#)Giải :

a) Đặt \(\frac{x}{3}=\frac{y}{7}=k\Rightarrow\hept{\begin{cases}x=3k\\y=7k\end{cases}}\)

\(\Rightarrow3k.7k=84\Rightarrow21k^2=84\Rightarrow k^2=4\Rightarrow\orbr{\begin{cases}k=2\\k=-2\end{cases}}\)

Nếu k = 2 \(\Rightarrow\hept{\begin{cases}\frac{x}{3}=2\\\frac{y}{7}=2\end{cases}\Rightarrow\hept{\begin{cases}x=6\\y=14\end{cases}}}\)

Nếu k = -2 \(\Rightarrow\hept{\begin{cases}\frac{x}{3}=-2\\\frac{y}{7}=-2\end{cases}\Rightarrow\hept{\begin{cases}x=-6\\y=-14\end{cases}}}\)

Vậy ...

M đâu vậy

Lộn. CM: AHK vuông cân

\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)

\(\Leftrightarrow\)\(\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\)

\(\Leftrightarrow\) \(\left(x+3\right).\left(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)

\(\Leftrightarrow\) \(x+3=0\) ( Vì \(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\ne0\) )

\(\Leftrightarrow\) \(x=-3\)

Vậy x = -3

\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)

\(\Rightarrow\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\)

\(\Rightarrow\left(x+3\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)

\(\Rightarrow x+3=0\Leftrightarrow x=-3\)

\(a,x^2-4x+6\)

\(=x^2-2.2.x+2^2+2\)

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

\(\left(x-2\right)^2\ge0\Rightarrow\left(x-2\right)^2+2\ge2\)

\(\Rightarrow\)biểu thức nhận giá trị dương với mọi x

\(b,x^2+5x+10\)

\(=x^2+2.\frac{5}{2}.x+\left(\frac{5}{2}\right)^2+\frac{15}{4}\)

\(=\left(x+\frac{5}{2}\right)^2+\frac{15}{4}\)

\(\left(x+\frac{5}{2}\right)^2\ge0\Rightarrow\left(x+\frac{5}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}\)

\(\Rightarrow\)biểu thức luôn nhận giá trị dương với mọi x

\(c,4x^2-4xy+2y^2+3\)

\(=\left(2x\right)^2-2.2x.y+y^2+y^2+3\)

\(=\left(2x-y\right)^2+y^2+3\)

\(\hept{\begin{cases}\left(2x+y\right)^2\ge0\\y^2\ge0\end{cases}\Rightarrow\left(2x+y\right)^2+y^2\ge0}\)

\(\Rightarrow\left(2x+y\right)^2+y^2+3\ge3\)

\(\Rightarrow\)biểu thức luôn nhận giá trị dương với mọi x

\(a,x^2-4x+6=x^2-4x+4+2=\left(x-2\right)^2+2\)

\(\left(x-2\right)^2\ge0\Rightarrow\left(x-2\right)^2+2\ge2>0\)

\(b,x^2+5x+10=x^2+2\cdot\frac{5}{2}\cdot x+\frac{25}{4}+\frac{15}{4}=\left(x+\frac{5}{2}\right)^2+\frac{15}{4}\)

\(\left(x+\frac{5}{2}\right)^2\ge0\Rightarrow\left(x+\frac{5}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\)

a) |3x - 2| - 5 = 7

=> |3x - 2| = 7 + 5

=> |3x - 2| = 12

=> \(\orbr{\begin{cases}3x-2=12\\3x-2=-12\end{cases}}\)

=> \(\orbr{\begin{cases}3x=14\\3x=-10\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{14}{3}\\x=-\frac{10}{3}\end{cases}}\)

b) |3x - 1| - |x + 2| = 0

=> |3x - 1| = |x + 2|

=> \(\orbr{\begin{cases}3x-1=x+2\\3x-1=-x-2\end{cases}}\)

=> \(\orbr{\begin{cases}2x=3\\4x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{4}\end{cases}}\)

thanks bn

\(a;0,25-\frac{1}{2}\left|1,5-x\right|=2,5\)

\(\Leftrightarrow\frac{1}{2}\left|1,5-x\right|=0,25-2,5\)

\(\Leftrightarrow\frac{1}{2}\left|1,5-x\right|=-2,25\)

\(\Leftrightarrow\left|1,5-x\right|=-2,25\cdot2=-4,5\)

Mà \(\left|1,5-x\right|\ge0\)Nên suy ra |1,5-x|=-4,5 là vô lý

\(b;\left|x+\frac{1}{6}\right|\cdot0,75+\frac{1}{4}=2\frac{1}{3}\)

\(\Leftrightarrow\left|x+\frac{1}{6}\right|\cdot\frac{3}{4}=\frac{7}{3}-\frac{1}{4}\)

\(\Leftrightarrow\left|x+\frac{1}{6}\right|\cdot\frac{3}{4}=\frac{25}{12}\)

\(\Leftrightarrow\left|x+\frac{1}{6}\right|=\frac{25}{12}\cdot\frac{4}{3}\)

\(\Leftrightarrow\left|x+\frac{1}{6}\right|=\frac{25}{9}\Leftrightarrow x+\frac{1}{6}=\pm\frac{25}{9}\)

TH1:\(x+\frac{1}{6}=\frac{25}{9}\)

\(\Leftrightarrow x=\frac{25}{9}-\frac{1}{6}=\frac{47}{18}\)

TH2:\(x+\frac{1}{6}=-\frac{25}{9}\)

\(\Leftrightarrow x=-\frac{25}{9}-\frac{1}{6}=\frac{-53}{18}\)

Vậy \(x=\frac{47}{18};-\frac{53}{18}\)

a) 3x(x + 2) + 4x(-2x + 3) + (2x - 3)(3x + 1)

= 3x² + 6x - 8x² + 12x + 6x² + 2x - 9x - 3

= (3x² - 8x² + 6x²) + (6x + 12x + 2x - 9x) - 3

= x³ + 11x - 3

b) (x² + 1)(x² - x + 2) - (x² - 1)(x² + x - 2)

= x⁴ - x³ + 3x² - x + 2 - x⁴ - x³ + 3x² + x - 2

= (x⁴ - x⁴) + (-x³ - x³) + (3x² + 3x²) + (-x + x) + (2 - 2)

= -2x³ + 6x²

c) (-2x - 3)² + (3x + 2)² + (4x + 1)

= 4x² + 12x + 9 + 9x² + 12x + 4 + 4x + 1 

= (4x² + 9x²) + (12x + 12x + 4x) + (9 + 4 + 1)

= 13x² + 28x + 14

\(\left|2x\right|-\left|-2,5\right|=\left|-7,5\right|\forall x>0\)

\(\Rightarrow2x-2,5=7,5\)

\(\Rightarrow2x=10\)

\(\Rightarrow x=5\)

\(|2x|-|-2,5|=|-7,5|\)

\(\Rightarrow|2x|-2=7,5\)

\(\Rightarrow|2x|=10\)

\(\Rightarrow\orbr{\begin{cases}2x=10\\2x=-10\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}}\)

\(KL:x\in\left\{\pm5\right\}\)

Theo đề bài ta có :

góc ABD = góc DBC

mà AB // Dy nên :

góc ABD = góc BDy

góc DBC = góc ADB

vì Bx // Et nên :

góc BDE = góc DEt

góc DBC = góc tEC

=> góc tEC = góc DEt

=> Et là tia phân giác của góc CED

đây giải có khi sai nên trước khi chép vào cân nhắc kĩ nhé

bạn ơi bạn biết vẽ hình ko