Tim GTLN
A=\(\sqrt{\chi}\)+8/2\(\sqrt{\chi}\)+1
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a: \(B=1-\sqrt{\left(x-1\right)^2+1}\)
(x-1)^2+1>=1
=>\(\sqrt{\left(x-1\right)^2+1}>=1\)
=>\(B< =0\)
Dấu = xảy ra khi x=1
b:
ĐKXĐ: -(x+2)^2+2>=0
=>-(x+2)^2>=2
=>(x+2)^2<=2
=>\(-\sqrt{2}-2< =x< =\sqrt{2}-2\)
\(-x^2+4x-2=-\left(x^2-4x+2\right)\)
\(=-\left(x^2-4x+4-2\right)=-\left(x-2\right)^2+2< =2\)
=>\(0< =\sqrt{4x-x^2-2}< =\sqrt{2}\)
=>1<=C<=căn 2+1
\(C_{max}=\sqrt{2}+1\Leftrightarrow x=2\)
Giải PT hở b?
ĐK : \(\)\(\left\{{}\begin{matrix}x\ge0\\1-x\ge0\\x\left(1-x\right)\ge0\end{matrix}\right.\Rightarrow}0\le x\le1\)
(0=<x=<1)
đặt \(\sqrt{x}=a;\sqrt{1-x}=b\left(a,b\ge0\right)\\ \Rightarrow\left\{{}\begin{matrix}1+\dfrac{2}{3}ab=a+b\\a^2+b^2=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b-\dfrac{2}{3}ab=1\\\left(a+b\right)^2-2ab=1\end{matrix}\right.\\ \left(ab=P\ge0;a+b=S\ge0\right)\\ \Rightarrow\left\{{}\begin{matrix}S-\dfrac{2}{3}P=1\\S^2-2P=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}S=1+\dfrac{2}{3}P\\1+\dfrac{4}{3}P+\dfrac{4}{9}P^2-2P=1\end{matrix}\right.\Rightarrow}}\\ P=\left[{}\begin{matrix}\dfrac{3}{2}\Rightarrow S=2\left(TM\right)\Rightarrow a,b\in\varnothing\\0\Rightarrow S=1\left(TM\right)\Rightarrow\left[{}\begin{matrix}a=1b=0\left(TM\right)\Rightarrow x=1\left(TM\right)\\a=0;b=1\left(TM\right)\Rightarrow x=0\left(TM\right)\end{matrix}\right.\end{matrix}\right.\)
vậy tập nghiệm của PT là:
x=1 hoặc x=0
hic mik giải 1 hồi mak bị lỗi r, nhg chủ yeus đặt căn x vs căn 1-x lak a vs b, sau đó tính tổng vs tích = hệ PT r tìm dc th
Lần sau bạn chú ý viết đầy đủ đề.
1.
\(\sqrt{9+4\sqrt{5}-\sqrt{9-4\sqrt{5}}}=\sqrt{9+4\sqrt{5}-\sqrt{5-2\sqrt{4.5}+4}}\)
\(=\sqrt{9+4\sqrt{5}-\sqrt{(\sqrt{5}-\sqrt{4})^2}}=\sqrt{9+4\sqrt{5}-(\sqrt{5}-\sqrt{4})}\)
\(=\sqrt{9+4\sqrt{5}-\sqrt{5}+2}=\sqrt{11+3\sqrt{5}}\)
2.
\(\sqrt{8-2\sqrt{7}-\sqrt{8+2\sqrt{7}}}=\sqrt{8-2\sqrt{7}-\sqrt{7+2\sqrt{7}+1}}\)
\(=\sqrt{8-2\sqrt{7}-\sqrt{(\sqrt{7}+1)^2}}\)
\(=\sqrt{8-2\sqrt{7}-\sqrt{7}-1}=\sqrt{7-3\sqrt{7}}\)
a)√x−2+12√4x−8=√9x−18−2
=>√x−2+12√4(x−2)=√9(x−2)−2
=>√x−2+12√22(x−2)=√32(x−2)−2
=>√x−2+12.2√(x−2)=3√(x−2)−2
=>√x−2+24√(x−2)=3√(x−2)−2
=>√x−2+24√(x−2)-3√(x−2)=-2
=>√x−2(1+24-3)=-2
=>22√x−2=-2
=>√x−2=-2/22
=>√x−2=-1/11
=>x−2=1/121
=>x=1/121+2=243/121
b)√(3x−1)2=5
=>|3x−1|=5
=>3x−1=5 hoặc 3x−1=-5
=>3x=6 hoặc 3x=-4
=>x=2 hoặc x=-4/3
1) ĐKXĐ: \(16x^2-25\ge0\)
\(\Leftrightarrow x^2\ge\dfrac{25}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{4}\\x\le-\dfrac{5}{4}\end{matrix}\right.\)
2) ĐKXĐ: \(4x^2-49\ge0\Leftrightarrow x^2\ge\dfrac{49}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{7}{2}\\x\le-\dfrac{7}{2}\end{matrix}\right.\)
3) ĐKXĐ: \(8-x^2\ge0\Leftrightarrow x^2\le8\)
\(\Leftrightarrow-2\sqrt{2}\le x\le2\sqrt{2}\)
4) ĐKXĐ: \(x^2-12\ge0\Leftrightarrow x^2\ge12\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge2\sqrt{3}\\x\le-2\sqrt{3}\end{matrix}\right.\)
5) ĐKXĐ: \(x^2+4\ge0\left(đúng\forall x\right)\)
1) \(=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
2) \(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}\)
3) \(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\)
5) \(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)
6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)
7) \(=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
\(\sqrt{29-4\sqrt{7}}=\sqrt{\left(2\sqrt{7}\right)^2-2.2\sqrt{7}.1+1^2}=\sqrt{\left(2\sqrt{7}-1\right)^2}=\left|2\sqrt{7}-1\right|\)
\(=2\sqrt{7}-1\)
\(\sqrt{19+6\sqrt{2}}=\sqrt{\left(3\sqrt{2}\right)^2+2.3\sqrt{2}.1+1^2}=\sqrt{\left(3\sqrt{2}+1\right)^2}=\left|3\sqrt{2}+1\right|\)
\(=3\sqrt{2}+1\)
\(\sqrt{28-6\sqrt{3}}=\sqrt{\left(3\sqrt{3}\right)^2-2.3\sqrt{3}.1+1^2}=\sqrt{\left(3\sqrt{3}-1\right)^2}=\left|3\sqrt{3}-1\right|\)
\(=3\sqrt{3}-1\)
\(\sqrt{46-6\sqrt{5}}=\sqrt{\left(3\sqrt{5}\right)^2-2.3\sqrt{5}.1+1^2}=\sqrt{\left(3\sqrt{5}-1\right)^2}=\left|3\sqrt{5}-1\right|\)
\(=3\sqrt{5}-1\)
\(\sqrt{49+8\sqrt{3}}=\sqrt{\left(4\sqrt{3}\right)^2+2.4\sqrt{3}.1+1^2}=\sqrt{\left(4\sqrt{3}+1\right)^2}=\left|4\sqrt{3}+1\right|\)
\(=4\sqrt{3}+1\)
\(\sqrt{32-8\sqrt{7}}=\sqrt{\left(2\sqrt{7}\right)^2-2.2\sqrt{7}.2+2^2}=\sqrt{\left(2\sqrt{7}-2\right)^2}=\left|2\sqrt{7}-2\right|\)
\(=2\sqrt{7}-2\)
\(\sqrt{29-4\sqrt{7}}=2\sqrt{7}-1\)
\(\sqrt{19+6\sqrt{2}}=3\sqrt{2}+1\)
\(\sqrt{28-6\sqrt{3}}=3\sqrt{3}-1\)
\(\sqrt{46-6\sqrt{5}}=3\sqrt{5}-1\)
\(\sqrt{49+8\sqrt{3}}=4\sqrt{3}+1\)
\(\sqrt{32-8\sqrt{7}}=2\sqrt{7}-2\)
\(\sqrt{x}\)lớn hơn bằng 0
=>\(\frac{8}{2}\sqrt{x}\) lớn hơn hặc bằng 0
=>\(\sqrt{x}\)+\(\frac{8}{2}\sqrt{x}\)lớn hơn bằng 0
=> \(\sqrt{x}\)\(+\frac{8}{2}\sqrt{x^{ }}\)+1 lớn hơn hoặc bằng 1
Vậy Amax =1 <=> x=0