Bài 1:

Tam giác MNP có: \(\widehat{M}=40^o;\widehat{N}=100^o\)

Tổng số đo 3 góc của 1 tam giác là 180^o, ta được:

\(\widehat{M}+\widehat{N}+\widehat{P}=180^o\\ \Leftrightarrow40^o+100^o+\widehat{P}=180^o\\ \Leftrightarrow140^o+\widehat{P}=180^o\\ \Leftrightarrow\widehat{P}=180^o-140^o=40^o\)

Vì: \(\widehat{M}=\widehat{P}=40^o\) => Tam giác MNP là tam giác cân tại N (ĐPCM)

Vẽ hình ạ

sao ko có nộp bài···

\(\left(\dfrac{1}{2}\right)^x+\left(\dfrac{1}{2}\right)^{x+2}=\dfrac{1}{128}\\ \left(\dfrac{1}{2}\right)^x.\left[1+\left(\dfrac{1}{2}\right)^2\right]=\dfrac{1}{128}\\ \left(\dfrac{1}{2}\right)^x.\dfrac{5}{4}=\dfrac{1}{128}\\ \left(\dfrac{1}{2}\right)^x=\dfrac{1}{128}:\dfrac{5}{4}=\dfrac{1}{128}.\dfrac{4}{5}=\dfrac{4}{640}=\dfrac{1}{160}\)

Thầy thấy số lẻ quá....

`#3107.101107`

\(\left(\dfrac{1}{2}\right)^x+\left(\dfrac{1}{2}\right)^{x+2}=\dfrac{1}{128}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^x\cdot\left[1+\left(\dfrac{1}{2}\right)^2\right]=\dfrac{1}{128}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^x\cdot\left(1+\dfrac{1}{4}\right)=\dfrac{1}{128}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^x\cdot\dfrac{5}{4}=\dfrac{1}{128}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^x=\dfrac{1}{128}\div\dfrac{5}{4}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^x=\dfrac{1}{160}\)

Bạn xem lại đề.

Ta có: \(\left\{{}\begin{matrix}\left|0,25x-1\right|\ge0\forall x\\\left|3-2y\right|\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left|0,25x-1\right|+\left|3-2y\right|\ge0\forall x,y\)

Mà: \(\left|0,25x-1\right|+\left|3-2y\right|=0\)

nên: \(\left\{{}\begin{matrix}0,25x-1=0\\3-2y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}0,25x=1\\2y=3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=\dfrac{3}{2}\end{matrix}\right.\)

Vậy: \(x=4;y=\dfrac{3}{2}\).