Question

[Bùi Hạnh Nguyên]

Câu 1:Chứng minh ; Ax//By

Câu 2 :Biết\(\widehat{xOn}+\widehat{ONM}+\widehat{NMy}=360^o\).Chứng minh Ox//My

Answer 1

[Nobi Nobita]

Câu 1: 

a) \(\left(x-\frac{1}{2}\right)^3=-27\) \(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(-3\right)^3\)

\(\Leftrightarrow x-\frac{1}{2}=-3\)\(\Leftrightarrow x=-\frac{5}{2}\)

Vậy \(x=-\frac{5}{2}\)

b( \(2x^2+x=0\)\(\Leftrightarrow x\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\2x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)

Vậy \(x=0\)hoặc \(x=-\frac{1}{2}\)

c) \(5^{x+2}=625\)\(\Leftrightarrow5^{x+2}=5^4\)

\(\Leftrightarrow x+2=4\)\(\Leftrightarrow x=2\)

Vậy \(x=2\)

d) \(\frac{7^x+7^{x+1}+7^{x+2}+7^{x+3}}{2^2.5^2.7^2}=2^2\)

\(\Leftrightarrow7^x+7^{x+1}+7^{x+2}+7^{x+3}=2^2.2^2.5^2.7^2\)

\(\Leftrightarrow7^x+7^x.7+7^x.7^2+7^x.7^3=\left(2.2.5.7\right)^2\)

\(\Leftrightarrow7^x+7^x.7+7^x.49+7^x.343=140^2\)

\(\Leftrightarrow7^x.\left(1+7+49+343\right)=19600\)

\(\Leftrightarrow7^x.400=19600\)

\(\Leftrightarrow7^x=49=7^2\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

Answer 2

[Nobi Nobita]

Câu 2:

a) \(C=1+4+4^2+4^3+.......+4^{48}\)

\(\Rightarrow4C=4+4^2+4^3+4^4+........+4^{49}\)

\(\Rightarrow4C-C=4^{49}-1\)

\(\Rightarrow3C=4^{49}-1\)

\(\Rightarrow C=\frac{4^{49}-1}{3}\)

b) Ta có: \(3C+1=4^{49}-1+1=4^{49}=4^{7.7}=\left(4^7\right)^7⋮4^7\)( đpcm )

c) \(C=1+4+4^2+4^3+........+4^{48}\)

\(=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+........+\left(4^{46}+4^{47}+4^{48}\right)\)

\(=\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+........+4^{46}.\left(1+4+4^2\right)\)

\(=\left(1+4+4^2\right).\left(1+4^3+....+4^{46}\right)\)

\(=\left(1+4+16\right).\left(1+4^3+........+4^{46}\right)\)

\(=21.\left(1+4^3+.....+4^{46}\right)⋮21\)