Đặt \(\left(4a;5b;-6c\right)=\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}x+y+z=-5\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=25\\\frac{xy+yz+zx}{xyz}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+z^2+2\left(xy+yz+zx\right)=25\\xy+yz+zx=0\end{matrix}\right.\)

\(\Rightarrow x^2+y^2+z^2=25\) hay \(16a^2+25b^2+36c^2=25\)

M = \(\frac{a^4-16}{a^4-4a^3+8a^2-16a+16}\)

=> M = \(\frac{\left(a^2+4\right)\left(a^2-4\right)}{\left(a^4-4a^3+4a^2\right)+\left(4a^2-16a+16\right)}\)

M = \(\frac{\left(a-2\right)\left(a+2\right)\left(a^2+4\right)}{a^2\left(a^2-4a+4\right)+4\left(a^2-4a+4\right)}\)

M = \(\frac{\left(a-2\right)\left(a+2\right)\left(a^2+4\right)}{\left(a^2+4\right)\left(a^2-4a+4\right)}\)

M = \(\frac{\left(a-2\right)\left(a+2\right)\left(a^2+4\right)}{\left(a^2+4\right)\left(a-2\right)^2}\)

M = \(\frac{a+2}{a-2}\)

a)7x² + 6x - 13

= 7x² - 7x + 13x - 13

= (7x² - 7x) + (13x - 13)

= 7x(x - 1) + 13(x - 1)

= (7x + 13)(x - 1)

b) 15(x - y) - 25x + 25y

= 15(x - y) - (25x - 25y)

= 15(x - y) - 25(x - y)

= (15 - 25)(x - y)

= -10(x - y)

c) 36 - 4a² + 20ab - 25b²

= 36 - (4a² - 20ab + 25b²)

= 6² - (2a - 5b)²

= (6 - 2a + 5b)(6 + 2a - 5b)

= (5b - 2a + 6)(2a - 5b + 6)

nhớ tik mik nha. Ak có j ko hiểu thì cứ bình luận ở dưới mik giải cho

a, \(7x^2+6x-13\)

=\(7x^2-7x+13x-13\)

=\(\left(7x^2-7x\right)+\left(13x-13\right)\)

=\(7x\left(x-1\right)+13\left(x-1\right)\)

=\(\left(x-1\right)\left(7x+13\right)\)

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(H^2=(6a-5b)^2\leq [(2a)^2+(-5b)^2](3^2+1^2)=10(4a^2+25b^2)\)

\(\leq 10.10=100\)

\(\Rightarrow H\leq 10\)

Vậy $H_{\max}=10$. Giá trị này đạt tại \(\left\{\begin{matrix} 4a^2+25b^2=10\\ \frac{2a}{3}=-\frac{5b}{1}\end{matrix}\right.\Leftrightarrow x=\frac{3}{2}; y=-\frac{1}{5}\)

heyzo tv

ủa cháu ghi lộn thành lớp 1, sự thật là cháu lớp 4 ròi    ahihi  :D

Bạn rút ra \(2a=\frac{5b+1}{3}\)

Sau đó thế vào \(4a^2+25b^2=\left(2a\right)^2+\left(5b\right)^2\)

Được : \(\frac{50b^2+10b+1}{9}=\frac{2\left[\left(5b^2\right)+5b\right]+1}{9}\)

=\(\frac{2\left[\left(5b^2\right)+2\cdot\frac{5}{2}b^{ }+\frac{25}{4}-\frac{25}{4}\right]+1}{9}\)

=\(\frac{2\left[5b+\frac{25}{2}\right]^2-\frac{23}{2}}{9}\ge\frac{-\frac{23}{2}}{9}=\frac{-23}{18}\)

Dấu = khi b=-5/2 và a=-23/12

\(=\left(16a^2-16a^2+40a-25\right)^2-\left(9a^2-9a^2+30a-25\right)^2\)

\(=\left(40a-25\right)^2-\left(30a-25\right)^2\)

\(=\left(40a-25-30a+25\right)\left(40a-25+30a-25\right)\)

\(=10a\left(70a-50\right)\)

\(=100a\left(7a-5\right)\)

Đặt x = 2a; y = -5b.

Áp dụng đẳng thức Bunhiacopski ta có:

\(\left(3x+y\right)^2\le\left(x^2+y^2\right)\left(9+1\right)\Rightarrow x^2+y^2\ge\frac{1}{10}\)

Hay: \(4a^2+25b^2\ge\frac{1}{10}\)

Dấu "=" xảy ra \(\Leftrightarrow\frac{3}{x}=\frac{1}{y}\Leftrightarrow3y=x\Leftrightarrow-15b=2a\Leftrightarrow6a=-45b\)

\(\Leftrightarrow b=-\frac{1}{50};a=\frac{3}{20}\)