.Tổng tất cả các góc:

\(120^o+140^o+100^o=360^o\)

Đây đúng với định lí song song.Khi đó dễ dàng có được

\(AB//DE\)

Theo đề bài ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\)

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

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)

Hay \(a=b=c\)

Thay vào bài toán:

\(\left(2a+70b+1945c\right)^{2018}=\left(2a+70a+1945a\right)^{2018}=2017a^{2018}\)

Lại có:

\(2017^{2018}.a^{39}.b^{13}.b^{1975}=2017^{2018}.a^{39}.a^{13}.a^{1975}=2017^{2018}.a^{2018}=2017a^{2018}\)Ta có đpcm

Áp dụng định lí tổng \(3\) góc trong 1 tam giác bằng \(180^o\) ta có:

\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\Leftrightarrow\widehat{B}+\widehat{C}=180^o-\widehat{A}=180^o-60^o=120^o\)

Mà \(\widehat{B}=2\widehat{C}\Leftrightarrow\dfrac{\widehat{B}}{2}=\dfrac{\widehat{C}}{1}\)

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

\(\dfrac{\widehat{B}}{2}=\dfrac{\widehat{C}}{1}=\dfrac{\widehat{B}+\widehat{C}}{2+1}=\dfrac{120^o}{3}=40^o\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=40^o.2=80^o\\\widehat{C}=40^o.1=40^o\end{matrix}\right.\)

\(\left(2ab-1\right)^2+\left(3bc-2\right)^2+\left(4ac-3\right)^2\ge0\forall x;b;c\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}2ab=1\\3bc=2\\4ac=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}ab=\dfrac{1}{2}\\bc=\dfrac{2}{3}\\ac=\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left(abc\right)^2=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}=\dfrac{1}{4}\)

\(\Rightarrow\left[{}\begin{matrix}abc=\dfrac{1}{2}\\abc=-\dfrac{1}{2}\end{matrix}\right.\)

Đến đây tự tìm được \(a;b;c\)

Đặt:

\(\dfrac{x}{3}=\dfrac{y}{7}=t\Leftrightarrow\left\{{}\begin{matrix}x=3t\\y=7t\end{matrix}\right.\)

Hay \(3t.7t=84\Leftrightarrow21t^2=84\Leftrightarrow t^2=4\Leftrightarrow t=\pm2\)

Với \(t=2\) thì \(\left\{{}\begin{matrix}x=2.3=6\\y=2.4=17\end{matrix}\right.\)

Với \(t=-2\) thì \(\left\{{}\begin{matrix}x=-2.3=-6\\y=-2.7=-14\end{matrix}\right.\)

\(A=\left|x-1\right|+\left|x-3\right|+\left|x-5\right|+\left|x-7\right|=8\)

\(A=\left|x-1\right|+\left|x-7\right|+\left|x-3\right|+\left|x-5\right|\)

\(A=\left|x-1\right|+\left|7-x\right|+\left|x-3\right|+\left|5-x\right|\)

\(A\ge \left|x-1+7-x\right|+\left|x-3+5-x\right|\)

\(A\ge6+2=8\)
Dấu "=" xảy ra khi: \(3\le x\le5\)

\(\left|x\right|+\left|y\right|\ge\left|x+y\right|\)

\(\Rightarrow\left(\left|x\right|+\left|y\right|\right)^2\ge\left|x+y\right|^2\)

\(\Rightarrow x^2+2\left|xy\right|+y^2\ge x^2+2xy+y^2\)

\(\Rightarrow2\left|xy\right|\ge2xy\left(luôn-đúng\right)\)

\(\left(x-1\right)\left(x-2\right)\left(x-5\right)\left(x-6\right)\)

\(=\left(x-1\right)\left(x-6\right)\left(x-2\right)\left(x-5\right)\)

\(=\left(x^2-6x-x+6\right)\left(x^2-5x-2x+10\right)\)

\(=\left(x^2-7x+6\right)\left(x^2-7x+10\right)\)

\(=\left(x^2-7x+8-2\right)\left(x^2-7x+8+2\right)\)

\(=\left(x^2-7x+8\right)^2-4\ge-4\)

Dấu "=" xảy ra khi:

\(x^2-7x+8=0\)

Đề là ntn:

\(A=49\left(\dfrac{1}{2.9}+\dfrac{1}{9.16}+\dfrac{1}{16.23}+...+\dfrac{1}{65.72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=7.\dfrac{35}{72}:\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=\dfrac{245}{72}:\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=\dfrac{735}{72}-\dfrac{7}{36}=\dfrac{735}{72}-\dfrac{14}{36}=\dfrac{721}{36}\)

Đặt:

\(\dfrac{a}{b}=\dfrac{c}{d}=t\Leftrightarrow\left\{{}\begin{matrix}a=bt\\c=dt\end{matrix}\right.\)

\(\dfrac{2a-5b}{3a}=\dfrac{2bt-5b}{3bt}=\dfrac{b\left(2t-5\right)}{3bt}=\dfrac{2t-5}{3t}\)

\(\dfrac{2c-5d}{3c}=\dfrac{2dt-5d}{3dt}=\dfrac{d\left(2t-5\right)}{3dt}=\dfrac{2t-5}{3t}\)

Ta có đpcm