\(\Leftrightarrow4\left(x^2+x-2\right)-\left(4x^2+11x-3\right)=2x-2\)

\(\Leftrightarrow4x^2+4x-8-4x^2-11x+3=2x-2\)

=>-7x-5=2x-2

=>-9x=3

hay x=-1/3

mk chịu

Dùng bđt Cô si giùm mig ạ

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

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

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

\(A=\left(2x-1\right)^2+4\ge4\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)

\(A=3x^2+\left(x-2\right)^2+1=4x^2-4x+5=\left(2x-1\right)^2+4\)

Vì \(\left(2x-1\right)^2\ge0\Rightarrow A\ge4\)

Dấu ''='' xảy ra \(\Leftrightarrow x=\frac{1}{2}\)

Vậy \(Min_A=4\Leftrightarrow x=\frac{1}{2}\)

Để pt có 2 nghiệm dương:

\(\left\{{}\begin{matrix}\Delta'=\left(m-3\right)^2-\left(m-1\right)\ge0\\x_1+x_2=-2\left(m-3\right)>0\\x_1x_2=m-1>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2-7m+10\ge0\\m< 3\\m>1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m\ge5\\m\le2\end{matrix}\right.\\m< 3\\m>1\end{matrix}\right.\)

\(\Rightarrow1< m\le2\)

\(x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)

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

Bài 1

\(a,\frac{x}{6}=\frac{y}{-8}\)

=> đề thiếu :))

\(b,\frac{x}{2}=\frac{y}{5}\)và \(x+y=35\)

ADTC dãy tỉ số bằng nhau ta cs

\(\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=\frac{35}{7}=5\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=5\\\frac{y}{5}=5\end{cases}\Rightarrow\hept{\begin{cases}x=10\\y=25\end{cases}}}\)

Bài 2 là bài đơn giản :)) e tự lm nha 1.1