a) Ta có : 6x² - 13x + 6 = 6x² - 9x - 4x + 6 = 3x(2x - 3) - 2(2x - 3) = (3x - 2)(2x - 3)

b) Ta có: 6x² + 7x - 3 = 6x² + 9x - 2x - 3 = 3x(2x + 3) - (2x + 3) = (3x - 1)(2x + 3)

\(a,6x^2-13x+6\)

\(=6x^2-9x-4x+6\)

\(=3x\cdot\left(2x-3\right)-x\cdot\left(2x-3\right)\)

\(=\left(2x-3\right)\cdot\left(3x-x\right)\)

\(=\left(2x-3\right)\cdot2x\)

\(b,6x^2+7x-3\)

\(=6x^2-2x+9x-3\)

\(=2x\cdot\left(3x-1\right)+3\cdot\left(3x-1\right)\)

\(=\left(3x-1\right)\cdot\left(2x+3\right)\)

\(\frac{19}{x+y}+\frac{19}{y+z}+\frac{19}{z+x}=\frac{133}{10}\)

\(\Rightarrow19\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{133}{10}\)

\(\Rightarrow\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{7}{10}\)

\(\frac{7x}{y+z}+\frac{7y}{z+x}+\frac{7z}{x+y}=\frac{133}{10}\)

\(\Rightarrow7\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=\frac{133}{10}\)

\(\Rightarrow\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=\frac{19}{10}\)

\(\Rightarrow\left(\frac{x}{y+z}+1+\frac{y}{z+x}+1+\frac{z}{x+y}+1\right)=\frac{19}{10}+3\)

\(\Rightarrow\left(\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{x+y}\right)=\frac{49}{10}\)

\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)=\frac{49}{10}\)

\(\Rightarrow\left(x+y+z\right).\frac{7}{10}=\frac{49}{10}\)

\(\Rightarrow x+y+z=7\)

Vậy x + y + z = 7

de la gi vay ban

phan h da thuc thanh nhan tu

Xét \(\Delta ABC\)có :

N là trung điểm AB

P là trung điểm AC

=> NP là đường trung bình 

=> NP//BC

Xét \(\Delta ABC\)ta có :

M là trung điểm BC

N là trung điểm AB

=> NM là đường trung bình 

=> NM//AC, NM =\(\frac{AC}{2}\)

Xét \(\Delta AHC\)có :

HP là trung tuyến 

=> HP = AP = PC = \(\frac{AC}{2}\)

=> NM = HP( \(=\frac{AC}{2}\))

Xét tứ giác NPMH có :

NP//HM ( NP//BC ; N , M \(\in\)BC)

NM = PH 

=> NPMH là hình thang cân 

                                                         Bài giải

Ta có : \(\frac{2}{7}x\left(3y-1\right)-\frac{2}{7}y\left(3y-1\right)\)

\(=\frac{2}{7}\left(3y-1\right)\left(x-y\right)\)

Đề thiếu nha bạn !

\(B=2x^2-4x-8=2\left(x^2-2x-4\right)\)

\(=2\left(x^2-2x+1-5\right)\)

\(=2\left[\left(x-1\right)^2-5\right]\)

\(=2\left(x-1\right)^2-10\ge-10\)

Vậy \(B_{min}=-10\Leftrightarrow x-1=0\Leftrightarrow x=1\)

\(F=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)

Đặt \(x^2+5x+4=t\)

\(\RightarrowĐT=t\left(t+2\right)=t^2+2t+1-1\)

\(=\left(t+1\right)^2-1\ge-1\)

hay \(\left(x^2+5x+5\right)^2-1\ge-1\)

Vậy \(F_{min}=-1\Leftrightarrow x^2+5x+5=0\)

\(\Leftrightarrow x^2+5x+\frac{25}{4}-\frac{5}{4}=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{5}{2}=\sqrt{\frac{5}{4}}\\x+\frac{5}{2}=-\sqrt{\frac{5}{4}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{5}{4}}-\frac{5}{2}\\x=-\sqrt{\frac{5}{4}}-\frac{5}{2}\end{cases}}\)

\(G=4x-x^2=-\left(x^2-4x+4-4\right)\)

\(=-\left[\left(x-2\right)^2-4\right]=-\left(x-2\right)^2+4\le4\)

Vậy \(G_{max}=4\Leftrightarrow x-2=0\Leftrightarrow x=2\)

\(H=25-x-5x^2=-5\left(x^2+\frac{x}{5}-5\right)\)

\(=-5\left(x^2+2x.\frac{1}{10}+\frac{1}{100}-\frac{501}{100}\right)\)

\(=-5\left[\left(x+\frac{1}{10}\right)^2-\frac{501}{100}\right]\)

\(=-5\left(x+\frac{1}{10}\right)^2+\frac{101}{20}\le\frac{101}{2}\)

Vậy \(H_{max}=\frac{101}{2}\Leftrightarrow x+\frac{1}{10}=0\Leftrightarrow x=-\frac{1}{10}\)

B = 2\(x^2\) - 4\(x\) - 8

B = 2(\(x^2\) - 2\(x\) + 4)  - 16

B = 2(\(x-2\))² - 16 

Vì (\(x-2\))² ≥ 0 ∀ \(x\) ⇒ 2(\(x-2\))² ≥ 0 ∀ \(x\)

⇒ 2(\(x-2\))²  - 16 ≥ -16 ∀ \(x\)

Dấu bằng xảy ra khi  (\(x-2\))² = 0 ⇒ \(x-2=0\) ⇒ \(x=2\)

Vậy B_min = -16 khi \(x=2\)

Tìm min của C biết:

C = \(x^2\) - 2\(xy\) + 2y² + 2\(x\) - 10y + 17

C = (\(x^2\) - 2\(xy\) + y²) + 2(\(x\) - y) + y² - 8y + 16 + 1

C = (\(x\) - y)² + 2(\(x\) - y) + 1  + (y² - 8y + 16) 

C = (\(x-y+1\))² + (y - 4)² 

Vì (\(x\) - y + 1)² ≥ 0 ∀ \(x;y\); (y - 4)² ≥ 0 ∀ y

Dấu bằng xảy ra khi: \(\left\{{}\begin{matrix}x-y+1=0\\y-4=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x-y+1=0\\y=4\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x-4+1=0\\y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=-1+4\\y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)

Vậy C_min = 0 khi (\(x;y\)) = (3; 4)