7^x+1 + 7^x . 42= 7^27 nhanh lên nha mik dag cần gấp
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Gọi hai góc kề bù là \(\widehat{aOc}\) và \(\widehat{cOb}\)
sau đó lần lượt gọi Ox và Oy là 2 tia phân giác của 2 góc
Từ đó ta có:
\(\widehat{xOy}=\dfrac{1}{2}\left(\widehat{aOb}+\widehat{bOc}\right)=\dfrac{1}{2}\cdot180^o=90^o\)
=>\(Ox\perp Oy\) (đpcm)
Gọi \(A_1;A_2\) lần lượt là 2 góc phân giác trong của góc A
Gọi \(A_3;A_4\) lần lượt là 2 góc phân giác ngoài của góc A
Góc kề bù của A \(\Rightarrow A_{trong}+A_{ngoài}=180^o\)
mà \(A_1=A_2\left(trong\right);A_3=A_4\left(ngoài\right)\)
\(\Rightarrow2A_2+2A_4=180^o\Rightarrow A_2+A_4=90^o\)
\(\Rightarrow dpcm\)
Do góc xoz =60o
mà Om là tia pgiac của \(\widehat{zox}\)
=>\(\widehat{zOm}=\widehat{mOx}=\dfrac{60}{2}=30^o\)
Ta có: \(\widehat{yOz}+\widehat{xOz}=100^o\) (do 2 góc kề bù)
=> \(\widehat{yOz}=100^o-\widehat{xOz}\\ =100^o-60^o=40^o\)
Mà On là tia phân giác \(\widehat{yOz}\)
=>\(\widehat{yOn}=\widehat{nOz}=\widehat{yOz}:2=40^o:2=20^o\)
\(\Rightarrow\widehat{mOn}=\widehat{nOz}+\widehat{zOm}=20^o+30^o=50^o\)
Vậy góc mOn=50o
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Lời giải:
Nếu $n$ lẻ thì $n+7$ chẵn
$\Rightarrow (n+4)(n+7)$ chẵn
Nếu $n$ chẵn thì $n+4$ chẵn
$\Rightarrow (n+4)(n+7)$ chẵn
Vậy $(n+4)(n+7)$ luôn là số chẵn với mọi $n$
\(...\Rightarrow7^{x+1}+7^x.7.6=7^{27}\)
\(\Rightarrow7^{x+1}+7^{x+1}.6=7^{27}\)
\(\Rightarrow7^{x+1}.\left(1+6\right)=7^{27}\)
\(\Rightarrow7^{x+1}.7=7^{27}\)
\(\Rightarrow7^{x+2}=7^{27}\Rightarrow x+2=27\Rightarrow x=25\)