Tính \(\sqrt{24-x^2}+\sqrt{8-x^2}\) biết \(\sqrt{24-x^2}-\sqrt{8-x^2}=2\)
Tính \(\sqrt{24-x^2}+\sqrt{8-x^2}\)Biết \(\sqrt{24-x^2}-\sqrt{8-x^2}=2\)
a) Tính \(\sqrt{24-x^2}+\sqrt{8-x^2}\) biết \(\sqrt{24-x^2}-\sqrt{8-x^2}\)= 2
b) Tính \(\sqrt{25-x^2}+\sqrt{15-x^2}\) biết \(\sqrt{25-x^2}-\sqrt{15-x^2}\)= 2
Tính \(A=\sqrt{24-x^2}+\sqrt{8-x^2}\) biết \(\sqrt{24-x^2}-\sqrt{8-x^2}=2\)
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 \(A=8\).
1/ Tính \(A=\sqrt{24-x^2}+\sqrt{8-x^2}\)biết \(\sqrt{24-x^2}-\sqrt{8-x^2}=2\)
2/ Giải phương trình: \(\sqrt{x-2}+\sqrt{x-5}=\sqrt{x+3}\)
Help me!!!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.
1/ Tìm x biết \(\sqrt{x+\sqrt{2x-1}}\)+ \(\sqrt{x-\sqrt{2x-1}}\)= \(\sqrt{2}\)
2/ Tính \(\sqrt{24-x^2}+\sqrt{8-x^2}\) biết \(\sqrt{24-x^2}-\sqrt{8-x^2}\)=2
help me!!!
\(\sqrt{24+8\sqrt{9-x^2}}=x+2\sqrt{3-x}+4\)
tìm x biết
\(\sqrt{24+8\sqrt{9-x^2}}=x+2\sqrt{3-x}+4\)
Tính tích:
\(A=\frac{3}{\sqrt[2]{2}}x\frac{8}{\sqrt[2]{3}}x\frac{15}{\sqrt[2]{4}}x\frac{24}{\sqrt[2]{5}}x....x\frac{899}{\sqrt[2]{30}}\)
Giải phương trình:
\(\sqrt{24+8\sqrt{9-x^2}}=x+2\sqrt{3-x}+4\)
\(\sqrt{24+8\sqrt{9-x^2}}=x+2\sqrt{3-x}+4\) \(\left(Đk:-3\le x\le3\right)\)
\(\sqrt{4\left(x+3\right)+8\sqrt{9-x^2}+4\left(3-x\right)}=x+2\sqrt{3-x}+4\)
\(\sqrt{\left(2\sqrt{x+3}+2\sqrt{3-x}\right)^2}=x+2\sqrt{3-x}+4\)
\(2\sqrt{x+3}+2\sqrt{3-x}=x+2\sqrt{3-x}+4\)
\(2\sqrt{x+3}=x+4\)
\(4\left(x+3\right)=x^2+8x+14\)
\(x^2+4x+2=0\)
\(\Delta=16-8=8\)
\(\Delta>0\)=> phương trình có 2 nghiệm phân biệt
\(\left[{}\begin{matrix}x=\dfrac{-4+2\sqrt{2}}{2}=-2+\sqrt{2}\\x=\dfrac{-4-2\sqrt{2}}{2}=-2-\sqrt{2}\end{matrix}\right.\)