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a) \(413.\left(413-26\right)+169=413^2-2.13.413+13^2=\left(413-13\right)^2=160000\)
b) \(\left(625^2+3\right).\left(25^4-3\right)-5^{16}+10\)
\(=\left(5^8+3\right)\left(5^8-3\right)-5^{16}+10\)
\(=5^{16}-9-5^{16}+10=1\)
c) \(\frac{41^2+39^2+8^2.39}{41^2-39^2}=\frac{\left(41+39\right)^2}{\left(41-39\right)\left(41+39\right)}=\frac{41+39}{41-39}=\frac{80}{2}=40\)
a)\(\frac{59^3-41^3}{18}+59.41=\frac{\left(59-41\right)\left(59^2+59.41+41^2\right)}{18}+59.41\)
\(=\frac{18.\left(59^2+59.41+41^2\right)}{18}+59.41=59^2+59.41+41^2+59.41=59^2+2.59.41+41^2=\left(59+41\right)^2=100^2\)
=10000
Pythagorean theorem: \(AD=\sqrt{BD^2-AB^2}=4\) (cm)
\(\Rightarrow BC=AD=4\left(cm\right)\)
\(CC'=\sqrt{BC'^2-BC^2}=4\sqrt{2}\)
The lateral surface area: \(2CC'.\left(BC+AB\right)=56\sqrt{2}\left(cm^2\right)\)
lần sau bạn ghi rõ đề bài nhé
a, \(A=x^2+10x+39=x^2+2.5.x+25+14=\left(x+5\right)^2+14\ge14\)
Dấu ''='' xảy ra khi \(x=-5\)
Vậy GTNN A là 14 khi x = -5
b, \(B=25x^2-70x+1000=\left(5x\right)^2-2.5x.7+49+951\)
\(=\left(5x-7\right)^2+951\ge951\)
Dấu ''='' xảy ra khi \(x=\frac{7}{5}\)
Vậy GTNN B là 951 khi x = 7/5
c, \(C=49x^2+64x+100=\left(7x\right)^2+2.7x.\frac{32}{7}+\frac{1024}{49}+\frac{3876}{49}\)
\(=\left(7x+\frac{32}{7}\right)^2+\frac{3876}{49}\ge\frac{3876}{49}\)
Dấu ''='' xảy ra khi \(x=-\frac{32}{49}\)
Vậy GTNN C là 3876/49 khi x = -32/49
d, \(D=5x^2+13x+41=5\left(x^2+\frac{13}{5}x+\frac{41}{5}\right)\)
\(=5\left(x^2+2.\frac{6,5}{5}.x+\frac{169}{100}+\frac{651}{100}\right)=5\left(x+\frac{6,5}{5}\right)^2+\frac{651}{20}\ge\frac{651}{20}\)
Dấu ''='' xảy ra khi \(x=-\frac{6,5}{2}\)
Vậy GTNN D là 651/20 khi x = -6,5/2
\(\frac{x-5}{45}-1+\frac{x-7}{43}-1=\frac{x-9}{41}-1+\frac{x-11}{39}-1\)
\(\Leftrightarrow\frac{x-50}{45}+\frac{x-50}{43}=\frac{x-50}{41}+\frac{x-50}{39}\)
\(\Leftrightarrow\left(x-50\right)\left(\frac{1}{45}+\frac{1}{43}-\frac{1}{41}-\frac{1}{39}\right)=0\)
\(\Leftrightarrow x-50=0\) (do \(\frac{1}{45}+\frac{1}{43}-\frac{1}{41}-\frac{1}{39}\ne0\))
\(\Rightarrow x=50\)
\(41^2+82\times59+59^2=41^2+2\times41\times59+59^2=\left(41+59\right)^2=100^2=10000\)
BÀI NÀY MÌNH ĐƯA VỀ HẰNG ĐẲNG THỨC THỨ 1 NHA : ( A + B ) ^2
41^2 + 89 . 59 . 59^2
= 41^2 + 2 . 41 . 59 + 59^2
= ( 41 + 59 ) ^2 = 100^2 = 1000
\(=\frac{41^2-39}{41^2+39^2+2.41.39}\)
\(=\frac{41^2-39}{41^2+2.41.39+39^2}\)
\(=\frac{41^2-39}{\left(41+39\right)^2}\)
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