Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\text{a) ĐKXĐ: }x\ge\sqrt{3}\)
\(\sqrt{x^2-3}\le x^2-3\)
\(\Leftrightarrow\left(\sqrt{x^2-3}\right)^2\le\left(x^2-3\right)^2\)
\(\Leftrightarrow x^2-3\le x^4-6x^2+9\)
\(\Leftrightarrow x^2-3-x^4+6x^2-9\le0\)
\(\Leftrightarrow-x^4+7x^2-12\le0\)
\(\Leftrightarrow-x^2+4x^2+3x^2-12\le0\)
\(\Leftrightarrow\left(-x^4+4x^2\right)+\left(3x^2-12\right)\le0\)
\(\Leftrightarrow-x^2\left(x^2-4\right)+3\left(x^2-4\right)\le0\)
\(\Leftrightarrow\left(x^2-4\right)\left(3-x^2\right)\le0\)
\(\text{Đến đây EZ rồi}\)
a,\(\sqrt{x^2-3}\le x^2-3\)
\(\Leftrightarrow x^2-3\le x^4-6x^2+9\)
\(\Leftrightarrow x^4-6x^2-x^2+12\ge0\)
\(\Leftrightarrow x^4-7x^2+12\ge0\)
\(\Leftrightarrow x^4-\frac{2.7}{2}.x^2+\frac{49}{4}-\frac{1}{4}\ge0\)
\(\Leftrightarrow\left(x^2-\frac{7}{2}\right)^2\ge\frac{1}{4}\)
\(\Leftrightarrow x^2-\frac{7}{2}\ge\frac{1}{2}\Leftrightarrow x^2\ge4\)
\(\Leftrightarrow x\le-2\)và \(x\ge2\)
KL:
b,\(\sqrt{x^2-6x+9}>x-6\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}>x-6\)
\(\Leftrightarrow|x-3|>x-6\)
Với x\(\ge\)3 phương trình <=>x-3>x-6 (luôn đúng)
Với x<3 phương trình <=> 3-x>x-6 <=>x<9/2 <=>x<4,5
KL:
a) ĐKXĐ : \(\orbr{\begin{cases}x\ge\sqrt{3}\\x\le-\sqrt{3}\end{cases}}\)
\(\sqrt{x^2-3}=x^2-3\)
\(\Leftrightarrow\sqrt{x^2-3}=\sqrt{x^2-3}\cdot\sqrt{x^2-3}\)
\(\Leftrightarrow\sqrt{x^2-3}-\sqrt{x^2-3}\cdot\sqrt{x^2-3}=0\)
\(\Leftrightarrow\sqrt{x^2-3}\left(1-\sqrt{x^2-3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2-3}=0\\\sqrt{x^2-3}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-3=0\\x^2-3=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x\in\left\{\pm\sqrt{3}\right\}\\x\in\left\{\pm2\right\}\end{cases}}\)( thỏa mãn )
b) ĐKXĐ : \(x\le6\)
\(\sqrt{x^2-6x+9}=6-x\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=6-x\)
\(\Leftrightarrow\left|x-3\right|=6-x\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=6-x\\x-3=x-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=9\\0x=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x\in\varnothing\end{cases}}\)( thỏa mãn )
\(A=\frac{15\sqrt{x}-11}{x-\sqrt{x}+3\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{45\sqrt{x}-11}{\left(\sqrt{x}+3\right)(\sqrt{x}-1)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{45\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{37\sqrt{x}-5x-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)
Thay \(x=\frac{1}{9}\) vào A ta có:
\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)
2. \(B=...\)
\(B=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{4x+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{\sqrt{x}+3}{-6}\)
Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)
hay \(P\le-\frac{1}{2}\)
Dấu "=" xảy ra <=> x=0
\(A=\sqrt{2-\sqrt{3}}.\left(\sqrt{6}+\sqrt{2}\right)\)
\(A=\sqrt{2-\sqrt{3}}.\sqrt{2}.\left(\sqrt{3}+1\right)\)
\(A=\sqrt{2.2-2\sqrt{3}}.\left(\sqrt{3}+1\right)\)
\(A=\sqrt{4-2\sqrt{3}}.\left(\sqrt{3}+1\right)\)
\(A=\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1}.\left(\sqrt{3}+1\right)\)
\(A=\sqrt{\left(\sqrt{3}-1\right)^2}.\left(\sqrt{3}+1\right)\)
\(A=\left|\sqrt{3}-1\right|.\left(\sqrt{3}+1\right)\)
\(A=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\) ( vi \(\sqrt{3}-1>0\) )
\(A=\left(\sqrt{3}\right)^2-1^2\)
\(A=3-1\)
\(A=2\)
vay \(A=2\)