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+)\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)= 2
\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\sqrt{\left(x-1+1\right)^2}+\sqrt{\left(x-1-1\right)^2}=2\)
\(\sqrt{x^2}+\sqrt{\left(x-2\right)^2}=2\)
\(x+x-2=2\)
\(2x=4\)
\(x=2\)
+) Hình như sai đâu bài chỗ \(\sqrt{x+3+4\sqrt{x+1}}\)
\(\)
Giải pt :
1
a. ĐKXĐ : \(x\ge4\)
Ta có :
\(\sqrt{x+3}-\sqrt{x-4}=1\\ \Leftrightarrow\sqrt{x+3}=1+\sqrt{x-4}\\ \Leftrightarrow x+3=x-3+2\sqrt{x-4}\\ \Leftrightarrow6=2\sqrt{x-4}\)
\(\Leftrightarrow3=\sqrt{x-4}\\ \Leftrightarrow x-4=9\)
\(\Leftrightarrow x=13\) (TM ĐKXĐ)
Vậy \(S=\left\{13\right\}\)
b.ĐKXĐ : \(-3\le x\le10\)
Ta có :
\(\sqrt{10-x}+\sqrt{x+3}=5\\ \Leftrightarrow13+2\sqrt{-x^2+7x+30}=25\\ \Leftrightarrow\sqrt{-x^2+7x+30}=6\\ \Leftrightarrow-x^2+7x+30=36\\ \Leftrightarrow-x^2+7x-6=0\\ \Leftrightarrow-x^2+x+6x-6=0\\ \Leftrightarrow-x\left(x-1\right)+6\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(6-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(TMĐKXĐ\right)\\x=6\left(TMĐKXĐ\right)\end{matrix}\right.\)
Vậy \(S=\left\{1;6\right\}\)
\(2,\)
\(a,\sqrt{x^2-4x+3}=3\)
\(\Rightarrow x^2-4x+3=9\)
\(\Rightarrow x^2-4x-6=0\)
\(\Rightarrow\left(x-2\right)^2=10\)
\(\Rightarrow\orbr{\begin{cases}x-2=\sqrt{10}\\x-2=-\sqrt{10}\end{cases}\Rightarrow\orbr{\begin{cases}x=2+\sqrt{10}\\x=2-\sqrt{10}\end{cases}}}\)
1/ \(C=\frac{x+9}{10\sqrt{x}}=\frac{\sqrt{x}}{10}+\frac{9}{10\sqrt{x}}\ge2.\frac{3}{10}=0,6\)
Đạt được khi x = 9
2/ \(E=\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=x-3\sqrt{x}+2\)
\(=\left(x-\frac{2.\sqrt{x}.3}{2}+\frac{9}{4}\right)-\frac{1}{4}\)
\(=\left(\sqrt{x}-\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Vậy GTNN là \(-\frac{1}{4}\)đạt được khi \(x=\frac{9}{4}\)
Không có GTLN nhé
\(x+2=3\sqrt{1-x^2}+\sqrt{1+x}\)
\(ĐKXĐ:-1\le x\le1\)
\(x+2=3\sqrt{1-x}\sqrt{1+x}+\sqrt{1+x}\)
\(\left(3\sqrt{1-x}\sqrt{1+x}-\frac{3}{2}\right)+\left(\sqrt{1+x}-x-\frac{1}{2}\right)=0\)
\(\frac{9\left(1-x\right)\left(1+x\right)-\frac{9}{4}}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1+x-\left(x+\frac{1}{2}\right)^2}{\sqrt{1+x}+x+\frac{1}{2}}=0\)
\(\frac{9-9x^2-\frac{9}{4}}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1+x-x^2-x-\frac{1}{4}}{\sqrt{1+x}+x+\frac{1}{2}}=0\)
\(\frac{\frac{27}{4}-9x^2}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{\frac{3}{4}-x^2}{\sqrt{1+x}+x+\frac{1}{2}}=0\)
\(\frac{9\left(\frac{3}{4}-x^2\right)}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{\frac{3}{4}-x^2}{\sqrt{1+x}+x+\frac{1}{2}}=0\)
\(\left(\frac{3}{4}-x^2\right)\left(\frac{9}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1}{\sqrt{1+x}+x+\frac{1}{2}}\right)=0\)
\(\orbr{\begin{cases}\frac{3}{4}-x^2=0\\\frac{9}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1}{\sqrt{1+x}+x+\frac{1}{2}}=0\left(KTM\right)\end{cases}< =>x=\frac{\sqrt{3}}{2}\left(TM\right)}\)
\(\frac{9}{3\sqrt{1-x}\sqrt{1+x}+\frac{3}{2}}+\frac{1}{\sqrt{1+x}+x+\frac{1}{2}}>0\)nên pt ktm