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1/\(\sqrt{24-x^2}-\sqrt{8-x^2}=2\)
\(\Rightarrow2A=\left(\sqrt{24-x^2}+\sqrt{8-x^2}\right)\left(\sqrt{24-x^2}-\sqrt{8-x^2}\right)\)
\(\Leftrightarrow2A=16\Rightarrow A=8\)
2/ ĐKXĐ : \(x\ge5\)
\(\sqrt{x-2}+\sqrt{x-5}=\sqrt{x+3}\)
\(\Rightarrow\left(\sqrt{x-2}+\sqrt{x-5}\right)^2=x+3\)
\(\Leftrightarrow2x+2\sqrt{x-2}.\sqrt{x-5}-7=x+3\)
\(\Rightarrow2\sqrt{x-2}.\sqrt{x-5}=10-x\)
\(\Leftrightarrow4\left(x-2\right)\left(x-5\right)=x^2-20x+100\)
\(\Leftrightarrow3x^2-8x-60=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-\frac{10}{3}\end{cases}}\)
Vì \(x\ge5\) nên x = 6 thỏa mãn đề bài.
Ta có: \(\left(\sqrt{24-x^2}+\sqrt{8-x^2}\right)\left(\sqrt{24-x^2}-\sqrt{8-x^2}\right)=\left(\sqrt{24-x^2}^2-\sqrt{8-x^2}^2\right)\)
\(\Rightarrow\left(\sqrt{24-x^2}+\sqrt{8-x^2}\right)2=24-x^2-\left(8-x^2\right)\)
\(\Rightarrow2A=16\)
\(\Rightarrow A=8\)
Vậy \(A=8\).
5.
\(\Leftrightarrow x^2+7-\left(x+4\right)\sqrt{x^2+7}+4x=0\)
Đặt \(\sqrt{x^2+7}=t>0\)
\(\Rightarrow t^2-\left(x+4\right)t+4x=0\)
\(\Delta=\left(x+4\right)^2-16x=\left(x-4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{x+4+x-4}{2}=x\\t=\frac{x+4-x+4}{2}=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+7}=x\left(x\ge0\right)\\\sqrt{x^2+7}=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+7=x^2\left(vn\right)\\x^2+7=16\end{matrix}\right.\)
Câu 6 bạn coi lại đề
4.
ĐKXĐ: ...
Đặt \(\sqrt{x+3}=a\ge0\)
\(\Rightarrow x+a=\sqrt{5x^2-a^2}\)
\(\Rightarrow x^2+2ax+a^2=5x^2-a^2\)
\(\Rightarrow2x^2-ax-a^2=0\)
\(\Rightarrow\left(x-a\right)\left(2x+a\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=x\\a=-2x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+3}=x\left(x\ge0\right)\\\sqrt{x+3}=-2x\left(x\le0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=x^2\left(x\ge0\right)\\x+3=4x^2\left(x\le0\right)\end{matrix}\right.\)
nhan lien hop va =8
What ??????????????????????
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Đặt \(S=\sqrt{24-x^2}+\sqrt{8-x^2}\)
\(\Leftrightarrow2S=\left(\sqrt{24-x^2}+\sqrt{8-x^2}\right)\left(\sqrt{24-x^2}-\sqrt{8-x^2}\right)\)
\(\Leftrightarrow2S=24-x^2-\left(8-x^2\right)=24-x^2-8+x^2\)
\(\Leftrightarrow2S=24-8-\left(x^2-x^2\right)=16\Leftrightarrow S=8\)
Ta có: \(\sqrt{24-x^2}-\sqrt{8-x^2}=2\)( * )\(\left(ĐK:x\le2\sqrt{2}\right)\)
\(\Leftrightarrow\frac{\left(\sqrt{24-x^2}-\sqrt{8-x^2}\right).\left(\sqrt{24-x^2}+\sqrt{8-x^2}\right)}{\sqrt{24-x^2}+\sqrt{8-x^2}}=2\)
\(\Leftrightarrow\frac{24-x^2-8+x^2}{\sqrt{24-x^2}+\sqrt{8-x^2}}=2\)
\(\Leftrightarrow\frac{16}{\sqrt{24-x^2}+\sqrt{8-x^2}}-2=0\)
\(\Leftrightarrow2.\left(\frac{8}{\sqrt{24-x^2}+\sqrt{8-x^2}}-1\right)=0\)
\(\Leftrightarrow\frac{8}{\sqrt{24-x^2}+\sqrt{8-x^2}}-1=0\)
\(\Leftrightarrow\frac{8}{\sqrt{24-x^2}+\sqrt{8-x^2}}=1\)
\(\Leftrightarrow8=\sqrt{24-x^2}+\sqrt{8-x^2}\)
\(\Leftrightarrow\left(\sqrt{24-x^2}-5\right)+\left(\sqrt{8-x^2}-3\right)=0\)
\(\Leftrightarrow\frac{24-x^2-25}{\sqrt{24-x^2}+5}+\frac{8-x^2-9}{\sqrt{8-x^2}+3}=0\)
\(\Leftrightarrow\frac{-1-x^2}{\sqrt{24-x^2}+5}+\frac{-1-x^2}{\sqrt{8-x^2}+3}=0\)
\(\Leftrightarrow-\left(x^2+1\right).\left(\frac{1}{\sqrt{24-x^2}+5}+\frac{1}{\sqrt{8-x^2}+3}\right)=0\)
+ TH1: \(x^2+1=0\)( ** )
Vì \(x^2\ge0\forall x\)\(\Rightarrow\)\(x^2+1>0\forall x\)mà \(x^2+1=0\)
\(\Rightarrow\)Phương trình ( ** ) vô nghiệm
+ TH2: \(\frac{1}{\sqrt{24-x^2}+5}+\frac{1}{\sqrt{8-x^2}+3}=0\)( *** )
Vì \(\hept{\begin{cases}\sqrt{24-x^2}\ge0\forall x\\\sqrt{8-x^2}\ge0\forall x\end{cases}}\)\(\Rightarrow\)\(\hept{\begin{cases}\sqrt{24-x^2}+5>0\forall x\\\sqrt{8-x^2}+3>0\forall x\end{cases}}\)
\(\Rightarrow\)\(\frac{1}{\sqrt{24-x^2}+5}+\frac{1}{\sqrt{8-x^2}+3}>0\)
mà \(\frac{1}{\sqrt{24-x^2}+5}+\frac{1}{\sqrt{8-x^2}+3}=0\)
\(\Rightarrow\)Phương trình ( *** ) vô nghiệm
Suy ra: Phương trình ( * ) vô nghiệm
Bạn arcobale_new CTV ơi
\(S=\sqrt{24-x^2}+\sqrt{8-x^2}\)
\(\Leftrightarrow2S=2.\left(\sqrt{24-x^2}+\sqrt{8-x^2}\right)=\sqrt{24-x^2}+\sqrt{8-x^2}+\sqrt{24-x^2}+\sqrt{8-x^2}\)