Ta có:\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1.\)

\(\Rightarrow a=b=c\)

\(\Rightarrow\frac{a^{49}\cdot b^{51}}{c^{100}}=\frac{a^{49+51}}{c^{100}}=\frac{a^{100}}{c^{100}}=1\)

a: 3x^2y(3x-4y)

=3x^2y*3x-3x^2y*4y

=9x^3y-12x^2y^2

c: (3x+5)(3xy-5x)

=9x^2y-15x^2+15xy-25x

\(x=\frac{\left(\sqrt{3}-1\right)\sqrt[3]{10+6\sqrt{3}}}{\sqrt{21+4\sqrt{5}}+3}=\frac{\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+1\right)}{2\sqrt{5}+1+3}=\frac{2}{2\sqrt{5}+4}=\sqrt{5}-2\)

\(A=\left(x^2-4x-2\right)^{2015}\)

Xong chắc thay vào nhưng để nguyên số mũ ko tính ra

Bài làm

       \(7+\left(\frac{7}{12}-\frac{1}{2}+3\right)-\left(\frac{1}{12}+5\right)\)

\(=\frac{84}{12}+\left(\frac{7}{12}-\frac{6}{12}+\frac{36}{12}\right)-\left(\frac{1}{12}+\frac{60}{12}\right)\)

\(=\frac{84}{12}+\frac{37}{12}-\frac{61}{12}\)

\(=\frac{60}{12}\)

\(=5\)

# Chúc bạn học tốt #

(1+3+5+7+...+2003+2005)×(125125×127-127127×125)

=(1+3+5+7+...+2003+2005)×0

=0

Ủng hộ nhé

( 1 + 3 + 5 + 7 + ... + 2003 + 2005 ) x 125125 x127 - 127127 x 125

= ( 1 + 3 + 5 + 7 + ... + 2003 + 2005 ) x ( 125 x 1001 x 127 ) - ( 127 x 1001 x 125 )

= ( 1 + 3 + 5 + 7 + ... + 2003 + 2005 ) x 0

= 0

 ( 1 + 3 + 5 + 7  + ... + 2003 + 2005 )x125125 x 127 -127127 x 125 )

= ( 1 + 3 + 5 + 7  + ... + 2003 + 2005 ) x 0

= 0

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TOI  MOI  HOC  LOP  4