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\(A=\dfrac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}=\dfrac{\left(\sqrt{x+3}\right)^2+2\sqrt{\left(x-3\right)\left(x+3\right)}}{2\left(\sqrt{x-3}\right)^2+\sqrt{\left(x-3\right)\left(x+3\right)}}\)
\(A=\dfrac{\sqrt{x+3}.\left(\sqrt{x+3}+2\sqrt{x-3}\right)}{\sqrt{x-3}.\left(2\sqrt{x-3}+\sqrt{x+3}\right)}=\dfrac{\sqrt{x+3}}{\sqrt{x-3}}\)
Không pít mk đang nghĩ đúng hay sai nhưng mà mk thấy đề bài với cái bước 1 k = nhau vì khi tìm đkxđ x\(\le\)-3 và x\(\ge3\) vì vậy nên nếu nhỏ hơn -3 thì x-3\(\le-6\) mà x-3\(\ge\)0 nên nhìn thật sự thấy mâu thuẫn lắm . TÓm lại khi mk bấm máy tính cái đề với x =-4 thì đc \(\dfrac{-\sqrt{7}}{7}\) và bấm máy tính cái bứoc 1 thì không ra .
p/s Mau bạn giải đáp giùm mk ~!
\(\frac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}=\frac{\left(\sqrt{x+3}\right)^2+2\sqrt{x+3}\sqrt{x-3}}{2.\left(\sqrt{x-3}\right)^2+\sqrt{x+3}\sqrt{x-3}}\)
\(=\frac{\sqrt{x+3}\left(\sqrt{x+3}+2\sqrt{x-3}\right)}{\sqrt{x-3}\left(2\sqrt{x-3}+\sqrt{x+3}\right)}=\frac{\sqrt{x+3}}{\sqrt{x-3}}\)
\(=\frac{\sqrt{x^2-9}}{x-3}\)
Với x<-3 ta có:
\(x+3+2\sqrt{x^2-9}=\sqrt{-\left(x+3\right)}.\sqrt{-\left(x+3\right)}+2\sqrt{-\left(x+3\right)}.\sqrt{3-x}\)
\(=\sqrt{-\left(x+3\right)}.\left(\sqrt{-\left(x+3\right)}+2\sqrt{3-x}\right)\)
\(6-2x+\sqrt{x^2-9}=\sqrt{3-x}\left(2\sqrt{3-x}+\sqrt{-\left(x+3\right)}\right)\)
Từ đó suy ra \(M=\frac{\sqrt{-\left(x+3\right)}}{\sqrt{3-x}}hayM=\sqrt{\frac{\left(x+3\right)}{\left(x-3\right)}}\)
Lời giải:
a)
\(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}+\frac{\sqrt{3}}{3}-\frac{2(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}\)
\(=\frac{2-\sqrt{3}}{4-3}+\frac{\sqrt{3}}{3}-\frac{2(3-\sqrt{3})}{3^2-3}\)
\(=2-\sqrt{3}+\frac{\sqrt{3}}{3}-\frac{3-\sqrt{3}}{3}=\frac{6-3\sqrt{3}}{3}+\frac{2\sqrt{3}-3}{3}=\frac{3-\sqrt{3}}{3}\)
b)
\(\sqrt{x-3+2\sqrt{x-4}}=\sqrt{(x-4)+2\sqrt{x-4}+1}=\sqrt{(\sqrt{x-4}+1)^2}=|\sqrt{x-4}+1|=\sqrt{x-4}+1\)
c)
\(\sqrt{2x+4\sqrt{2x-4}}=\sqrt{(2x-4)+2.2\sqrt{2x-4}+2^2}\)
\(=\sqrt{(\sqrt{2x-4}+2)^2}=|\sqrt{2x-4}+2|=\sqrt{2x-4}+2\)
d)
\(\sqrt{x-\sqrt{2x-1}}=\frac{1}{\sqrt{2}}\sqrt{2x-2\sqrt{2x-1}}=\frac{1}{\sqrt{2}}\sqrt{(2x-1)-2\sqrt{2x-1}+1}\)
\(=\frac{1}{\sqrt{2}}\sqrt{(\sqrt{2x-1}-1)^2}=\frac{|\sqrt{2x-1}-1|}{\sqrt{2}}\)
e)
\(\sqrt{x+6\sqrt{x-9}}-\sqrt{x-9}=\sqrt{(x-9)+2.3\sqrt{x-9}+3^2}-\sqrt{x-9}\)
\(=\sqrt{(\sqrt{x-9}+3)^2}-\sqrt{x-9}=|\sqrt{x-9}+3|-\sqrt{x-9}\)
\(=\sqrt{x-9}+3-\sqrt{x-9}=3\)
Lời giải:
a)
\(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}+\frac{\sqrt{3}}{3}-\frac{2(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}\)
\(=\frac{2-\sqrt{3}}{4-3}+\frac{\sqrt{3}}{3}-\frac{2(3-\sqrt{3})}{3^2-3}\)
\(=2-\sqrt{3}+\frac{\sqrt{3}}{3}-\frac{3-\sqrt{3}}{3}=\frac{6-3\sqrt{3}}{3}+\frac{2\sqrt{3}-3}{3}=\frac{3-\sqrt{3}}{3}\)
b)
\(\sqrt{x-3+2\sqrt{x-4}}=\sqrt{(x-4)+2\sqrt{x-4}+1}=\sqrt{(\sqrt{x-4}+1)^2}=|\sqrt{x-4}+1|=\sqrt{x-4}+1\)
c)
\(\sqrt{2x+4\sqrt{2x-4}}=\sqrt{(2x-4)+2.2\sqrt{2x-4}+2^2}\)
\(=\sqrt{(\sqrt{2x-4}+2)^2}=|\sqrt{2x-4}+2|=\sqrt{2x-4}+2\)
d)
\(\sqrt{x-\sqrt{2x-1}}=\frac{1}{\sqrt{2}}\sqrt{2x-2\sqrt{2x-1}}=\frac{1}{\sqrt{2}}\sqrt{(2x-1)-2\sqrt{2x-1}+1}\)
\(=\frac{1}{\sqrt{2}}\sqrt{(\sqrt{2x-1}-1)^2}=\frac{|\sqrt{2x-1}-1|}{\sqrt{2}}\)
e)
\(\sqrt{x+6\sqrt{x-9}}-\sqrt{x-9}=\sqrt{(x-9)+2.3\sqrt{x-9}+3^2}-\sqrt{x-9}\)
\(=\sqrt{(\sqrt{x-9}+3)^2}-\sqrt{x-9}=|\sqrt{x-9}+3|-\sqrt{x-9}\)
\(=\sqrt{x-9}+3-\sqrt{x-9}=3\)
Đặt \(a=\sqrt{x+3}\) , \(b=\sqrt{x-3}\).
Ta có : \(A=\frac{\left(x+3\right)+2\sqrt{\left(x-3\right)\left(x+3\right)}}{2\left(x-3\right)+\sqrt{\left(x-3\right)\left(x+3\right)}}=\frac{a^2+2ab}{2b^2+ab}\)
\(=\frac{a^2+2ab}{2b^2+ab}=\frac{a\left(a+2b\right)}{b\left(a+2b\right)}=\frac{a}{b}=\frac{\sqrt{x+3}}{\sqrt{x-3}}\)
a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}=4\)
Thay x = 4 => \(\sqrt{x}=2\) vào B ta được :
\(B=\frac{2+5}{2-3}=-7\)
b, Ta có : Với \(x\ge0;x\ne9\)
\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\)
Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
\(A=\dfrac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}\left(x>3\right)\\ A=\dfrac{\left(x+3\right)+2\sqrt{\left(x-3\right)\left(x+3\right)}}{2\left(x-3\right)+\sqrt{\left(x-3\right)\left(x+3\right)}}\\ A=\dfrac{\sqrt{x+3}\left(\sqrt{x+3}+2\sqrt{x-3}\right)}{\sqrt{x-3}\left(2+\sqrt{x+3}\right)}\)
Tới đây chịu rùi, hình như đề sai đk?
Bạn làm sai rồi đáp số là : \(\dfrac{\sqrt{x^2-9}}{x-3}\)