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 \(\perp\) )

A = 3.|1 - 2x| - 5

Ta có : \(\left|1-2x\right|\ge0\forall x\)

\(\Rightarrow3.\left|1-2x\right|-5\ge-5\forall x\)

\(\Rightarrow A\ge-5\forall x\)

Dấu "=" xảy ra

\(\Leftrightarrow3.\left|1-2x\right|=0\Leftrightarrow1-2x=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy ...

b,c tương tự

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 )

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

Ta có : \(\left\{{}\begin{matrix}\left|x-1\right|\ge x-1\forall x\\\left|2-x\right|\ge2-x\forall x\end{matrix}\right.\)

\(\Rightarrow\left|x-1\right|+\left|2-x\right|\ge x-1+2-x\forall x\)

\(\Rightarrow A\ge1\forall x\)

Dấu "=" xảy ra

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\2-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\le2\end{matrix}\right.\Leftrightarrow1\le x\le2\)

Vậy MinA = 1 khi \(1\le x\le2\)

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

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

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

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

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

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

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

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

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

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

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

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

a, Ta có : \(\widehat{BAD}+\widehat{BAC}+\widehat{CAE}=180^o\)

Mà : \(\widehat{BAC}=90^o\) (cmt)

\(\Rightarrow\widehat{BAD}+\widehat{CAE}=90^o\)

Lại có : \(\widehat{CAE}+\widehat{ACE}=90^o\) ( tổng 2 góc nhọn trog \(\Delta\) vuông )

\(\Rightarrow\widehat{DAB}=\widehat{ACE}\)

b, Xét \(\Delta DAB\) có :

\(\widehat{ADB}+\widehat{DAB}+\widehat{ABD}=180^o\) ( t/c tổng 3 góc 1 tam giác )

Xét \(\Delta ACE\) có :

\(\widehat{ACE}+\widehat{AEC}+\widehat{CAE}=180^o\) ( t/c tổng 3 góc 1 tam giác )

\(\Rightarrow\widehat{ADB}+\widehat{DAB}+\widehat{ABD}=\widehat{ACE}+\widehat{AEC}+\widehat{CAE}\)

Mà : \(\widehat{DAB}=\widehat{ACE}\left(cmt\right);\widehat{BDE}=\widehat{CEA}\left(=90^o\right)\)

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

Xét \(\Delta\) ABD và \(\Delta\) ACE có :

AB = AC (gt)

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

\(\widehat{DAB}=\widehat{ACE}\left(cmt\right)\)

\(\Rightarrow\Delta CAE=\Delta ABD\left(gcg\right)\)

=> AD = CE ( 2 cạnh tương ứng )

BD = AE ( 2 cạnh tương ứng )

Mà : DE = AD + AE

=> DE = BD + CE

Xét \(\Delta ABC\) có :

\(\widehat{BAC}+\widehat{ABC}+\widehat{ACB}=180^o\) ( t/c tổng 3 góc 1 tam giác)

Mà : \(\widehat{ABC}=70^o\left(gt\right)\)

\(\Rightarrow\widehat{BAC}+\widehat{ACB}=110^o\)

Vì : CK là tia p/g của \(\widehat{ACB}\) (gt)

=> \(\widehat{KCB}=\widehat{KCA}=\dfrac{\widehat{ABC}}{2}\) ( đ/l tia p/g của 1 góc ) (1)

Vì AK là tia p/g của \(\widehat{BAC}\) (gt)

=> \(\widehat{BAK}=\widehat{KAC}=\dfrac{\widehat{BAC}}{2}\) ( đ/l tia p/g của 1 góc ) (2)

Từ (1) và (2) => \(\widehat{KCA}+\widehat{KAC}=\dfrac{\widehat{BAC}}{2}+\dfrac{\widehat{ACB}}{2}\)

\(=\dfrac{\widehat{BAC}+\widehat{ACB}}{2}=\dfrac{110^o}{2}=55^o\)

Xét \(\Delta ACK\) có :

\(\widehat{KAC}+\widehat{AKC}+\widehat{KCA}=180^o\) ( t/c tổng 3 góc 1 tam giác )

Mà : \(\widehat{KCA}+\widehat{KAC}=55^o\left(cmt\right)\)

\(\Rightarrow\widehat{AKC}=125^o\)