so sánh \(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2}+\sqrt{2}}.\sqrt{2-\sqrt{2}+\sqrt{2}}\)với \(\sqrt{3}\)
Cảm ơn
so sánh\(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)và\(\sqrt{3}\)
rút gọn
1/\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
2.\(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
so sánh 1/ \(\sqrt{7}-\sqrt{6}với\sqrt{3}-\sqrt{2}\)
2/\(\sqrt{2005}+\sqrt{2007}với2\sqrt{2006}\)
3/\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}với\sqrt{2}+1\)
giải giúp mình đi mai là mình đi học rồi
Ta có:
bla bla ........
vậy đáp số là... quên mất rồi
So sánh 2 số: \(R=\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(S=\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
Ta có:
\(R=\)\(\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\)\(\dfrac{\sqrt{10}+3\sqrt{2}}{5+\sqrt{5}}+\dfrac{\sqrt{10}-3\sqrt{2}}{5-\sqrt{5}}\)
\(=\dfrac{4\sqrt{2}}{\sqrt{5}\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\dfrac{4\sqrt{2}}{4\sqrt{5}}=\sqrt{\dfrac{2}{5}}\)
Làm câu S tương tự như này rồi đối chiếu kết quả nha
So sánh 2 số: \(R=\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(S=\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
Giúp mình gấp với ạ. Cảm ơn trước
Rút gọn biểu thức \(P=(12-6\sqrt{3})\sqrt{\dfrac{3}{14-8\sqrt{3}}} -3\sqrt{2(1-\sqrt{-2\sqrt{3}+4})+2\sqrt{{4}+2\sqrt{3}}}\)
\(P=\left(12-6\sqrt{3}\right)\sqrt{\dfrac{3}{14-8\sqrt{3}}}-3\sqrt{2\left(1-\sqrt{-2\sqrt{3}+4}\right)+2\sqrt{4+2\sqrt{3}}}=\left(12-6\sqrt{3}\right)\sqrt{\dfrac{3\left(14+8\sqrt{3}\right)}{\left(14-8\sqrt{3}\right)\left(14+8\sqrt{3}\right)}}-3\sqrt{2\left(1-\sqrt{3-2\sqrt{3}+1}\right)+2\sqrt{3+2\sqrt{3}+1}}\)\(=\left(12-6\sqrt{3}\right)\sqrt{\dfrac{42+24\sqrt{3}}{4}}-3\sqrt{2\left(1-\sqrt{\left(\sqrt{3}-1\right)^2}\right)+2\sqrt{\left(\sqrt{3}+1\right)^2}}=\dfrac{6\sqrt{\left(2-\sqrt{3}\right)^2\left(42+24\sqrt{3}\right)}}{2}-3\sqrt{2\left(1-\sqrt{3}+1\right)+2\left(\sqrt{3}+1\right)}=3\sqrt{\left(7-4\sqrt{3}\right)\left(42+24\sqrt{3}\right)}-3\sqrt{2-2\sqrt{3}+2+2\sqrt{3}+2}=3\sqrt{294+168\sqrt{3}-168\sqrt{3}-288}-3\sqrt{6}=3\sqrt{6}-3\sqrt{6}=0\)
Rút gọn các biểu thức sau:
a) A=\(\sqrt{9x^4+6x^2-1}-\sqrt{9x^4}\)
b) B=\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
c) C=\(\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}+\sqrt{13-4\sqrt{3}}-\sqrt{22+12\sqrt{2}}\)
d) D=\(\sqrt{17-6\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
Giải nhanh hộ mình với
Cảm ơn trước nhé
câu đầu bạn xem lại đề đi nha
các phần còn lại
b)B=\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)=\(\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)=\(\sqrt{7}-1-\left(\sqrt{7}+1\right)=-2\)
c)tính từng căn nha
\(\sqrt{13-4\sqrt{3}}=\sqrt{12-2\sqrt{12}+1}=\sqrt{\left(\sqrt{12}-1\right)^2}=\sqrt{12}-1=2\sqrt{3}-1\)
\(\sqrt{22-12\sqrt{2}}=\sqrt{18-4\sqrt{18}+4}=\sqrt{\left(\sqrt{18}-2\right)^2}=\sqrt{18}-2=3\sqrt{2}-3\)
\(\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}=3\sqrt{2}-2\sqrt{3}\)
thay vào tính C đc C=2
d)có \(\sqrt{9+4\sqrt{2}}=\sqrt{8+2\sqrt{8}+1}=\sqrt{\left(\sqrt{8}+1\right)^2}=\sqrt{8}+1\)\(\Rightarrow6\sqrt{2+\sqrt{9+4\sqrt{2}}}=6\sqrt{2+\sqrt{8}+1}=6\sqrt{2+2\sqrt{2}+1}\)
=\(6\sqrt{\left(\sqrt{2}+1\right)^2}=6\left(\sqrt{2}+1\right)=6\sqrt{2}+6\)\(\Rightarrow D=\sqrt{17-6\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{17-6\sqrt{2}-6}=\sqrt{11-6\sqrt{2}}=\sqrt{9-6\sqrt{2}+2}\)
=\(\sqrt{\left(3-\sqrt{2}\right)^2}=3-\sqrt{2}\)
Tính:
\(\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{6}}+\frac{1}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{8}}+\frac{1}{\sqrt{8}-\sqrt{9}}\)
Giúp mình với nhé. Cảm ơn
1)so sánh 2 số sau M=\(\sqrt{18}-\sqrt{8}\) và N=\(\dfrac{5+\sqrt{5}}{\sqrt{5}+1}-\sqrt{6-2\sqrt{5}}\)
2)cho biểu thức A=\((\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}):(\dfrac{x-4}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}})\) với x>0,\(x\ne4\),\(x\ne9\)
câu 2 rút gọn A và tìm các giá trị nguyên của x để A nhận giá trị âm
1) So sánh:
N = \(\dfrac{5+\sqrt{5}}{\sqrt{5}+1}-\sqrt{6-2\sqrt{5}}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-\left(\sqrt{5}-1\right)=1\)
M = \(\sqrt{18}-\sqrt{8}\)
\(=3\sqrt{2}-2\sqrt{2}\)
\(=\sqrt{2}\)
Ta có: \(1=\sqrt{1}\)
Mà 1 < 2
\(\Rightarrow\sqrt{1}< \sqrt{2}\)
Hay 1 \(< \sqrt{2}\)
Vậy N < M
2) Với \(x>0;x\ne4;x\ne9\), ta có:
A = \(\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{x-4}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left[\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)
\(=\dfrac{x-3\sqrt{x}-2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{x-4-2\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x-3}\right)}\)
\(=\dfrac{-x-3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-2\sqrt{x}+2}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-2\sqrt{x}+2}\)
\(=\dfrac{-x}{x-2\sqrt{x}+2}\)
Giúp mình với mình cảm ơn trước!!! Rút gọn biểu thức nha:
\(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt[]{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\\ =\sqrt{2\left(2+\sqrt{2}\right)}.\sqrt{\left(2+\sqrt{2+\sqrt{2}}\right)\left(2-\sqrt{2+\sqrt{2}}\right)}\\ =\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{4-\left(2+\sqrt{2}\right)}\\ =\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\\ =\sqrt{2}.\sqrt{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}\\ =\sqrt{2}.\sqrt{4-2}\\ =2\)