\(\dfrac{4x-5}{5x+7}=\dfrac{4x-7}{5x+9}\)

\(\Rightarrow\left(4x-5\right)\left(5x+9\right)=\left(4x-7\right)\left(5x+7\right)\)

\(\Rightarrow20x^2-11x-45=20x^2+7x-49\)

\(\Rightarrow-11x-45=7x-49\)

\(\Rightarrow4=18x\Rightarrow x=\dfrac{9}{2}\)

Vậy...

\(\dfrac{2x-5}{4x+1}=\dfrac{3x-7}{6x-11}\)

\(\Rightarrow\left(2x-5\right)\left(6x-11\right)=\left(3x-7\right)\left(4x+1\right)\)

\(\Rightarrow12x^2-52x+55=12x^2-25x-7\)

\(\Rightarrow52x-55=25x+7\)

\(\Rightarrow27x=62\)

\(\Rightarrow x=\dfrac{62}{27}\)

Vậy...

a, không nhìn rõ

b, \(\dfrac{a+2\sqrt{a}+1}{a-1}\)

\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\)

a, \(A=m^2-6m+x^2-x+3\)

\(=x^2-6m+9+x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{25}{4}\)

\(=\left(m-3\right)^2+\left(x-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\ge\dfrac{-25}{4}\)

Dấu " = " khi \(\left\{{}\begin{matrix}\left(m-3\right)^2=0\\\left(x-\dfrac{1}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(MIN_A=\dfrac{-25}{4}\) khi m = 3, \(x=\dfrac{1}{2}\)

b, \(B=3x^2-6x+12=3\left(x^2-2x+4\right)\)

\(=3\left(x^2-2x+1+3\right)=3\left(x-1\right)^2+9\ge9\)

Dấu " = " khi \(3\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy MIN B = 9 khi x = 1

=> p^2 = (m-1)(m+n). => m+n thuộc ước dương của p^2 . mà p là số nguyên tố => m+n thuộc p,1,p^2. mà m+n> m-1=> m+n = p^2 => m-1 =1 => m=2=> p^2 = n+2(đpcm)

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{19}{9^2.10^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+...+\dfrac{19}{81.100}\)

\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{81}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

\(D=20-x^2\le20\)

Dấu " = " khi \(x^2=0\Rightarrow x=0\)

Vậy MAX D = 20 khi x = 0

Quan sát tình huống trò chuyện giữa bạn Cường và bạn Mai trong hình dưới đây:

Lan nói rằng Mai đã ăn nhiều bánh hơn Cường. Em có đồng ý không? Tại sao?

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-yz-xz+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-yz-xz-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)