Chứng minh \(\left(64.27\right)^6=12^{18}\)
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Chứng minh (64.27) mũ 6= 12 mũ 18
\(\left(64.27\right)^6=4^{3.6}.3^{3.6}=\left(4.3\right)^{18}=12^{18}\)
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chứng minh \(\left(64.27\right)^6.75^{18}=30^{36}\)
\(\left(64.27\right)^6.75^{18}=\left(2^6.3^3\right)^6.\left(3.5^2\right)^{18}=2^{36}.3^{36}.5^{36}=\left(2.3.5\right)^{36}=30^{36}\)
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Chứng minh rằng:
a)\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\) là số nguyên
b)\(\left(\sqrt{3}-1\right).\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(130-12\left\{\left[18-\left(6:x+3\right)\right]-3\right\}=10\)
cac ban giai giai ho minh nhe
12 { [ 18 - ( 6 : x + 3 ) ] - 3 } = 130 - 10 = 120
18 - ( 6 : x + 3 ) - 3 = 120 : 12 = 10
18 - ( 6 : x + 3 ) = 10 + 3 = 13
6 : x + 3 = 18 - 13 = 5
6 : x = 5 - 3 = 2
x = 6 : 2 = 3
Vậy x = 3
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130 - 12{[18 - (6 : x + 3)] - 3} = 10
12{[18 - (6 : x + 3)] - 3} = 130 - 10
12{[18 - (6 : x + 3)] - 3} = 120
18 - (6 : x + 3) - 3 = 120 : 12
15 - (6 : x + 3) = 10
6 : x + 3 = 15 - 10
6 : x + 3 = 5
6 : x = 5 - 3
6 : x = 2
x = 6 : 2
x = 3
Vậy x = 3
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130 - 12{[18 - (6 : x + 3)] - 3} = 10
12{[18 - (6 : x + 3)] - 3} = 130 - 10
12{[18 - (6 : x + 3)] - 3} = 120
18 - (6 : x + 3) - 3 = 120 : 12
15 - (6 : x + 3) = 10
6 : x + 3 = 15 - 10
6 : x + 3 = 5
6 : x = 5 - 3
6 : x = 2
x = 6 : 2
x = 3
Vậy x = 3
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Xem thêm câu trả lời
Chứng minh: (\(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}-\left(5\sqrt{\dfrac{1}{12}}+12\right)=-14,5\sqrt{2}\)
\(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}-\left(5\sqrt{\dfrac{1}{12}}+12\right)\)
\(=\left(2\sqrt{3}-6\sqrt{3}+2\sqrt{6}\right)\sqrt{6}-\left(\dfrac{5\sqrt{3}}{6}+12\right)\)
\(=6\sqrt{2}-18\sqrt{2}+12-\left(\dfrac{5\sqrt{3}+72}{6}\right)\)
\(=-12\sqrt{2}+12-\dfrac{5\sqrt{3}+72}{6}\)
\(=\dfrac{-72\sqrt{2}+72-5\sqrt{3}-72}{6}=\dfrac{5\sqrt{3}+72\sqrt{2}}{6}\simeq-18,4139\)
Ta có: \(-14,5\sqrt{2}\simeq-20,506\)
\(VT\ne VP\)
Đẳng thức không xảy ra
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Chứng minh đẳng thức:
\(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}-\left(4\sqrt{\frac{1}{2}}+12\right)=-14\sqrt{2}\)
Chứng minh các số sau đây là số nguyên:
b) B = \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(B=\left(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+11\right)\)
\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(=6-121=-115\) là số nguyên (đpcm)
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b) Ta có: \(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
=6-121=-115
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Chứng minh biểu thức sau nhận giá trị nguyên:
\(D=\left(\sqrt{3}-1\right)\sqrt{6+2.\sqrt{2.}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18}-\sqrt{128}}}}\)
Tính nhanh \(\frac{\left(-18\right)^2.\left(-16^2\right)-6^2.\left(-12\right)^3}{-6^5.4^2-\left(-18\right)^2.96}\)
