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~~~~~a)~~~~~
\(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\right)^2}\)
\(=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}-\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}\)
\(=2.\sqrt{\frac{1}{2}}=\sqrt{2}\)
*****b)*****
(Hình như đề có cái gì đó sai sai hả bạn?)
~~~~~c)~~~~~
\(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\)
\(=\left(3\sqrt{2}-2\sqrt{6}+\sqrt{6}-2\sqrt{2}\right)\sqrt{\left(\sqrt{\frac{1}{2}}+\sqrt{\frac{3}{2}}\right)^2}\)
\(=\left(\sqrt{2}-\sqrt{6}\right).\left(\sqrt{\frac{1}{2}}+\sqrt{\frac{3}{2}}\right)\)
\(=1+\sqrt{3}-\sqrt{3}-3\)
\(=-2\)
*****d)*****
\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(2\sqrt{2}+3\sqrt{5}\right)^2}\)
\(=2\sqrt{2}-\sqrt{5}-2\sqrt{2}-3\sqrt{5}\)
\(=-4\sqrt{5}\)
(Chúc bạn học tốt và tíck cho mìk vs nhé ~~~~~bạn xem lại câu b hộ mình luôn nha~~~~~!)
a) \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-2\sqrt{40}}-\sqrt{53+12\sqrt{10}}\)
\(=\sqrt{\left(\sqrt{8}\right)^2-2.\sqrt{8}.\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}\right)^2+2.3\sqrt{5}.2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{8}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)
\(=\left|\sqrt{8}-\sqrt{5}\right|-\left|3\sqrt{5}+2\sqrt{2}\right|\)
= √8 - √5 - 3√5 - 2√2 = -4√5
b) (1+√3-√2).(1+√3+√2)= [(1+√3)^2-(√2)^2] = 4+2√3-2=2+2√3
a) =sprt{13-=sprt{160}} - =sprt{53+4=sprt{90}}
= =sprt{(=sprt{8} - =sprt{5})2 } - =sprt{(=sprt{45} + =sprt{8})2 }
= =sprt{8} - =sprt{5} - =sprt{45} - =sprt{8}
= -3=sprt{5}
b) ( 1 + =sprt{3} - =sprt{2} )( 1+ =sprt{3} + =sprt{2} )
= ( 1 + =sprt{3} )2 - (=sprt{2})2
= 4 + 2=sprt{3} - 2
=2 + 2=sprt{3}
rút gọn dk \(\sqrt{a+1}+\sqrt{a-1}+\sqrt{a\left(a+1\right)}\)
ta có \(\frac{53}{9-2\sqrt{7}}=9+\sqrt{7}\)( cứ lm bt theo cách trục căn thức)
rồi thay vào
\(A=\frac{1}{\sqrt{x-1}-\sqrt{x}}+\frac{1}{\sqrt{x-1}+\sqrt{x}}+\frac{\sqrt{x^3}-x}{\sqrt{x}-1}\) \(ĐKXĐ:x\ne\pm1\)
\(=\frac{\sqrt{x-1}+\sqrt{x}+\sqrt{x-1}-\sqrt{x}}{\left(\sqrt{x-1}-\sqrt{x}\right)\left(\sqrt{x-1}+\sqrt{x}\right)}+\frac{x\sqrt{x}-x}{\sqrt{x}-1}\)
\(=\frac{2\sqrt{x-1}}{x-1-x}+\frac{x\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=x-2\sqrt{x-1}\)
Câu c mình ko làm được
1) Cách 1 :
\(M=\sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}\)
\(M=\sqrt{9-6\sqrt{2}+2}+\sqrt{9+6\sqrt{2}+2}\)
\(M=\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(3+\sqrt{2}\right)^2}\)
\(M=\left|3-\sqrt{2}\right|+\left|3+\sqrt{2}\right|\)
\(M=3-\sqrt{2}+3+\sqrt{2}=6\)
Cách 2 :
\(M=\sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}\)
\(\Rightarrow M^2=11-6\sqrt{2}+2\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}+11+6\sqrt{2}\)
\(\Leftrightarrow M^2=22+2.7=36\)
\(\Leftrightarrow M=6\left(\sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}>0\right)\)
2)
\(A=53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}\)
\(\Leftrightarrow A=53-20\sqrt{4+\sqrt{8-4\sqrt{2}+1}}\)
\(\Leftrightarrow A=53-20\sqrt{4+\sqrt{\left(2\sqrt{2}-1\right)^2}}\)
\(\Leftrightarrow A=53-20\sqrt{4+\left|2\sqrt{2}-1\right|}\)
\(\Leftrightarrow A=53-20\sqrt{4+2\sqrt{2}-1}\)
\(\Leftrightarrow A=53-20\sqrt{3+2\sqrt{2}}\)
\(\Leftrightarrow A=53-20\sqrt{2+2\sqrt{2}+1}\)
\(\Leftrightarrow A=53-20\left(\sqrt{2}+1\right)\)
\(\Leftrightarrow A=53-20\sqrt{2}-20=33-20\sqrt{2}\)
3)
\(M=\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(M=\sqrt{3-\sqrt{5}}.\left(3\sqrt{10}-3\sqrt{2}+5\sqrt{2}-\sqrt{10}\right)\)
\(M=\sqrt{3-\sqrt{5}}\left(2\sqrt{10}+2\sqrt{2}\right)\)
\(M=2\sqrt{2}.\sqrt{3-\sqrt{5}}\left(\sqrt{5}+1\right)\)
\(\Rightarrow M^2=8.\left(3-\sqrt{5}\right).\left(5+2\sqrt{5}+1\right)\)
\(\Leftrightarrow M^2=\left(24-8\sqrt{5}\right)\left(6+2\sqrt{5}\right)\)
\(\Leftrightarrow M^2=144+48\sqrt{5}-48\sqrt{5}-80\)
\(\Leftrightarrow M^2=64\Leftrightarrow M=8\left(\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right).\left(\sqrt{10}-\sqrt{2}\right)>0\right)\)
giải giúp mình bài này ới ạ mình đng cần gấp
Cho biểu thức
c=(căng x-2/căng x+2+căng x+2/căng x-2)nhân căng x+2/2 - 4 căng x/căng x-2
a)
\(P=\frac{\sqrt{a}}{\sqrt{a}+3}+\frac{2\sqrt{a}}{\sqrt{a}-3}-\frac{3a+9}{a-9}\)
\(P=\frac{\sqrt{a}}{\sqrt{a}+3}+\frac{2\sqrt{a}}{\sqrt{a}-3}-\frac{3a+9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}+\frac{\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\frac{3a+9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{a-3\sqrt{a}+3+3\sqrt{a}-3a-9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{-2a-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{-2a-3}{a-9}\)
b) Để \(P=\frac{1}{3}\Rightarrow\frac{-2a-3}{a-9}=\frac{1}{3}\)
\(\Rightarrow3\left(-2a-3\right)=a-9\)
\(\Rightarrow-6a-9=a-9\)
\(\Rightarrow-6a-a=-9+9\)
\(\Rightarrow-7a=0\left(L\right)\)
Vậy ko có gt của a để P=1/3 ( mk ko chắc.....)
