Question

Tính \(\sqrt{24-x^2}+\sqrt{8-x^2}\) biết \(\sqrt{24-x^2}-\sqrt{8-x^2}=2\)

Answer 1

nhan lien hop va =8

Answer 2

What ??????????????????????

Bn hỏi cái quái gì vậy?

Answer 3

Mình nhầm dấu bn ạ 

Xin lỗi nhiều nha

Answer 4

bạn phạm mình tâm nói đúng rồi !

Đặt \(S=\sqrt{24-x^2}+\sqrt{8-x^2}\)

\(\Leftrightarrow2S=\left(\sqrt{24-x^2}+\sqrt{8-x^2}\right)\left(\sqrt{24-x^2}-\sqrt{8-x^2}\right)\)

\(\Leftrightarrow2S=24-x^2-\left(8-x^2\right)=24-x^2-8+x^2\)

\(\Leftrightarrow2S=24-8-\left(x^2-x^2\right)=16\Leftrightarrow S=8\)

Answer 5

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

Answer 6

Bạn arcobale_new CTV ơi

       \(S=\sqrt{24-x^2}+\sqrt{8-x^2}\)

\(\Leftrightarrow2S=2.\left(\sqrt{24-x^2}+\sqrt{8-x^2}\right)=\sqrt{24-x^2}+\sqrt{8-x^2}+\sqrt{24-x^2}+\sqrt{8-x^2}\)