a)\(\left(2x-3\right)\left(x^2+2x-4\right)\)

\(=2x^3+4x^2-8x-3x^2-6x+12\)

\(=2x^3+x^2-14x+12\)

b) \(x\left(y-x\right)-y\left(x-y\right)\)

\(=x\left(y-x\right)+y\left(y-x\right)\)

\(=\left(y-x\right)\left(x+y\right)\)

\(A=2x^2+4x-1\)

\(\Leftrightarrow A=2\left(x^2+2x-\dfrac{1}{2}\right)\)

\(\Leftrightarrow A=2\left[\left(x^2+2x+1\right)-\dfrac{3}{2}\right]\)

\(\Leftrightarrow A=2\left[\left(x+1\right)^2-\dfrac{3}{2}\right]\)

Vậy GTNN của \(A=\dfrac{-3}{2}\) khi \(x=-1\)

bài 2: tìm x

a)\(x^2+y^2-2x+4y+5=0\)

\(\Leftrightarrow x^2+y^2-2x+4y+1+4=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy x=1; y=-2

b)\(5x^2+9y^2-12xy-6x+9=0\)

\(\Leftrightarrow\left(4x^2-12xy+9y^2\right)+\left(x^2-6x+9\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2.3-3.y=0\\x=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=3\end{matrix}\right.\)

Vậy x=2; y=3

làm ơn làm phước tick cho mình lên 180 điểm với

\(0,3x=1,2y\) \(\Leftrightarrow\dfrac{x}{1,2}=\dfrac{y}{0,3}\)và \(x+y=50\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\dfrac{x}{1,2}=\dfrac{y}{0,3}=\dfrac{x+y}{1,2+0,3}=\dfrac{50}{1,5}=\dfrac{100}{3}\)

\(\dfrac{x}{1,2}=\dfrac{100}{3}\Rightarrow x=\dfrac{100}{3}.1,2=40\)

\(\dfrac{y}{0,3}=\dfrac{100}{3}\Rightarrow y=\dfrac{100}{3}.0,3=10\)

Vậy x=40; y=10

\(A=2+2^2+2^3+2^4+......+2^{20}\)

\(\Leftrightarrow A=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+......+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)

\(\Leftrightarrow A=\left(2+2^2+2^3+2^4\right)+2^4\left(2+2^2+2^3+2^4\right)+......+2^{16}\left(2+2^2+2^3+2^4\right)\)

\(\Leftrightarrow A=30+2^4.30+......+2^{16}.30\)

\(\Leftrightarrow A=30\left(1+2^4+2^8+2^{12}+2^{16}\right)\)

\(\Leftrightarrow A=........0\)

Vậy chữ số tận cùng của A=0. Vì N nhân với số tròn chục thì chữ số tận cùng sẽ bằng 0

d) \(\left(1+x\right)^2-4x\left(1-x^2\right)\)

\(=\left(1+x\right)^2-4x\left(1^2-x^2\right)\)

\(=\left(1+x\right)^2-4x\left(1-x\right)\left(1+x\right)\)

\(=\left(1+x\right)\left[\left(1+x\right)-4x\left(1-x\right)\right]\)

\(=\left(1+x\right)\left[1+x-4x+4x^2\right]\)

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

\(2x=3y=5z\)\(\Rightarrow\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{5}}\) và \(x+y+z=95\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{5}}=\dfrac{x+y+z}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{5}}=\dfrac{95}{\dfrac{31}{30}}=\dfrac{2850}{31}\)

\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{2850}{31}\Rightarrow x=\dfrac{1425}{31}\)

\(\dfrac{y}{\dfrac{1}{3}}=\dfrac{2850}{31}\Rightarrow y=\dfrac{950}{31}\)

\(\dfrac{z}{\dfrac{1}{5}}=\dfrac{2850}{31}\Rightarrow z=\dfrac{570}{31}\)

Vậy \(x=\dfrac{1425}{31}\) ; \(y=\dfrac{950}{31}\) ; \(z=\dfrac{570}{31}\)

= 36000                      tick nha

3)\(A=-x^2+2x-3\)

\(\Leftrightarrow A=-x^2+2x-1-2\)

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

\(\Leftrightarrow A=-\left(x-1\right)^2-2\)

Vậy GTLN của A=-2 khi x=1