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\(A=\sqrt{14+6\sqrt{5}}+\sqrt{14-6\sqrt{5}}\)
\(A=\sqrt{9+6\sqrt{5}+5}+\sqrt{9-6\sqrt{5}+5}\)
\(A=\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(A=3+\sqrt{5}+3-\sqrt{5}=6\)
b) \(B=\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(B=\sqrt{3-4\sqrt{3}+4}-\sqrt{3+4\sqrt{3}+4}\)
\(B=\sqrt{\left(\sqrt{3}-2\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(B=2-\sqrt{3}-\sqrt{3}-2=-2\sqrt{3}\)
Câu a tách 14 thành 5+9 . Có hằng đẳng thức
Câu b tương tự tách 7 thành 4+ 3 nhé
a) \(\sqrt{14-6\sqrt{5}}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
b, c) tương tự câu a.
d) \(\left(3-\sqrt{2}\right)\sqrt{11+6\sqrt{2}}\)
\(=\left(3-\sqrt{2}\right)\sqrt{\left(3+\sqrt{2}\right)^2}\)
\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)
\(=9-2\)
\(=7\)
e) \(\sqrt{11-6\sqrt{2}+\sqrt{3-2\sqrt{2}}}\)
\(=\sqrt{11-6\sqrt{2}+\sqrt{\left(1-\sqrt{2}\right)^2}}\)
\(=\sqrt{11-6\sqrt{2}+\sqrt{2}-1}\)
\(=\sqrt{10-5\sqrt{2}}\)
Ta có :
a)\(\left(2\sqrt{5}-\sqrt{7}\right)\left(2\sqrt{5}-\sqrt{7}\right)=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)
b)\(\left(5\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{3}-5\sqrt{2}\right)=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
c)\(\sqrt{9+4\sqrt{5}}=\sqrt{2^2+2.2.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2+\sqrt{5}\right|=2+\sqrt{5}\)
Bài 1:
\(A=\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\sqrt{3}+2\sqrt{7}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}\)
\(B=\dfrac{9\sqrt{3}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}=\dfrac{9\sqrt{3}+9\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
\(=\dfrac{18\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
\(C=\sqrt{5-2\sqrt{6}}=\sqrt{3-2\sqrt{6}+2}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)
Bài 2:
\(\left(\sqrt{12}+3\sqrt{15}+4\sqrt{135}\right)\sqrt{3}\)
\(=6+9\sqrt{5}+36\sqrt{5}\)
\(=6+45\sqrt{5}\)
g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
1/\(\sqrt{8-2\sqrt{15}}-\sqrt{21-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(2\sqrt{5}-1\right)^2}\)
Bạn tự làm tiếp
2/ \(\frac{4}{\sqrt{7-4\sqrt{3}}}-\frac{4}{7-4\sqrt{3}}=\frac{4}{\sqrt{\left(2-\sqrt{3}\right)^2}}-\frac{4}{\left(2-\sqrt{3}\right)^2}=\frac{4}{2-\sqrt{3}}-\frac{4}{\left(2-\sqrt{3}\right)^2}\)
\(=\frac{8-4\sqrt{3}-4}{\left(2-\sqrt{3}\right)^2}=\frac{4-4\sqrt{3}}{\left(2-\sqrt{3}\right)^2}\) đến đây ko rút gọn được nữa, nghi bạn chép sai đề.
Tử số của phân số thứ hai là 4 hay 1 vậy?
3/ \(\frac{\sqrt{8+2\sqrt{15}}-\sqrt{4-2\sqrt{3}}}{\sqrt{6-2\sqrt{5}}}=\frac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{\left(\sqrt{5}-1\right)^2}}=\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{3+\sqrt{5}}{2}\)
4/ \(\frac{10}{\sqrt{\left(\sqrt{5}-2\right)^2}}-\frac{12}{\sqrt{\left(3+\sqrt{5}\right)^2}}+\frac{20}{\sqrt{\left(\sqrt{5}-1\right)^2}}=\frac{10}{\sqrt{5}-2}-\frac{12}{3+\sqrt{5}}+\frac{20}{\sqrt{5}-1}\)
\(=\frac{10\left(\sqrt{5}+2\right)}{1}-\frac{12\left(3-\sqrt{5}\right)}{4}+\frac{20\left(\sqrt{5}+1\right)}{4}=16+18\sqrt{5}\)
\(\frac{10}{\sqrt{5}-2.\sqrt{5}.2+4}-\frac{12}{\sqrt{\sqrt{5}+2.\sqrt{5}.3+9}}+\frac{20}{\sqrt{5-2.\sqrt{5}.1+1}}=\frac{10}{\left(\sqrt{5}-2\right)^2}-\frac{12}{\sqrt{\left(\sqrt{5}+3\right)^2}}+\frac{20}{\sqrt{\left(\sqrt{5}-1\right)^2}}=\frac{10}{\sqrt{5}-2}-\frac{12}{\sqrt{5}+3}+\frac{20}{\sqrt{5}-1}=\frac{10\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right).\left(\sqrt{5}+2\right)}-\frac{12.\left(\sqrt{5}-3\right)}{\left(\sqrt{5}+3\right).\sqrt{5}-3\left(\right)}+\frac{20.\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\frac{10\sqrt{5}-20}{5-4}-\frac{12\sqrt{5}-36}{5-9}+\frac{20\sqrt{5}+20}{5-1}\\=\frac{40\sqrt{5}-80+12\sqrt{5}+36+20\sqrt{5}+20}{4}=\\ 18\sqrt{5}-6\)
xin chào em mới học dưới lớp tám thôi khó quá không biết làm
a) \(A=\sqrt{10+\sqrt{99}}=\sqrt{10+3\sqrt{11}}=\frac{1}{\sqrt{2}}.\sqrt{20+6\sqrt{11}}\)
\(=\frac{1}{\sqrt{2}}.\sqrt{\left(3+\sqrt{11}\right)^2}=\frac{3+\sqrt{11}}{2}\)
b) \(B=\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(=3\sqrt{2}+\sqrt{3}-3\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
c) bn ktra lại đề
d) ĐK: \(x\ge0\)
\(\sqrt{x+1+2\sqrt{x}}=\sqrt{\left(\sqrt{x}+1\right)^2}=\sqrt{x}+1\)
e) đk: \(x\ge-1\)
\(\sqrt{2x+3+2\sqrt{x^2+3x+2}}=\sqrt{x+1+2\sqrt{\left(x+1\right)\left(x+2\right)}+x+2}\)
\(=\sqrt{\left(\sqrt{x+1}+\sqrt{x+2}\right)^2}=\sqrt{x+1}+\sqrt{x+2}\)
Giải:
\(\sqrt{6-2\sqrt{5}}+\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{5-2\sqrt{5}+1}+\sqrt{9-6\sqrt{5}+5}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(=\left|\sqrt{5}-1\right|+\left|3-\sqrt{5}\right|\)
\(=\sqrt{5}-1+3-\sqrt{5}\)
\(=2\)
thanks nhìu ạ