Lời giải:

Xét tam giác $BAM$ và $CDM$ có:

$BM=CM$

$AM=DM$

$\widehat{BMA}=\widehat{CMD}$ (đối đỉnh)

$\Rightarrow \triangle BAM=\triangle CDM$ (c.g.c)

$\Rightarrow AB=CD$ và $\widehat{BAM}=\widehat{CDM}$

Mà 2 góc này ở vị trí so le trong nên $AB\parallel CD$

$AB\perp AC$ nên $CD\perp AC\Rightarrow \widehat{DCA}=90^0$

Xét tam giác $BAC$ và $DCA$ có:

$\widehat{BAC}=\widehat{DCA}=90^0$

$BA=CD$ (cmt)

$AC$ chung

$\Rightarrow \triangle BAC=\triangle DCA$ (c.g.c)

$\Rightarrow BC=DA$

$\Rightarrow BC:2=DA:2\Rightarrow BM=AM$

$\Rightarrow MBA$ cân tại $M\Rightarrow \widehat{MBA}=\widehat{MAB}$ 

Hay $\widehat{ABC}=\widehat{BAD}$

Hình vẽ:

\(a\)) Đặt \(6x=10y=15z=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{k}{6}\\y=\dfrac{k}{10}\\z=\dfrac{k}{15}\end{matrix}\right.\) \(\Rightarrow\dfrac{k}{6}+\dfrac{k}{10}+\dfrac{k}{15}=90\)
\(\Leftrightarrow\dfrac{k}{3}=90\Leftrightarrow k=270\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{270}{6}=45\\y=\dfrac{270}{10}=27\\z=\dfrac{270}{15}=18\end{matrix}\right.\)
Vậy \(x=45;y=27;z=18\)
\(b\)) Đặt \(9x=3y=2z=q\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{q}{9}\\y=\dfrac{q}{3}\\z=\dfrac{q}{2}\end{matrix}\right.\) \(\Rightarrow\dfrac{q}{9}-\dfrac{q}{3}+\dfrac{q}{2}=50\)
\(\Rightarrow\dfrac{5q}{18}=50\) \(\Leftrightarrow q=180\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{180}{9}=20\\y=\dfrac{180}{3}=60\\z=\dfrac{180}{2}=90\end{matrix}\right.\)
Vậy \(x=20;y=60;z=90\)
\(c\)) Đặt \(2x=3y=-2z=r\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{r}{2}\\y=\dfrac{r}{3}\\z=-\dfrac{r}{2}\end{matrix}\right.\) \(\Rightarrow2\cdot\dfrac{r}{2}-3\cdot\dfrac{r}{3}+4\cdot\left(-\dfrac{r}{2}\right)=48\)
\(\Leftrightarrow-2r=48\) \(\Leftrightarrow r=-24\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-24}{2}=-12\\y=\dfrac{-24}{3}=-8\\z=-\dfrac{-24}{2}=12\end{matrix}\right.\)
Vậy \(x=-12;y=-8;z=12\)
\(d\)) Đặt \(\dfrac{x+1}{3}=\dfrac{y+2}{4}=\dfrac{z+3}{5}=u\)
\(\Rightarrow\left\{{}\begin{matrix}x=3u-1\\y=4u-2\\z=5u-3\end{matrix}\right.\) \(\Rightarrow3u-1+4u-2+5u-3=30\)
\(\Leftrightarrow12u=36\) \(\Leftrightarrow u=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\cdot3-1=8\\y=4\cdot3-2=10\\z=5\cdot3-3=12\end{matrix}\right.\)
Vậy \(x=8;y=10;z=12\)
\(e\)) Đặt \(\dfrac{x-1}{3}=\dfrac{x-2}{4}=\dfrac{z-3}{5}=p\)
\(\Rightarrow\left\{{}\begin{matrix}x=3p+1\\y=4p+2\\z=5p+3\end{matrix}\right.\) \(\Rightarrow3p+1+4p+2+5p+3=30\)
\(\Leftrightarrow12p=24\) \(\Leftrightarrow p=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\cdot2+1=7\\y=4\cdot2+2=10\\z=5\cdot2+3=13\end{matrix}\right.\)
Vậy \(x=7;y=10;z=13\)
\(g\)) \(\left\{{}\begin{matrix}\dfrac{x}{4}=\dfrac{y}{3}\\x:y=12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x:y=\dfrac{4}{3}\\x:y=12\end{matrix}\right.\) (Vô lí)
Vậy không có giá trị \(x,y\) thỏa mãn
\(h\)) Đặt \(-6x=-15y=10z=a\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{a}{6}\\y=-\dfrac{a}{15}\\z=\dfrac{a}{10}\end{matrix}\right.\) \(\Rightarrow\left(-\dfrac{a}{6}\right)\cdot\left(-\dfrac{a}{15}\right)\cdot\dfrac{a}{10}=240\)
\(\Leftrightarrow\dfrac{a^3}{900}=240\) \(\Leftrightarrow a^3=216000\) \(\Leftrightarrow a=60\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{60}{6}=-10\\y=-\dfrac{60}{15}=-4\\z=\dfrac{60}{10}=6\end{matrix}\right.\)
Vậy \(x=-10;y=-4;z=6\)
\(i\)) Đặt \(-18x=-12y=24z=s\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{s}{18}\\y=-\dfrac{s}{12}\\z=\dfrac{s}{24}\end{matrix}\right.\) \(\Rightarrow\left(-\dfrac{s}{18}\right)\cdot\left(-\dfrac{s}{12}\right)\cdot\dfrac{s}{24}=576\)
\(\Leftrightarrow\dfrac{s^3}{5184}=576\) \(\Leftrightarrow s^3=2985984\) \(\Leftrightarrow s=144\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{144}{18}=-8\\y=-\dfrac{144}{12}=-12\\z=\dfrac{144}{24}=6\end{matrix}\right.\)
Vậy \(x=-8;y=-12;z=6\)
\(k\)) \(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{2}=\dfrac{z}{5}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2y}{3}\\z=\dfrac{5y}{2}\end{matrix}\right.\)\(\Rightarrow\dfrac{2y}{3}+y+\dfrac{5y}{2}=50\)
\(\Leftrightarrow\dfrac{25y}{6}=50\) \(\Leftrightarrow y=12\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2\cdot12}{3}=8\\z=\dfrac{5\cdot12}{2}=30\end{matrix}\right.\)
Vậy \(x=8;y=12;z=30\)
\(l\)) \(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\2y=3z\end{matrix}\right.\) \(\Rightarrow x=z=\dfrac{2y}{3}\)\(\Rightarrow\dfrac{2y}{3}+y+\dfrac{2y}{3}=49\)
\(\Leftrightarrow\dfrac{7y}{3}=49\) \(\Leftrightarrow y=21\)
\(\Rightarrow x=z=\dfrac{2\cdot21}{3}=14\)
Vậy \(x=14;y=21;z=14\).

Đặt \(\dfrac{a+b}{3}=\dfrac{b+c}{4}=\dfrac{c+a}{5}=t\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=3t\\b+c=4t\\c+a=5t\end{matrix}\right.\)
\(\Rightarrow\left(a+b\right)+\left(b+c\right)+\left(c+a\right)=3t+4t+5t\)
\(\Leftrightarrow2\left(a+b+c\right)=12t\)
\(\Leftrightarrow a+b+c=6t\)
+ \(\left\{{}\begin{matrix}a+b=3t\\a+b+c=6t\end{matrix}\right.\) \(\Rightarrow3t+c=6t\) \(\Leftrightarrow c=3t\)
+ \(\left\{{}\begin{matrix}b+c=4t\\a+b+c=6t\end{matrix}\right.\) \(\Rightarrow a+4t=6t\) \(\Leftrightarrow a=2t\)
+ \(\left\{{}\begin{matrix}c+a=5t\\a+b+c=6t\end{matrix}\right.\) \(\Rightarrow b+5t=6t\) \(\Leftrightarrow b=t\)
Thay \(a=2t;b=t;c=3t\) vào \(M\) ta được
\(M=10\cdot2t+t-7\cdot3t+2017=20t+t-21t+2017=2017\)
Vậy \(M=2017\)

 

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{3x-2y}{4}=\dfrac{4y-3z}{2}=\dfrac{2z-4x}{3}=\dfrac{12x-8y}{16}=\dfrac{8y-6z}{4}\)

\(=\dfrac{6z-12x}{9}=\dfrac{12x-8y+8y-6z+6z-12x}{16+4+9}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-2y}{4}=0\\\dfrac{4y-3z}{2}=0\\\dfrac{2z-4x}{3}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}3x=2y\\4y=3z\\2z=4x\end{matrix}\right.\)

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

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2y}{6}=\dfrac{3z}{12}=\dfrac{x-2y+3z}{2-6+12}=\dfrac{8}{8}=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.1=2\\y=3.1=3\\z=4.1=4\end{matrix}\right.\)

Ta có: \(\dfrac{3x-2y}{4}=\dfrac{4y-3z}{2}=\dfrac{2z-4x}{3}\)

hay \(\dfrac{12x-8y}{16}=\dfrac{8y-6z}{4}=\dfrac{6z-12x}{9}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{12x-8y}{16}=\dfrac{8y-6z}{4}=\dfrac{6z-12x}{9}=\dfrac{12x-8y+8y-6z+6z-12x}{16+4+9}=\dfrac{0}{29}=0\)

Do đó:

\(\dfrac{3x-2y}{4}=0\Rightarrow3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\left(1\right)\)

\(\dfrac{4y-3z}{2}=0\Rightarrow4y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\left(2\right)\)

\(\dfrac{2z-4x}{3}=0\Rightarrow2z=4x\Rightarrow\dfrac{z}{4}=\dfrac{x}{2}\left(3\right)\)

Từ (1), (2) và (3) suy ra: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}=\dfrac{x-2y+3z}{2-6+12}=\dfrac{8}{8}=1\)

Do đó:

\(\dfrac{x}{2}=1\Rightarrow x=2.1=2\)

\(\dfrac{y}{3}=1\Rightarrow y=3.1=3\)

\(\dfrac{z}{4}=1\Rightarrow z=4.1=4\)

Vậy x = 2; y = 3; z = 4.

\(#NqHahh\) 

Lời giải:
Theo bài ra ta có:

$\frac{x}{3}=\frac{y}{5}; \frac{y}{4}=\frac{z}{5}$

$\Rightarrow \frac{x}{12}=\frac{y}{20}=\frac{z}{25}$

Áp dụng TCDTSBN:

$\frac{x}{12}=\frac{y}{20}=\frac{z}{25}=\frac{x+y+z}{12+20+25}=\frac{456}{57}=8$

$\Rightarrow x=12.8=96; y=20.8=160; z=25.8=200$

Lời giải:
a. Xét tam giác $BAD$ và $BED$ có:

$\widehat{BAD}=\widehat{BED}=90^0$

$BD$ chung

$\widehat{ABD}=\widehat{EBD}$ (do $BD$ là phân giác $\widehat{ABC}$)

$\Rightarrow \triangle BAD=\triangle BED$ (ch-gn)

$\Rightarrow AB=EB$

b.

$AD< AC$ (do $D$ nằm giữa $A,C$)

$AC< BC$ (do $BC$ là cạnh huyền)

$\Rightarrow AD< BC$

c.

Tam giác $ABH$ vuông tại $H$ nên $BH< BA$ (do $BA$ là cạnh huyền)

Mà $BA=BE$ nên $BH< BE(1)$

Tam giác $ABC$ vuông tại $A$

$\Rightarrow BA< BC$. Mà $BA=BE$ nên $BE< BC(2)$

Từ $(1); (2)\Rightarrow BH< BE< BC$

Hình vẽ:

Xét + \(\left|x-1\right|+\left|x-1996\right|\)
\(=\left|x-1\right|+\left|1996-x\right|\ge\left|x-1+1996-x\right|=1995\)
Dấu \("="\) xảy ra \(\Leftrightarrow\left(x-1\right)\left(1996-x\right)\ge0\)
\(\Rightarrow1\le x\le1996\)
       + \(\left|x-2\right|+\left|x-1995\right|\)
\(=\left|x-2\right|+\left|1995-x\right|\ge\left|x-2+1995-x\right|=1993\)
Dấu \("="\) xảy ra \(\Leftrightarrow\left(x-2\right)\left(1995-x\right)\ge0\)
\(\Rightarrow2\le x\le1995\)
           \(...\)
        + \(\left|x-997\right|+\left|x-998\right|\)
\(=\left|x-997\right|+\left|998-x\right|\ge\left|x-997+998-x\right|=1\)
Dấu \("="\) xảy ra \(\Leftrightarrow\left(x-997\right)\left(998-x\right)\ge0\)
\(\Rightarrow997\le x\le998\)
Do đó nên
\(\left(\left|x-1\right|+\left|x-1996\right|\right)+\left(\left|x-2\right|+\left|x-1995\right|\right)+...+\left(\left|x-997\right|+\left|x-998\right|\right)\ge1995+1993+...+1\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+...+\left|x-1996\right|\ge\left(1+1995\right)\left[\left(1995-1\right):2+1\right]:2=996004\)
Dấu \("="\) xảy ra \(\Leftrightarrow997\le x\le998\)
Vậy giá trị nhỏ nhất của \(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+...+\left|x-1996\right|\) là \(996004\) khi \(997\le x\le998\)

Lời giải:

Xét tam giác $ABD$ và $ACD$ có:

$AD$ chung

$\widehat{ADB}=\widehat{ADC}=90^0$

$AB=AC$ (do $ABC$ cân tại $A$)

$\Rightarrow \triangle ABD=\triangle ACD$ (ch-cgv)

$\Rightarrow \widehat{BAD}=\widehat{CAD}$

$\Rightarrow AD$ là phân giác $\widehat{BAC}$

Cũng từ tam giác bằng nhau trên 

$\Rightarrow BD=DC$

$\Rightarrow D$ là trung điểm $BC$

Hình vẽ: