Tim x=\(\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+.....}}}}\)
Moi nguoi thu giai bai nay di
Ai nhanh mik cho 3 tick
\(\frac{2\sqrt{2}}{\sqrt{x+1}}+\sqrt{x}=\sqrt{x+9}\)
moi nguoi lam nhanh bai nay ho minh
(=) \(2\sqrt{2}+\sqrt{x.\left(x+1\right)}=\sqrt{\left(x+1\right)\left(x+9\right)}\)(nhân cả 2 vế cho \(\sqrt{x+1}\) ).
(=) \(8+4\sqrt{2x\left(x+1\right)}+x\left(x+1\right)=\left(x+1\right)\left(x+9\right)\) \(\Leftrightarrow4\sqrt{2x^2+2x}=x^2+10x+9-x^2-x-8\)
(=) \(4\sqrt{2x^2+2x}=9x+1\) (=) \(16\left(2x^2+2x\right)=81x^2+18x+1\)(=) \(0=49x^2-14x+1\)
(=)\(\left(7x-1\right)^2=0\) (=) \(x=\frac{1}{7}\)
\(\frac{2\sqrt{2}}{\sqrt{x+1}}+\sqrt{x}=\sqrt{x+9}\)
moi nguoi lam nhanh bai nay ho minh
tim \(x=\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+....}}}}}\)
\(x=\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+....}}}}}\)
\(\Rightarrow x^2=5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+...}}}}\)
\(\Rightarrow x^4=25+10\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+....}}}}+13+\sqrt{5+\sqrt{13+\sqrt{5+...}}}\)
\(\Leftrightarrow x^4=38+10x^2+x\)
\(\Leftrightarrow x^4-10x^2-x-38=0\)
giải ra tìm x xong
Thực Hiện Phép Tính Sau :
\(\text{a)}\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(\text{b)}\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)
\(\text{c)}\sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\)
Ai giúp mk vs. Mk tick cho
\(a,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(\sqrt{20}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)
\(=\sqrt{1}=1\)
b,c
\(\sqrt{13+4\sqrt{3}}=\sqrt{13+2\sqrt{12}}=\sqrt{12}+1=2\sqrt{3}+1\)
=>BT=\(\sqrt{5-\left(2\sqrt{3}+1\right)}+\sqrt{3+\left(2\sqrt{3}+1\right)}\)
\(=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
c,\(=\sqrt{1+\sqrt{3+2\sqrt{3}+1}}+\sqrt{1-\sqrt{3-\left(2\sqrt{3}-1\right)}}\)
\(=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
tìm y :
\(y=\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+...}}}}}\)
bạn nào giúp mình , mình tick cho
Nhận thấy: \(y>2\)
Ta xét:
\(y^2=5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+...}}}}\)
\(\Rightarrow\) \(\left(y^2-5\right)^2=13+y\)
\(\Leftrightarrow\) \(y^4-10y^2-y+12=0\)
\(\Leftrightarrow\) \(\left(y^4-9y^2\right)-\left(y^2-9\right)-\left(y-3\right)=0\)
\(\Leftrightarrow\) \(\left[\left(y+3\right)\left(y+1\right)\left(y-1\right)-1\right]\left(y-3\right)=0\)
Mà \(y>2\) nên \(\left[\left(y+3\right)\left(y+1\right)\left(y-1\right)-1\right]>0\)
Do đó, ta dễ dàng suy ra \(y-3=0\) hay \(y=3\)
Giai phương trình \(2\sqrt{x-1}+3\sqrt{5-x}=2\sqrt{13}\)
tìm y :
\(y=\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+...}}}}}\)
các bạn giúp mình nhé mình tick cho
Tìm x biết x = \(\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+...}}}}}}\)
\(x=\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+...}}}}}}\)
\(\Leftrightarrow x=\sqrt{5+\sqrt{13+x}}\) (\(x\ge0\))
\(\Leftrightarrow x^2=5+\sqrt{13+x}\)
\(\Leftrightarrow x^2-9=\sqrt{13+x}-4\)
\(\Leftrightarrow\left(x-3\right).\left(x+3\right)=\dfrac{x-3}{\sqrt{13+x}+4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=\dfrac{1}{\sqrt{x+13}+4}\left(∗\right)\end{matrix}\right.\)
Xét (*) ta có VT \(\ge3\) (1)
mà \(VP=\dfrac{1}{\sqrt{x+13}+4}\le\dfrac{1}{4}\) (2)
Từ (1) và (2) dễ thấy (*) vô nghiệm
Hay x = 3
Rút gọn các biểu thức sau
a) \(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
b) \(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}}\)
Giúp mình nhanh nha, xong mình tick cho :v
Cho sửa phần mẫu số của câu trên thành \(\sqrt{6}+\sqrt{2}\)
\(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-|2\sqrt{3}+1|}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{4+2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+|\sqrt{3}-1|}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{\sqrt{2}.\sqrt{4+2\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{3}+1}{\sqrt{3}+1}=1\)