\(9x^2-6x+5\\ =9\left(x^2-\dfrac{2}{3}x+\dfrac{5}{9}\right)\)

\(=9\).(\(x^2-2.x\).\(\dfrac{1}{3}\)\(+\dfrac{1}{9}+\dfrac{4}{9}\))

\(=9\left(x-\dfrac{1}{3}\right)^2+\dfrac{4}{9}\le\dfrac{4}{9}\)

Dấu "=" xảy ra khi \(9\left(x-\dfrac{1}{3}\right)^2=0\Leftrightarrow\left(x-\dfrac{1}{3}\right)^2=0\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)\(\)

\(\left(2x-133\right):3=33\\ 2x-133=99\\ 2x=232\\ x=116\)

\(x-7=-10\\ x=-3\\ x-5=-100\\ x=-95\\ x+100=-1\\ x=-101\\ x-8=-20\\ x=-12\\ x+8=-22\\ x=-30\)

\(a:b=114\\ \Rightarrow6a:b=114.6=684\)

Mình làm nhầm

\(a:b=114\\ \Rightarrow a:\left(6b\right)=\left(a:b\right):6=114:6=19\)

\(\left(-10\right)-\left(-10\right)=\left(-10\right)+10=0\)

Sao bạn lại đăng câu hỏi lên Toán vậy?

\(Ta\) \(có\):

\(\dfrac{x}{y}=\dfrac{-10}{5}=k=-2\\ \Rightarrow x=k.y=-2.y\)

Vậy khi x=5 thì y=\(\dfrac{-5}{2}\)

x,y,z tỉ lệ thuận với 3,5,7

\(\Rightarrow\) \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y-z}{3+5-7}=\dfrac{25}{1}=25\)

\(\dfrac{x}{3}=25\Rightarrow x=75\)

\(\dfrac{y}{5}=25\Rightarrow y=125\)

\(\dfrac{z}{7}=25\Rightarrow z=175\)