*****~~~~~~~~~~*****

 \(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{6+\sqrt{6}}{\sqrt{6}+1}\)

\(=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{\sqrt{6}\left(\sqrt{6}+1\right)}{\sqrt{6}+1}\)

\(=\sqrt{3}+\sqrt{6}\)

\(=\sqrt{3}\left(1+\sqrt{2}\right)\)

*****~~~~~~~~~~*****

\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\)

\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)

\(=\sqrt{3}+2+\sqrt{2}\)

(Chúc bạn học tốt nha!)

\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\frac{1}{2-\sqrt{3}}\)

\(=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\frac{\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)

\(=\sqrt{3}-2-\sqrt{2}=-2\)

dòng cuối là \(\sqrt{3}-2-\sqrt{3}=-2\)nhá

         Bài làm :

Ta có :

\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\frac{1}{2-\sqrt{3}}\)

\(=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{4}\right)}{\sqrt{5}-2}-\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)

\(=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\frac{2+\sqrt{3}}{4-3}\)

\(=\sqrt{3}-\left(2+\sqrt{3}\right)\)

\(=\sqrt{3}-2-\sqrt{3}\)

\(=-2\)

a)

\(\left(3-\sqrt{15}\right)\sqrt{4+\sqrt{15}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{5+2\sqrt{15}+3}}{\sqrt{2}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}{\sqrt{2}}\\ =\left(\sqrt{9}-\sqrt{15}\right)\cdot\dfrac{\left|\sqrt{5}+\sqrt{3}\right|}{\sqrt{2}}\)

\(=\sqrt{3}\left(\sqrt{3}-\sqrt{5}\right)\cdot\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\) (vì \(\sqrt{5}+\sqrt{3}>0\))

\(=\sqrt{3}\cdot\dfrac{3-5}{\sqrt{2}}\\ =\sqrt{3}\cdot\dfrac{-2}{\sqrt{2}}\\ =\sqrt{3}\cdot\dfrac{-\sqrt{4}}{\sqrt{2}}\\ =-\sqrt{6}\)

b)

\(\sqrt{29-12\sqrt{5}}-\sqrt{24-8\sqrt{5}}\\ =\sqrt{20-2\cdot3\cdot2\sqrt{5}+9}-\sqrt{20-2\cdot2\cdot2\sqrt{5}+4}\\ =\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(2\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}-3\right|-\left|2\sqrt{5}-2\right|\)

\(=2\sqrt{5}-3-\left(2\sqrt{5}-2\right)\) (vì \(2\sqrt{5}-3>0;2\sqrt{5}-2>0\))

\(=2\sqrt{5}-3-2\sqrt{5}+2\\ =-1\)

a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)

\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)

\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)

\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)

\(=-\left(2-5\right)\)

\(=-\left(-3\right)\)

\(=3\)

b) Ta có:

\(x^2-x\sqrt{3}+1\) 

\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)

\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)

Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên

\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)

Dấu "=" xảy ra:

\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)

Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)

a)

\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)

\(\sqrt{\frac{3\sqrt{5}+1}{2\sqrt{5}-3}}\left(\sqrt{10}-\sqrt{2}\right)\)

\(=\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}\)

?? :v

\(\sqrt{\frac{3\sqrt{5}+1}{2\sqrt{5}-3}}\left(\sqrt{10}-\sqrt{2}\right)\)

\(=\sqrt{\frac{\left(3\sqrt{5}+1\right)\left(2\sqrt{5}+3\right)}{\left(2\sqrt{5}-3\right)\left(2\sqrt{5}+3\right)}}.\left(\sqrt{10}-\sqrt{2}\right)\)

\(=\sqrt{3+\sqrt{5}}.\sqrt{2}.\left(\sqrt{5}-1\right)\)

\(=\sqrt{6+2\sqrt{5}}.\left(\sqrt{5}-1\right)\)

\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=4\)

\(\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}+\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}-\sqrt{5}}\)

\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)^2}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}+\frac{\left(\sqrt{6}+\sqrt{5}\right)^2}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)

\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)^2+\left(\sqrt{6}+\sqrt{5}\right)^2}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)

\(=\frac{11-2\sqrt{30}+11+2\sqrt{30}}{\left(\sqrt{6}\right)^2-\left(\sqrt{5}\right)^2}\)

\(=\frac{22}{1}=22\)

\(\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}+\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}-\sqrt{5}}\)

\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)+\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}\)

\(=\frac{\left(\sqrt{6}-\sqrt{5}\right)^2+\left(\sqrt{6}+\sqrt{5}\right)^2}{\sqrt{6}^2+\sqrt{5}^2}\)

\(=\sqrt{6}^2-2\sqrt{6}.\sqrt{5}+\sqrt{5}^2+\sqrt{6}^2+2\sqrt{6}.\sqrt{5}+\sqrt{5}^2\)

\(=6+5+6+5=22\)

Sai chỗ nào vậy mấy bạn ? Chỉ giúp mk với ! Ko chỉ mà k sai hơi kì hen!

\(a,=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}}{4}=\dfrac{\sqrt{5}}{2}\\ b,=\sqrt{\left(3-\sqrt{5}\right)^2}+\left|2-\sqrt{5}\right|=3-\sqrt{5}+\sqrt{5}-2=1\\ c,=\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}-\dfrac{-\sqrt{3}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}=\sqrt{5}-\sqrt{3}+\sqrt{3}=\sqrt{5}\)

Bạn tham khảo lời giải tại đây:

Câu hỏi của khanhhuyen6a5 - Toán lớp 9 | Học trực tuyến

\(a,=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\\ b,=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\\ c,=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)