Tự vẽ hình

Vì H là trung điểm của BC (gt)

=> BH = HC

Xét ΔAHB và ΔAHC có :

AH chung

BH = HC (cmt)

AB = AC (gt)

=> ΔAHB = ΔAHC ( c.c.c)

=> \(\widehat{AHB}=\widehat{AHC}\) ( 2 góc t/ứ )

Mà : \(\widehat{AHB}+\widehat{AHC}=180^o\) ( 2 góc kề bù )

=> \(\widehat{AHB}=90^o\) => AH \(\perp\) BC ( đ/n 2 đg thẳng \(\perp\) )

à mình biết bài này đáp án là 216 mét vuông bạn hỏi diện tích chứ gì

Vì : \(DE\perp BC\left(gt\right)\Rightarrow\widehat{BED}=90^o\) (đ/n 2 đg thẳng \(\perp\) )

\(\Delta ABC\) vuông tại A => \(\widehat{BAC}=90^o\) (đ/n △ vuông )

Xét \(\Delta ABD\) vuông và \(\Delta EBD\) vuông có :

BD chung

\(\widehat{BAD}=\widehat{BED}\) (= 90^o)

\(\Rightarrow\Delta ABD=\Delta EBD\) ( cạnh huyền-góc nhọn)

\(\Rightarrow\widehat{ADB}=\widehat{EDB}\) ( 2 góc t/ứ )

Mà : \(\widehat{ADF}=\widehat{EDC}\) ( 2 góc đối dỉnh )

\(\Rightarrow\widehat{ADB}+\widehat{ADF}=\widehat{EDB}+\widehat{EDC}\Rightarrow\widehat{BDF}=\widehat{BDC}\)

Có : BD là tia p/g của \(\widehat{ABC}\)

\(\Rightarrow\widehat{ABD}=\widehat{DBC}\)

Xét \(\Delta BFD\) và \(\Delta BCD\) có :

BD chung

\(\widehat{ABD}=\widehat{DBC}\left(cmt\right)\)

\(\widehat{BDF}=\widehat{BDC}\left(cmt\right)\)

\(\Rightarrow\Delta BFD=\Delta BCD\left(g.c.g\right)\)

=> DF = DC ( 2 cạnh t/ứ )

a) |2x - 1| - 3 = 5

|2x-  1| = 8

TH1: 2x - 1 = 8

2x=  9

x = 9/2

TH2: 2x -1 = -8

2x = -7

x = -7/2 

\(2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\)

\(=1-\dfrac{1}{2n+1}\Rightarrow A=\left(1-\dfrac{1}{2n+1}\right)\cdot\dfrac{1}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2n+1}< \dfrac{1}{2}\)

Vậy A < \(\dfrac{1}{2}\)

Vì : \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) tồn tại => a,b,c khác 0

Ta có : \(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)

\(\Rightarrow\dfrac{a\left(bz-cy\right)}{a^2}=\dfrac{b\left(cx-az\right)}{b^2}=\dfrac{c\left(ay-bx\right)}{c^2}\)

\(\Rightarrow\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}\)

Áp dụng t/c dãy tỉ số bằng = nhau ta có :

\(\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}\)

\(=\dfrac{\left(abz-abz\right)+\left(bcx-bcx\right)+\left(acy-acy\right)}{a^2+b^2+c^2}=\dfrac{0}{a^2+b^2+c^2}=0\)

Do đó : \(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}=0\)

\(\Rightarrow\left\{{}\begin{matrix}bz-cy=0\\cx-az=0\\ay-bx=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}bz=cy\\cx=az\\ay=bx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{y}{b}=\dfrac{z}{c}\\\dfrac{x}{a}=\dfrac{z}{c}\\\dfrac{x}{a}=\dfrac{y}{b}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)

\(\left(4x-1\right)^2=\left(1-4x\right)^4\)

\(\Rightarrow\left(4x-1\right)^2=\left(4x-1\right)^4\)

\(\Rightarrow\left(4x-1\right)^2\left[1-\left(4x-1\right)^2\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(4x-1\right)^2=0\\1-\left(4x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x-1=0\\\left(4x-1\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x=1\\4x-1\in\left\{1;-1\right\}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x\in\left\{\dfrac{1}{2};0\right\}\end{matrix}\right.\)

Chtt ko co cau tra loi bn oi!!!! Neu bn biet thi giai chi tiet gium minh , minh tick cho nha Tuan !!!!!!!!

vì b + b = b => b = 0 (0+0=0)