqrt{1-x}\))\(\left(2+2\sqrt{1-x^2}\right)=8\)(1)(đk: \(-1\le x\le1\))
đặt \(\sqrt{1+x}+\sqrt{1-x}\) =a (\(a\ge0\)
=> \(a^2=2+2\sqrt{1-x^2}\)
khi đó
(1)\(\Leftrightarrow a^3=8\Leftrightarrow a=\sqrt{8}=2\) (tm)
=>\(\sqrt{1+x}+\sqrt{1-x}\) =2
\(\Leftrightarrow2+2\sqrt{1-x^2}=4\)
\(\Leftrightarrow2\sqrt{1-x^2}=2\)
\(\Leftrightarrow\sqrt{1-x^2}=1\Leftrightarrow1-x^2=1\)
\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)(tm)
vậy x=0 là nghiệm của phương trình
1) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\\ =\left(\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}\right):\left(\frac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\\ =\frac{\sqrt{x}-2}{\sqrt{x}+1}:\frac{x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{\sqrt{x}-2}{\sqrt{x}+1}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
b) \(P=\frac{\sqrt{x}-1}{\sqrt{x}+2}< 0\)
Dễ thấy \(\sqrt{x}+2\ge2>0\forall x\ge0\)
Nên để \(P< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)
Vậy với \(0\le x< 1\)thì P<0
1) ĐKXĐ: \(x>0;x\ne4;x\ne9\)
(*lười lắm, ko chép lại đề nha :V*)
\(P=\frac{\left(2+\sqrt{x}\right)^2+\sqrt{x}\left(2-\sqrt{x}\right)+4x+2\sqrt{x}-4}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{2\sqrt{x}-\left(\sqrt{x}+3\right)}{\sqrt{x}\left(2-\sqrt{x}\right)}\\ =\frac{4+4\sqrt{x}+x+2\sqrt{x}-x+4x+2\sqrt{x}-4}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\\ =\frac{4x+8\sqrt{x}}{2+\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}-3}\\ =\frac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\cdot\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{4x}{\sqrt{x}-3}\)
2) Để P>0 thì
\(\frac{4x}{\sqrt{x}-3}>0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x>0\\\sqrt{x}-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4x< 0\\\sqrt{x}-3< 0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\\sqrt{x}>3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\\sqrt{x}< 3\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 9\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>9\\x< 0\left(ktm\right)\end{matrix}\right.\)
Vậy với \(x>9\) thì \(P>0\).
Chúc bạn học tốt nha
.
Bạn giải thêm cho mk câu này đi
c) tìm giá trị của x để P = -1
\(\left\{{}\begin{matrix}4x^2-4xy+4y^2=4\\x^2+xy+2y^2=4\end{matrix}\right.\)
\(\Rightarrow3x^2-5xy+2y^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(3x-2y\right)=0\Rightarrow\left[{}\begin{matrix}y=x\\y=\frac{3}{2}x\end{matrix}\right.\)
Thay vào pt đầu:
- Với \(y=x\Rightarrow x^2-x^2+x^2=1\Rightarrow x^2=1\Rightarrow x=...\)
- Với \(y=\frac{3}{2}x\Rightarrow x^2-\frac{3}{2}x^2+\left(\frac{3}{2}x\right)^2=1\Leftrightarrow x^2=\frac{4}{7}\Rightarrow x=...\)
A=\(\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right)\):\(\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)Đk x>0 x#0 x#1
=\(\frac{x-1}{\sqrt{x}\left(\sqrt{x-1}\right)}\):\(\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{\sqrt{x}+1}{\left(\sqrt{x-1}\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{1}{\sqrt{x}-1}\)
=\(\frac{\sqrt{x}+1}{\sqrt{x}}.\sqrt{x}-1\)
=\(\frac{x-1}{\sqrt{x}}\)
Ta có 3+\(2\sqrt{2}=\left(\sqrt{2}+1\right)^2\)(thay và A ta dc
=>\(\frac{3+2\sqrt{2}-1}{\sqrt{2}+1}\)
= \(\frac{2\sqrt{2}+2}{\sqrt{2}+1}\)
=2
mk nhầm....\(\frac{x-1}{\sqrt{x}}>0\)=> \(x-1>0\Rightarrow x>1\)
mk làm r nhé
ĐKXĐ: ...
\(\Leftrightarrow\left(\frac{x^2}{x+1}\right)^2+\frac{x^2}{x+1}-12=0\)
Đặt \(\frac{x^2}{x+1}=t\)
\(\Rightarrow t^2+t-12=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{x^2}{x+1}=3\\\frac{x^2}{x+1}=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-3x-3=0\\x^2+4x+4=0\end{matrix}\right.\) \(\Rightarrow\) bấm casio