\(B=x\left(2x-1\right)=2x^2-x=2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{1}{8}=2\left(x-\dfrac{1}{4}\right)^2-\dfrac{1}{8}\ge-\dfrac{1}{8}\)

\(minB=-\dfrac{1}{8}\Leftrightarrow x=\dfrac{1}{4}\)

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

\(minC=-\dfrac{4}{3}\Leftrightarrow x=-\dfrac{2}{3}\)

`B=x(2x-1)`

`=2x(x-1/2)`

`=2(x^2-1/2x)`

`=2(x^2-1/2x+1/16)-1/8`

`=2(x-1/4)^2-1/8>=-1/8`

Dấu "=" xảy ra khi `x=1/4`

`C=x(3x+4)`

`=3x(x+4/3)`

`=3(x^2+4/3x)`

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

`=3(x+2/3)^2-4/3>=-4/3`

Dấu "=" xảy ra khi `x=-2/3`

(a) Điều kiện : \(x\ne-1.\)

Ta có : \(P=\dfrac{x^4+x}{x^2-x+1}+1-\dfrac{2x^2+3x+1}{x+1}\)

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

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

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

\(=x^2-x.\)

Vậy : Với mọi \(x\ne-1\) thì \(P=x^2-x.\)

(b) Ta có : \(P=x^2-x\)

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

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

Vậy : \(MinP=-\dfrac{1}{4}.\) Dấu đẳng thức xảy ra khi và chỉ khi \(x=\dfrac{1}{2}.\)

Tìm GTTN của biểu thức :

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

Ta có : \(\left(3x+4\right)^4\ge0\)

\(\Rightarrow A\ge-5\)

\(MinA=-5khix=\frac{-4}{3}\)

Học tốt!

A = |x - 2| + |2x - 3| + |3x - 4|

= |x - 2| + |2x - 3| + |4 - 3x|

\(\ge\)|x - 2 + 2x - 3 + 4 - 3x| = 1

vậy MIN_A = 1

Có ai ko giúp mk với@@@@22222

dễ mà!!!

phân tích ra pạn!!!

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GTNN LÀ -10

a.

\(A=\left(x^4+y^2+1-2x^2y+2x^2-2y\right)+2\left(y^2-2y+1\right)+2026\)

\(A=\left(x^2-y+1\right)^2+2\left(y-1\right)^2+2026\ge2026\)

\(A_{min}=2026\) khi \(\left(x;y\right)=\left(0;1\right)\)

b.

Đặt \(x-1=t\Rightarrow x=t+1\)

\(\Rightarrow A=\dfrac{3\left(t+1\right)^2-8\left(t+1\right)+6}{t^2}=\dfrac{3t^2-2t+1}{t^2}=\dfrac{1}{t^2}-\dfrac{2}{t}+3=\left(\dfrac{1}{t}-1\right)^2+2\ge2\)

\(A_{min}=2\) khi \(t=1\Rightarrow x=2\)

\(A=\dfrac{3x^2-8x+6}{x^2-2x+1}=\dfrac{3x^2-8x+6}{\left(x-1\right)^2}=\dfrac{2\left(x-1\right)^2+\left(x-2\right)^2}{\left(x-1\right)^2}=2+\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge2\)

Dấu \("="\Leftrightarrow x=2\)

Với mọi x t có :

\(\left|3x-1\right|\ge0\)

\(\Leftrightarrow2\left|3x-1\right|\ge0\)

\(\Leftrightarrow2\left|3x-1\right|-4\ge-4\)

\(\Leftrightarrow A\ge-4\)

Dấu "=" xảy ra khi : \(\left|3x-1\right|=0\)

\(\Leftrightarrow x=\frac{1}{3}\)

Vậy \(A_{Min}=-4\Leftrightarrow x=\frac{1}{3}\)