tính \(\sqrt{A}\)
a)A= 13 - 2\(\sqrt{42}\)
b)A= 46 + 6\(\sqrt{5}\)
c)A= 12 - 3\(\sqrt{15}\)
cứu tui
Những câu hỏi liên quan
Tính \(\sqrt{A}\) biết A bằng:
a/ 13-\(12\sqrt{42}\)
b/ \(46-6\sqrt{5}\)
c/ \(12-3\sqrt{15}\)
Bài 1: Rút gọn
a) \(\sqrt{13-2\sqrt{42}}\)
b) \(\sqrt{46+6\sqrt{5}}\)
c) \(\sqrt{12-3\sqrt{15}}\)
d) \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
e) \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)
\(\sqrt{13-2\sqrt{42}}=\sqrt{6-2\sqrt{6}.\sqrt{7}+7}=\sqrt{\left(\sqrt{6}-\sqrt{7}\right)^2}=\left|\sqrt{6}-\sqrt{7}\right|=\sqrt{7}-\sqrt{6}\)
\(\sqrt{46+6\sqrt{5}}=\sqrt{45+6\sqrt{5}+1}=\sqrt{3^2.5+6\sqrt{5}+1}=\sqrt{3^2.5+2.3.\sqrt{5}+1^2}=\sqrt{\left(3.\sqrt{5}+1\right)^2}=3\sqrt{5}+1\)
\(\sqrt{12-3\sqrt{15}}=\sqrt{3}\sqrt{4-\sqrt{15}}=\sqrt{\frac{3}{2}}.\sqrt{8-2\sqrt{15}}=\sqrt{\frac{3}{2}}.\sqrt{3-2\sqrt{15}+5}=\sqrt{\frac{3}{2}}.\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{\frac{3}{2}}.\left(\sqrt{5}-\sqrt{3}\right)\)
\(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{3-2\sqrt{15}+5}-\sqrt{8+2\sqrt{15}}=\sqrt{3-2\sqrt{3}\sqrt{5}+5}-\sqrt{3+2\sqrt{3}\sqrt{5}+5}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{5}-\sqrt{3}-\sqrt{3}-\sqrt{5}=-2\sqrt{3}\)
\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{\frac{1}{2}}\left(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\right)=\sqrt{\frac{1}{2}}\left(\sqrt{1+2\sqrt{5}+5}-\sqrt{1-2\sqrt{5}+5}\right)=\sqrt{\frac{1}{2}}\left(\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\right)=\sqrt{\frac{1}{2}}\left(1+\sqrt{5}-\sqrt{5}+1\right)=\sqrt{\frac{1}{2}}.2=\sqrt{\frac{4}{2}}=\sqrt{2}\)
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Tính \(\sqrt{A}\) biết
\(A=13-2\sqrt{42}\)
\(A=46+6\sqrt{5}\)
\(A=12-3\sqrt{15}\)
A=13−2√42
=7-2\(\sqrt{6}.\sqrt{7}\)+6
=\(\left(\sqrt{7}-\sqrt{6}\right)^2\)
=>\(\sqrt{A}=\sqrt{7}-\sqrt{6}\)
A=46+6√5
=45+2.\(3\sqrt{5}\)+1
=(\(3\sqrt{5}+1\))2
=>\(\sqrt{A}=3\sqrt{5}+1\)
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Bài 1: TínhAsqrt{46-6sqrt{5}}-sqrt{29-12sqrt{5}}Bsqrt{13-sqrt{160}-sqrt{53+4sqrt{90}}}Csqrt{15-6sqrt{6}}+sqrt{35-12sqrt{6}}Dsqrt{5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}E sqrt{4-sqrt{7}}+sqrt{4+sqrt{7}}F sqrt{3+sqrt{11+6sqrt{2}}}-sqrt{5+2sqrt{6}}Gsqrt{5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}Bài 2: so sánha) sqrt{24}+sqrt{45} và 12b) sqrt{37}-sqrt{15} và 2c) sqrt{16} và sqrt{15}timessqrt{17}d) 8 và sqrt{15}+sqrt{17}
Đọc tiếp
Bài 1: Tính
A=\(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
B=\(\sqrt{13-\sqrt{160}-\sqrt{53+4\sqrt{90}}}\)
C=\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
D=\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
E= \(\sqrt{4-\sqrt{7}}+\sqrt{4+\sqrt{7}}\)
F= \(\sqrt{3+\sqrt{11+6\sqrt{2}}}-\sqrt{5+2\sqrt{6}}\)
G=\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
Bài 2: so sánh
a) \(\sqrt{24}+\sqrt{45}\) và 12
b) \(\sqrt{37}-\sqrt{15}\) và 2
c) \(\sqrt{16}\) và \(\sqrt{15}\times\sqrt{17}\)
d) 8 và \(\sqrt{15}+\sqrt{17}\)
Bài 2 :
a,\(\sqrt{24}+\sqrt{45}< \sqrt{25}+\sqrt{49}=5+7=12=>\sqrt{24}+\sqrt{45}< 12\)
b. \(\sqrt{37}-\sqrt{15}>\sqrt{36}-\sqrt{16}=6-4=2=>\sqrt{37}-\sqrt{15}>2\)
c, \(\sqrt{15}.\sqrt{17}>\sqrt{15}.\sqrt{16}>\sqrt{16}=>\sqrt{15}.\sqrt{17}>\sqrt{16}\)
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Bài 1:
a) sqrt{13-2sqrt{42}}
b) sqrt{46+6sqrt{5}}
c) sqrt{12-3sqrt{15}}
d) sqrt{11+sqrt{96}}
Bài 2:
a) Asqrt{6+2sqrt{2}sqrt{3-sqrt{4+2sqrt{3}}}}
b) Bsqrt{5}-sqrt{3-sqrt{29-12sqrt{5}}}
c) Csqrt{3-sqrt{5}}left(sqrt{10}+sqrt{2}right)left(3+sqrt{5}right)
d) Dsqrt{4+sqrt{7}}-sqrt{4-sqrt{7}}
e) Esqrt{8+2sqrt{10+2sqrt{5}}}+sqrt{8-2sqrt{10+2sqrt{5}}}
g) Gsqrt{x+2sqrt{x-1}}+sqrt{x-2sqrt{x-1}}
h) H4x-sqrt{9x^2-12x+4}
i) frac{sqrt{7}-sqrt{2}}{sqrt{7}+sqrt{2}}+frac{sqrt{7}+sqrt{2}}{sqrt{7}-sqrt...
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Bài 1:
a) \(\sqrt{13-2\sqrt{42}}\)
b) \(\sqrt{46+6\sqrt{5}}\)
c) \(\sqrt{12-3\sqrt{15}}\)
d) \(\sqrt{11+\sqrt{96}}\)
Bài 2:
a) \(A=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
b) \(B=\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
c) \(C=\sqrt{3-\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
d) \(D=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
e) \(E=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
g) \(G=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
h) \(H=4x-\sqrt{9x^2-12x+4}\)
i) \(\frac{\sqrt{7}-\sqrt{2}}{\sqrt{7}+\sqrt{2}}+\frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}}\)
Bài 1:
a)
\(\sqrt{13-2\sqrt{42}}=\sqrt{6+7-2\sqrt{6.7}}=\sqrt{(\sqrt{7}-\sqrt{6})^2}=|\sqrt{7}-\sqrt{6}|=\sqrt{7}-\sqrt{6}\)
b)
\(\sqrt{46+6\sqrt{5}}=\sqrt{46+2\sqrt{45}}=\sqrt{45+1+2\sqrt{45.1}}=\sqrt{(\sqrt{45}+1)^2}=\sqrt{45}+1\)
\(=3\sqrt{5}+1\)
c)
\(\sqrt{12-3\sqrt{15}}=\sqrt{\frac{24-6\sqrt{15}}{2}}=\sqrt{\frac{24-2\sqrt{135}}{2}}=\sqrt{\frac{15+9-2\sqrt{15.9}}{2}}\)
\(=\sqrt{\frac{(\sqrt{15}-\sqrt{9})^2}{2}}=\frac{\sqrt{15}-\sqrt{9}}{\sqrt{2}}=\frac{\sqrt{15}-3}{\sqrt{2}}\)
d)
\(\sqrt{11+\sqrt{96}}=\sqrt{11+2\sqrt{24}}=\sqrt{8+3+2\sqrt{8.3}}\)
\(=\sqrt{(\sqrt{8}+\sqrt{3})^2}=\sqrt{8}+\sqrt{3}\)
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Bài 1:
a)
\(\sqrt{13-2\sqrt{42}}=\sqrt{6+7-2\sqrt{6.7}}=\sqrt{(\sqrt{7}-\sqrt{6})^2}=|\sqrt{7}-\sqrt{6}|=\sqrt{7}-\sqrt{6}\)
b)
\(\sqrt{46+6\sqrt{5}}=\sqrt{46+2\sqrt{45}}=\sqrt{45+1+2\sqrt{45.1}}=\sqrt{(\sqrt{45}+1)^2}=\sqrt{45}+1\)
\(=3\sqrt{5}+1\)
c)
\(\sqrt{12-3\sqrt{15}}=\sqrt{\frac{24-6\sqrt{15}}{2}}=\sqrt{\frac{24-2\sqrt{135}}{2}}=\sqrt{\frac{15+9-2\sqrt{15.9}}{2}}\)
\(=\sqrt{\frac{(\sqrt{15}-\sqrt{9})^2}{2}}=\frac{\sqrt{15}-\sqrt{9}}{\sqrt{2}}=\frac{\sqrt{15}-3}{\sqrt{2}}\)
d)
\(\sqrt{11+\sqrt{96}}=\sqrt{11+2\sqrt{24}}=\sqrt{8+3+2\sqrt{8.3}}\)
\(=\sqrt{(\sqrt{8}+\sqrt{3})^2}=\sqrt{8}+\sqrt{3}\)
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Bài 2:
a)
\(A=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3+1+2\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{(\sqrt{3}+1)^2}}}=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{(\sqrt{3}-1)^2}}=\sqrt{6+2(\sqrt{3}-1)}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}=\sqrt{(\sqrt{3}+1)^2}=\sqrt{3}+1\)
b)
\(B=\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}\)
\(=\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-\sqrt{9})^2}}=\sqrt{5}-\sqrt{3-(\sqrt{20}-\sqrt{9})}\)
\(=\sqrt{5}-\sqrt{6-2\sqrt{5}}=\sqrt{5}-\sqrt{5+1-2\sqrt{5}}=\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}\)
\(=\sqrt{5}-(\sqrt{5}-1)=1\)
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Xem thêm câu trả lời
Rút gọn
A=\(\sqrt{13+4\sqrt{10}}\)
B= \(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
C= \(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{7}}\)
\(A=\sqrt{13+4\sqrt{10}}=\sqrt{13+2\sqrt{40}}=\sqrt{8+2.\sqrt{5}.\sqrt{8}+5}=\sqrt{\left(\sqrt{8}+\sqrt{5}\right)^2}=\sqrt{8}+\sqrt{5}\)
\(B=\sqrt{46-6\sqrt{5}}=\sqrt{46-2\sqrt{45}}=\sqrt{\left(\sqrt{45}-1\right)^2}=\sqrt{45}-1=3\sqrt{5}-1\)
\(C=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{7}}\)
\(C=-\sqrt{3}-\sqrt{2}+\dfrac{\sqrt{5}+\sqrt{3}}{2}-\dfrac{\sqrt{7}+\sqrt{5}}{2}\)
\(C=-\sqrt{3}-\sqrt{2}+\dfrac{\sqrt{3}-\sqrt{7}}{2}\)
\(C=\dfrac{-2\sqrt{3}-2\sqrt{2}+\sqrt{3}-\sqrt{7}}{2}=\dfrac{-\sqrt{3}-2\sqrt{2}-\sqrt{7}}{2}\)
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Bài 1: tính sqrt{A}a) A46+6sqrt{5}b) A12-3sqrt{15}Bài 2: Rút gọna) A sqrt{6+2sqrt{2}cdotsqrt{3}-sqrt{4+2sqrt{3}}}b) B sqrt{5}-sqrt{3sqrt{29}-12sqrt{5}}c) C sqrt{x+2sqrt{x-1}}+sqrt{x-2sqrt{x-1}}Bài 3: Giải Phương Trìnha) sqrt{9-x^2}-3sqrt{3-x}0b)5sqrt{3x-1}sqrt{75}c)3sqrt{1-3x}sqrt{12}Bài 4: Giải Bất Phương Trìnha) 3sqrt{x-1}gesqrt{12}b) 2sqrt{2x+1}lesqrt{24}c) sqrt{left(x-1right)left(x+2right)}gesqrt{10}
Đọc tiếp
Bài 1: tính \(\sqrt{A}\)
a) \(A=46+6\sqrt{5}\)
b) \(A=12-3\sqrt{15}\)
Bài 2: Rút gọn
a) A= \(\sqrt{6+2\sqrt{2}\cdot\sqrt{3}-\sqrt{4+2\sqrt{3}}}\)
b) B= \(\sqrt{5}-\sqrt{3\sqrt{29}-12\sqrt{5}}\)
c) C= \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
Bài 3: Giải Phương Trình
a) \(\sqrt{9-x^2}-3\sqrt{3-x}=0\)
b)\(5\sqrt{3x-1}=\sqrt{75}\)
c)\(3\sqrt{1-3x}=\sqrt{12}\)
Bài 4: Giải Bất Phương Trình
a) \(3\sqrt{x-1}\ge\sqrt{12}\)
b) \(2\sqrt{2x+1}\le\sqrt{24}\)
c) \(\sqrt{\left(x-1\right)\left(x+2\right)}\ge\sqrt{10}\)
có ai giúp tui left(sqrt{6}+sqrt{2}right)left(sqrt{3}-2right)sqrt{3}+2frac{2sqrt{8}-sqrt{12}}{sqrt{18}-sqrt{48}}-frac{sqrt{5}+sqrt{27}}{sqrt{30}+sqrt{162}}2sqrt{5}-sqrt{125}-sqrt{80}+sqrt{605}sqrt{8sqrt{3}-2sqrt{25sqrt{12}}+8sqrt{sqrt{192}}}2. cho Asqrt{15a^2-8asqrt{15}+16}a. Rút gọn Ab. Tính A khi asqrt{frac{3}{5}}+sqrt{frac{5}{3}}
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có ai giúp tui
\(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{3}+2\)
\(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
\(\sqrt{8\sqrt{3}-2\sqrt{25\sqrt{12}}+8\sqrt{\sqrt{192}}}\)
2. cho A=\(\sqrt{15a^2-8a\sqrt{15}+16}\)
a. Rút gọn A
b. Tính A khi a=\(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\)
2) \(A=\sqrt{15a^2-8a\sqrt{15}+16}\\ =\sqrt{\left(a\sqrt{15}-4\right)^2}\)
b) Khi a=\(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\) thì
\(A=\sqrt{\left[\left(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\right)\sqrt{15}-4\right]^2}\)
\(=\sqrt{\left[\left(3+5\right)-4\right]^2}\)
\(=\sqrt{4^2}\)
\(=4\)
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a) A=\(\sqrt{\left(4-\sqrt{15}\right)^2+\sqrt{15}}\)
b) B=\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}\)
c) C=\(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
d)D=\(\sqrt{29+12\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)
a: Sửa đề: \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(=4-\sqrt{15}+\sqrt{15}=4\)
b: \(A=2-\sqrt{3}+\sqrt{3}-1=1\)
c: \(C=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
d: Sửa đề: \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=2\sqrt{5}+3-2\sqrt{5}+3\)
=6
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a) \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(A=\left|4-\sqrt{15}\right|+\sqrt{15}\)
\(A=4-\sqrt{15}+\sqrt{15}\)
\(A=4\)
b) \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)}\)
\(B=\left|2-\sqrt{3}\right|+\left|1-\sqrt{3}\right|\)
\(B=2-\sqrt{3}-1+\sqrt{3}\)
\(B=1\)
c) \(C=\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(C=\sqrt{\left(3\sqrt{5}\right)^2-2\cdot3\sqrt{15}\cdot2+2^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\sqrt{5}\cdot2+2^2}\)
\(C=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(C=\left|3\sqrt{5}-2\right|-\left|3\sqrt{5}+2\right|\)
\(C=3\sqrt{5}-2-3\sqrt{5}-2\)
\(C=-4\)
d) \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(D=\sqrt{\left(2\sqrt{5}\right)^2+2\cdot2\sqrt{5}\cdot3+3^2}-\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot3+3^3}\)
\(D=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(D=\left|2\sqrt{5}+3\right|-\left|2\sqrt{5}-3\right|\)
\(D=2\sqrt{5}+3-2\sqrt{5}+3\)
\(D=6\)
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