không có dữ kiện nào làm sao mà tính

a) Xét ΔBDC có: BM = CM (gt); BD // ME (gt).

=> DE = CE.

Xét ΔAME có: AI = MI (gt); ID // ME (do BD // ME).

=> AD = DE = DC/2 (đpcm).

b) Xét ΔBDC có: BM = CM (gt); DE = CE (cm ở câu a).

=> ME là đường trung bình của ΔBDC.

=> \(ME=\dfrac{1}{2}BD\) (a)

Xét ΔAME có: AI = MI (gt); AD = DE (cm ở câu a).

=> ID là đường trung bình ΔAME.

=> \(ID=\dfrac{1}{2}ME\) (b)

Từ (a) và (b) suy ra \(\dfrac{BD}{DI}=4\)

lên bài giảng điện tử violet đó

đề sai

đề bài?

a) \(\left(2x-1\right)^3=-27\)

\(\Leftrightarrow\left(2x-1\right)^3=\left(-3\right)^3\)

\(\Leftrightarrow2x-1=-3\)

\(\Leftrightarrow2x=-3+1\)

\(\Leftrightarrow2x=-2\)

\(\Leftrightarrow x=-1\)

b) \(\left(x-3\right)^{10}=\left(x-3\right)^{30}\)

\(\Leftrightarrow\left(x-3\right)^{10}=\left[\left(x-3\right)^{10}\right]^3\)

\(\Leftrightarrow x-3=\left(x-3\right)^3\)

giải ra x = 4 ; x = 2; x = 3

\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=x^4+9x^3+23x^2+15x+7x^3+63x^2+161x+105+15x^4\\ +135x^3+345x^2+225x+105x^3+945x^2+2415x+1575+15\)

\(=16x^4+256x^3+1376x^2+2816x+1695\)

\(=16x^3\left(x+16\right)+32x\left(43x+88\right)+1695\)

......

\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=x^4+9x^3+23x^2+15x+7x^3+63x^2+161x+105+15\)

\(=x^4+16x^3+86x^2+176x+120\)

\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)

\(\left|x\right|-\left|2x-3\right|=x-1\)

\(\Leftrightarrow\left|x\right|-\left|2x-3\right|+\left|2x-3\right|=x-1+\left|2x-3\right|\)

\(\Leftrightarrow\left|x\right|=\left|2x-3\right|+x-1\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\left|2x-3\right|+x-1\\x=-\left|2x-3\right|-x+1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=\left|2x-3\right|+x-1+1\\x-1=-\left|2x-3\right|-x+1-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=\left|2x-3\right|+x\\x-1=-\left|2x-3\right|-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1-x=\left|2x-3\right|+x-x\\x-1+x=-\left|2x-3\right|-x+x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}1=\left|2x-3\right|\\2x-1=-\left|2x-3\right|\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|2x-3\right|=1\Rightarrow\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\\2x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\\\left|2x-3\right|=-2x+1\Rightarrow\left[{}\begin{matrix}2x-3=-2x+1\\2x-3=2x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x+2x=1+3\\2x-2x=-1+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x=4\left(giốngth1\right)\\x\in\varnothing\end{matrix}\right.\end{matrix}\right.\)

Vậy x = 2 hoặc x = 1.