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a: ĐKXĐ: x>=0
b: ĐKXĐ: x-1>0 và -(x2-x-6)>=0
=>x>1 và (x-3)(x+2)<=0
=>x>1 và -2<=x<=3
=>1<x<=3
1: ĐKXĐ: 6-3x>=0 và x<>3
=>x<=2
2: ĐKXĐ: 3-2x>0
=>2x<3
hay x<3/2
3: ĐKXĐ: x>=0
\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
a: \(P=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
c: Để \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\) là số nguyên thì \(\sqrt{x}+1-2⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1\in\left\{1;2\right\}\)
=>x=0
1) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ge-1\end{matrix}\right.\)
2) \(A=2^2+\left(3\sqrt{2}\right)^2+2.2.3\sqrt{2}-12\sqrt{2}=4+18+12\sqrt{2}-12\sqrt{2}=22\)\(B=\sqrt{4+3+4\sqrt{3}-\sqrt{3}=\sqrt{7+3\sqrt{3}}}\)
3) a) \(A=\dfrac{x\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{x\sqrt{x}-2x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)b) Ta có :
\(x=3+2\sqrt{2}=\left(\sqrt{2}\right)^2+2.1.\sqrt{2}+1^2=\left(\sqrt{2}+1\right)^2\)Thay x vào A ta đc : \(A=\sqrt{x}-1=\sqrt{\left(\sqrt{2}+1\right)^2}-1=\sqrt{2}+1-1=\sqrt{2}\)4) a)
\(\sqrt{9x-27}+\sqrt{x-3}-\dfrac{1}{2}\sqrt{4x-12}=7\Leftrightarrow3\sqrt{x-3}+\sqrt{x-3}-\dfrac{1}{2}.2.\sqrt{x-3}=7\Leftrightarrow3\sqrt{x-3}=7\Leftrightarrow x-3=\dfrac{49}{9}\Leftrightarrow x=\dfrac{76}{9}\)b)Đề chuyển thánh sinB=3/4 nha
Ta có: sin2B+cos2B=1=> cosB=\(\dfrac{\sqrt{7}}{4}\)
cosC=sinB=3/4
cai nay hinh nhu la co trong nang cao hat trien lo 8 thi phai cho
\(1a.Để:A=\dfrac{x}{x-2}+\sqrt{x-2}\) xác định thì :
\(\left\{{}\begin{matrix}x-2\ne0\\x-2\ge0\end{matrix}\right.\) \(\Leftrightarrow\) \(x>2\)
\(1b.Taco:B=\sqrt{-x^2+2x-1}=-\sqrt{\left(x-1\right)^2}\)
\(Để:B=\sqrt{-x^2+2x-1}=-\sqrt{\left(x-1\right)^2}\) xác định thì :
\(\left(x-1\right)^2\ge0\) ( luôn đúng )
KL.................
\(2.\sqrt{9x^2+6x+1}=\sqrt{11-6\sqrt{2}}\)
\(\Leftrightarrow\sqrt{\left(3x+1\right)^2}=\sqrt{9-2.3\sqrt{2}+2}=\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(\Leftrightarrow|3x+1|=|3-\sqrt{2}|=3-\sqrt{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=3-\sqrt{2}\\3x+1=\sqrt{2}-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2-\sqrt{2}}{3}\\x=\dfrac{\sqrt{2}-4}{3}\end{matrix}\right.\)
KL.............
\(3a.\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.2\sqrt{5}.3+9}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-|2\sqrt{5}-3|}}=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}=\sqrt{\sqrt{5}-|\sqrt{5}-1|}=\sqrt{1}=1\)
\(3b.\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+|2\sqrt{2}+1|}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{13+30|\sqrt{2}+1|}=\sqrt{43+30\sqrt{2}}=\sqrt{18+2.3\sqrt{2}.5+25}=\sqrt{\left(3\sqrt{2}+5\right)^2}=|3\sqrt{2}+5|=3\sqrt{2}+5\)
- ĐK \(x^2-8x+18\ge0\Rightarrow x^2-8x+16+2\ge0\)\(\Rightarrow\left(x-4\right)^2+2\ge2\forall x\)TXD : R
- ĐK \(9x^2-6x+1>0\Rightarrow\left(3x-1\right)^2>0\forall x\ne\frac{1}{3}\)\(\Rightarrow TXD=R|\left\{\frac{1}{3}\right\}\)
ĐKXĐ: \(\left\{{}\begin{matrix}9x-6>=0\\2x-1< >0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\x< >\dfrac{1}{2}\end{matrix}\right.\)
ĐKXĐ
\(\left\{{}\begin{matrix}9x-6\ge0\\2x-1\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9x\ge6\\2x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\x\ne\dfrac{1}{2}\end{matrix}\right.\)