\(C=\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
D= \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
Những câu hỏi liên quan
\(B=\left(2-\sqrt{3}\right).\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right).\sqrt{26-15\sqrt{3}}\)
\(C=\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3+\sqrt{5}}\)
Thu gọn B= \(21\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)^2-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)^2-15\sqrt{5}\)
Thu gọn A= \(\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
Sửa đề
\(A=\left(2-\sqrt{3}\right)\sqrt[3]{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{26-15\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{8-12\sqrt{3}+18-3\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\left(2+\sqrt{3}\right)\sqrt[3]{\left(2-\sqrt{3}\right)^3}\)
\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=0\)
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Bài 1: Rút gọn biểu thức
a) Asqrt{26+15sqrt{3}}
b) Bsqrt{4+sqrt{7}}-sqrt{4-sqrt{7}}-sqrt{2}
c) Csqrt{8-2sqrt{15}}-sqrt{8+2sqrt{15}}
d) Dleft(sqrt{6}-2right)left(5+sqrt{24}right)sqrt{5-sqrt{24}}
e) Eleft(sqrt{10}-sqrt{2}right)left(sqrt{3+sqrt{5}}right)
f) Fsqrt{sqrt{5}-sqrt{3-sqrt{29-12sqrt{5}}}}
g) Gleft(2-sqrt{3}right)sqrt{26+15sqrt{3}}-left(2+sqrt{3}right)sqrt{26-15sqrt{3}}
h) Hfrac{left(2+sqrt{3}right)sqrt{2-sqrt{3}}}{sqrt{2+sqrt{3}}}
Đọc tiếp
Bài 1: Rút gọn biểu thức
a) \(A=\sqrt{26+15\sqrt{3}}\)
b) \(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
c) \(C=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
d) \(D=\left(\sqrt{6}-2\right)\left(5+\sqrt{24}\right)\sqrt{5-\sqrt{24}}\)
e) \(E=\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{3+\sqrt{5}}\right)\)
f) \(F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
g) \(G=\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
h) \(H=\frac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\)
a)
\(A=\sqrt{26+15\sqrt{3}}=\sqrt{\frac{52+30\sqrt{3}}{2}}=\sqrt{\frac{27+25+2\sqrt{27.25}}{2}}\)
\(=\sqrt{\frac{(\sqrt{27}+\sqrt{25})^2}{2}}=\frac{\sqrt{27}+\sqrt{25}}{\sqrt{2}}=\frac{3\sqrt{3}+5}{\sqrt{2}}=\frac{3\sqrt{6}+5\sqrt{2}}{2}\)
b)
\(B\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)
\(=\sqrt{7+1+2\sqrt{7}}-\sqrt{7+1-2\sqrt{7}}-2\)
\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}-2=\sqrt{7}+1-(\sqrt{7}-1)-2=0\)
\(\Rightarrow B=0\)
c)
\(C=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{3+5-2\sqrt{3.5}}-\sqrt{3+5+2\sqrt{3.5}}\)
\(=\sqrt{(\sqrt{5}-\sqrt{3})^2}-\sqrt{(\sqrt{5}+\sqrt{3})^2}=(\sqrt{5}-\sqrt{3})-(\sqrt{5}+\sqrt{3})=-2\sqrt{3}\)
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d)
\(D=(\sqrt{6}-2)(5+2\sqrt{6})\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(2+3+2\sqrt{2.3})\sqrt{2+3-2\sqrt{2.3}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})^2\sqrt{(\sqrt{3}-\sqrt{2})^2}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})^2(\sqrt{3}+\sqrt{2})^2=\sqrt{2}[(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})]^2\)
\(=\sqrt{2}.1^2=\sqrt{2}\)
e)
\(E=(\sqrt{10}-\sqrt{2})\sqrt{3+\sqrt{5}}=(\sqrt{5}-1).\sqrt{2}.\sqrt{3+\sqrt{5}}\)
\(=(\sqrt{5}-1)\sqrt{6+2\sqrt{5}}=(\sqrt{5}-1)\sqrt{5+1+2\sqrt{5.1}}\)
\(=(\sqrt{5}-1)\sqrt{(\sqrt{5}+1)^2}=(\sqrt{5}-1)(\sqrt{5}+1)=4\)
f)
\(F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-3)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-(\sqrt{20}-3)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}}=\sqrt{\sqrt{5}-(\sqrt{5}-1)}=\sqrt{1}=1\)
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g) Áp dụng kết quả phần a):
\(G=(2-\sqrt{3}).\frac{3\sqrt{6}+5\sqrt{2}}{2}-(2+\sqrt{3}).\frac{3\sqrt{6}-5\sqrt{2}}{2}\)
\(=\frac{\sqrt{6}+\sqrt{2}}{2}-\frac{\sqrt{6}-\sqrt{2}}{2}=\sqrt{2}\)
h)
\(H=\frac{(2+\sqrt{3})\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}=\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\sqrt{\frac{(2-\sqrt{3})^2}{(2+\sqrt{3})(2-\sqrt{3})}}=\sqrt{\frac{(2-\sqrt{3})^2}{2^2-3}}=\sqrt{(2-\sqrt{3})^2}=2-\sqrt{3}\)
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C/m
\(\left(2-\sqrt{3}\right).\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right).\sqrt{26-15\sqrt{3}}=\sqrt{2}\)
1.Tính: a, sqrt{left(5-sqrt{3}right)^2}-sqrt{left(sqrt{3}-2right)^2}
b, Bleft(2-sqrt{3}right).sqrt{26+15sqrt{3}}-left(2+sqrt{3}right).sqrt{26-15sqrt{3}}
c, sqrt{sqrt{5}-sqrt{3-sqrt{29-12sqrt{5}}}}
d, Aleft(sqrt{10}-sqrt{2}right)sqrt{3+sqrt{5}}
2.Giải pt:
a,sqrt{x^2-3x-2}x-2
b,5sqrt{x-1}-sqrt{36x-36}+sqrt{9x-9}sqrt{8x+12}
c,sqrt{x}+sqrt{1-x}+sqrt{xleft(1-xright)}1
Đọc tiếp
1.Tính: a, \(\sqrt{\left(5-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
b, B=\(\left(2-\sqrt{3}\right).\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right).\sqrt{26-15\sqrt{3}}\)
c, \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
d, A=\(\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}\)
2.Giải pt:
a,\(\sqrt{x^2-3x-2}=x-2\)
b,\(5\sqrt{x-1}-\sqrt{36x-36}+\sqrt{9x-9}=\sqrt{8x+12}\)
c,\(\sqrt{x}+\sqrt{1-x}+\sqrt{x\left(1-x\right)}=1\)
Bài 1:
a: \(=\left|5-\sqrt{3}\right|-\left|\sqrt{3}-2\right|\)
\(=5-\sqrt{3}-2+\sqrt{3}=3\)
b; \(B=\dfrac{\left(2-\sqrt{3}\right)\cdot\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\cdot\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\cdot\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)
\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-6\sqrt{3}+10-9+5\sqrt{3}}{\sqrt{2}}\)
\(=\dfrac{20-18}{\sqrt{2}}=\sqrt{2}\)
c: \(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3+3-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}=1\)
d: \(A=\left(\sqrt{5}-1\right)\cdot\sqrt{6+2\sqrt{5}}\)
\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)
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Rút gọn biểu thức
M = \(\dfrac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}\)
N= \(\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
a) Ta có: \(M=\dfrac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}\)
\(=\dfrac{2\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}-2\sqrt{7}+3\sqrt{6}\)
\(=2\sqrt{7}+2\sqrt{6}-2\sqrt{7}+3\sqrt{6}\)
\(=5\sqrt{6}\)
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b) Ta có: \(N=\left(2-\sqrt{3}\right)\left(\sqrt{26+15\sqrt{3}}\right)-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{27+2\cdot3\sqrt{3}\cdot5+25}-\left(2+\sqrt{3}\right)\sqrt{27-2\cdot3\sqrt{3}\cdot5+25}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{\left(3\sqrt{3}+5\right)^2}-\left(2+\sqrt{3}\right)\sqrt{\left(3\sqrt{3}-5\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)
\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-\left(6\sqrt{3}-10+9-5\sqrt{3}\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}\)
\(=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
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\(\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}}\)
Ta có: \(\hept{\begin{cases}\left(2-\sqrt{3}\right)^2.\left(26+15\sqrt{3}\right)=2+\sqrt{3}\\\left(2+\sqrt{3}\right)^2.\left(26-15\sqrt{3}\right)=2-\sqrt{3}\end{cases}}\)
Sửa đề:
\(\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right)\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{3}+1-\sqrt{3}+1\right)\)
\(=\sqrt{2}\)
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thực hiện phép tính:a)left(sqrt{8}-3sqrt{2}+sqrt{10}right)left(sqrt{2}-3sqrt{0,4}right)b)left(15sqrt{50}+5sqrt{200}-3sqrt{450}right):sqrt{10}c)sqrt[3]{20+14sqrt{2}}+sqrt[3]{20-14sqrt{2}}d)sqrt{6+2sqrt{5}}+sqrt{6-2sqrt{5}}e)sqrt{11+6sqrt{2}}-sqrt{11-6sqrt{2}}f)sqrt[3]{5sqrt{2}+7}-sqrt[3]{5sqrt{2}-7}g)sqrt[3]{26+15sqrt{3}}-sqrt[3]{26-15sqrt{3}}
Đọc tiếp
thực hiện phép tính:a)\(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)\)
b)\(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
c)\(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
d)\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
e)\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
f)\(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
g)\(\sqrt[3]{26+15\sqrt{3}}-\sqrt[3]{26-15\sqrt{3}}\)
@.@ Trời ơi, nhiều thế ^^
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)=\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-\frac{3\sqrt{2}}{\sqrt{5}}\right)\)
\(=\left(\sqrt{2}.\sqrt{5}-\sqrt{2}\right)\left(\sqrt{2}-\frac{3\sqrt{2}}{\sqrt{5}}\right)=2\sqrt{5}-2-6+\frac{6}{\sqrt{5}}=\frac{16\sqrt{5}}{5}-8\)
b) \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}=\frac{75\sqrt{2}+50\sqrt{2}-45\sqrt{2}}{\sqrt{10}}=\frac{80\sqrt{2}}{\sqrt{10}}=\frac{80}{\sqrt{5}}=16\sqrt{5}\)c) \(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}\)
\(=2+\sqrt{2}+2-\sqrt{2}=4\)
d) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)}^2\)
\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
e) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
f)\(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}=1+\sqrt{2}-\sqrt{2}+1=2\)g) \(\sqrt[3]{26+15\sqrt{3}}-\sqrt[3]{26-15\sqrt{3}}=\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\sqrt[3]{\left(2-\sqrt{3}\right)^3}\)
\(=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}\)
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Thực hiện phép tính:a,left(sqrt{28}-2sqrt{14}+sqrt{7}right)cdotsqrt{7}+7sqrt{8}b,left(sqrt{8}-3sqrt{2}+sqrt{10}right)cdotleft(sqrt{2}-3sqrt{0.4}right)c,left(15sqrt{50}+5sqrt{200}-3sqrt{450}right):sqrt{10}d,sqrt{6+2sqrt{5}}+sqrt{6-2sqrt{5}}e,sqrt{11+6sqrt{2}}-sqrt{11-6sqrt{2}}f,sqrt[3]{5sqrt{2}+7}-sqrt[3]{5sqrt{2}-7}g,sqrt[3]{20+14sqrt{2}}+sqrt[3]{20-14sqrt{2}}h,sqrt[3]{26+15sqrt{3}}-sqrt[3]{26-15sqrt{3}}
Đọc tiếp
Thực hiện phép tính:
\(a,\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(b,\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\cdot\left(\sqrt{2}-3\sqrt{0.4}\right)\)
\(c,\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(d,\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(e,\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(f,\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(g,\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
\(h,\sqrt[3]{26+15\sqrt{3}}-\sqrt[3]{26-15\sqrt{3}}\)
g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath