tham khảo: https://hoc24.vn/cau-hoi/.2256230161739

a) ⇔ \(4x^2+4x-x-1=0\)

⇔ \(4x^2+3x-1=0\)

⇔ \(4x(x+1)-(x+1)=0\)

⇔ \((x+1)(4x-1)=0\)

⇒ \(\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy...

b) \(x^3-4x^2+4x=0\)

⇔ \(x^2(x-2)-2x(x-2)=0\)

⇔ \((x-2)(x^2-2x)=0\)

⇒ \(\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

Vậy...

c) \(x^2-3x+2=0\)

⇔ \(x(x-2)-(x-2)=0\)

⇔ \((x-1)(x-2)=0\)

⇒ \(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy...

\(M=4x^2-3x+\dfrac{1}{4x}+2011\)

\(M=4x^2-4x+1+x+\dfrac{1}{4x}+2011\)

\(M=\left(2x-1\right)^2+\left(x+\dfrac{1}{4x}\right)+2010\)

Vì \(\left(2x-1\right)^2\ge0\) và \(x>0\)

\(\Rightarrow\dfrac{1}{4x}>0\)

Lợi dụng BĐT Cauchy cho 2 số nguyên dương ta có:

\(x+\dfrac{1}{4x}\ge2\sqrt{x\dfrac{1}{4x}}=2.\dfrac{1}{2}=1\)

\(\Rightarrow M=\left(2x-1\right)^2+\left(x+\dfrac{1}{4x}\right)+2010\ge0+1+2010=2011\)

\(\Rightarrow M\ge2011\)

Dấu " = " xảy ra khi:

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\x=\dfrac{1}{4x}\\x>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x^2=\dfrac{1}{4}\\x>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\\x>0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(M_{min}=2011\) đạt được khi \(x=\dfrac{1}{2}\)

`#3107.101107`

a)

`x^2 + 6x + 10`

`= (x^2 + 2*x*3 + 3^2) + 1`

`= (x + 3)^2 + 1`

Vì `(x + 3)^2 \ge 0` `AA` `x`

`=> (x + 3)^2 + 1 \ge 1` `AA` `x`

Vậy, GTNN của bt là 1 khi `(x + 3)^2 = 0`

`<=> x + 3 = 0`

`<=> x = -3`

b)

`4x^2 - 4x + 5`

`= [(2x)^2 - 2*2x*1 + 1^2] + 4`

`= (2x - 1)^2 + 4`

Vì `(2x - 1)^2 \ge 0` `AA` `x`

`=> (2x - 1)^2 + 4 \ge 4` `AA` `x`

Vậy, GTNN của bt là `4` khi `(2x - 1)^2 = 0`

`<=> 2x - 1 = 0`

`<=> 2x = 1`

`<=> x = 1/2`

c)

`x^2 - 3x + 1`

`= (x^2 - 2*x*3/2 + 9/4) - 5/4`

`= (x - 3/2)^2 - 5/4`

Vì `(x - 3/2)^2 \ge 0` `AA` `x`

`=> (x - 3/2)^2 - 5/4 \ge -5/4` `AA` `x`

Vậy, GTNN của bt là `-5/4` khi `(x - 3/2)^2 = 0`

`<=> x - 3/2 = 0`

`<=> x = 3/2`

a: Ta có: \(\left(3x+5\right)^2-4x^2=0\)

\(\Leftrightarrow\left(3x+5+2x\right)\left(3x+5-2x\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)

x>o nhá

Áp dụng BĐT Cauchy ta có:

\(M=4x^2-3x+\dfrac{1}{4x}+2011\)

\(=\left(4x^2-4x+1\right)+\left(x+\dfrac{1}{4x}\right)+2010\)

= \(\left(2x-1\right)^2+\left(x+\dfrac{1}{4x}\right)+2010\)

\(\ge0+2\sqrt{x.\dfrac{1}{4x}}+2010\) = \(1+2010=2011\)

=> Dấu = xảy ra <=> \(2x=1\) => \(x=\dfrac{1}{2}\)

Vậy ........................................

Ta có \(A\left(x\right)=\dfrac{1}{3}x+1=0\Leftrightarrow x=-1:\dfrac{1}{3}=-3\)

\(B\left(x\right)=-\dfrac{3}{4}x+\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}\left(-\dfrac{3}{4}\right)=4\)

\(C=\left(2x-4\right)\left(x+1\right)=0\Leftrightarrow x=2;x=-1\)

\(D\left(x\right)-4x\left(x-2\right)=0\Leftrightarrow x=0;x=2\)

a) MTC = (x -2)(x + 2). Ta rút gọn được M = 1 x − 2  

b) Gợi ý:  x 2 + 5 x + 6 = ( x + 2 ) ( x + 3 ) ; x 2 + x − 12 = ( x − 3 ) ( x + 4 )

Ta có  N = ( x + 2 ) ( x + 3 ) ( x − 3 ) ( x + 4 ) : ( x + 2 ) 2 x ( x − 3 ) = x ( x + 3 ) ( x + 2 ) ( x + 4 )