Giups mik vs

a, \(A=\left(\frac{4}{2x+1}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)

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

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

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

a: \(A=2x^2-8x+1\)

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

\(=2\left(x^2-4x+4-\dfrac{7}{2}\right)\)

\(=2\left(x-2\right)^2-7>=-7\)

Dấu = xảy ra khi x=2

b: \(B=\left(x-3\right)^2+\left(x-1\right)^2\)

\(=x^2-6x+9+x^2-2x+1\)

\(=2x^2-8x+10\)

\(=2x^2-8x+8+2\)

\(=2\left(x-2\right)^2+2>=2\)

Dấu = xảy ra khi x=2

\(A=0,6+\left|\dfrac{1}{2}-x\right|\\ Vì:\left|\dfrac{1}{2}-x\right|\ge\forall0x\in R\\ Nên:A=0,6+\left|\dfrac{1}{2}-x\right|\ge0,6\forall x\in R\\ Vậy:min_A=0,6\Leftrightarrow\left(\dfrac{1}{2}-x\right)=0\Leftrightarrow x=\dfrac{1}{2}\)

\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\\ Vì:\left|2x+\dfrac{2}{3}\right|\ge0\forall x\in R\\ Nên:B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\forall x\in R\\ Vậy:max_B=\dfrac{2}{3}\Leftrightarrow\left|2x+\dfrac{2}{3}\right|=0\Leftrightarrow x=-\dfrac{1}{3}\)

\(A=\left|x-3\right|+\left|y+3\right|+2016\)

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

\(\left|y+3\right|\ge0\)

\(\Rightarrow\left|x-3\right|+\left|y+3\right|+2016\ge2016\)

Dấu ''='' xảy ra khi \(x-3=y+3=0\)

\(x=3;y=-3\)

\(MinA=2016\Leftrightarrow x=3;y=-3\)

\(\left(x-10\right)+\left(2x-6\right)=8\)

\(x-10+2x-6=8\)

\(3x=8+10+6\)

\(3x=24\)

\(x=\frac{24}{3}\)

x = 8

\(A=\left(x-1\right)\left(2x-1\right)\left(2x^2-3x-1\right)+2017\)

\(=\left(2x^2-3x+1\right)\left(2x^2-3x-1\right)+2017\)

\(=\left(2x^2-3x\right)^2-1+2017\)

\(=\left(2x^2-3x\right)^2+2016\ge2016\)

\(\Leftrightarrow2x^2-3x=0\Leftrightarrow x\left(2x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)

Vậy \(A_{min}=2016\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)

ai thấy mình làm đúng thì k cho mình nha!

A=\(\left(2x^2-3x+1\right)\left(2x^2-3x-1\right)+2017\)

ĐẶT \(2x^2-3x=t\)

\(\Leftrightarrow\left(t+1\right)\left(t-1\right)+2017\)

\(\Leftrightarrow t^2-1+2017\)

\(\Leftrightarrow t^2+2016\ge2016\left(do.t^2\ge0\right)\)

DẤU ''='' XẢY RA KHI VÀ CHỈ KHI \(t^2=0\Leftrightarrow2x^2-3x=0\Leftrightarrow x\left(2x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=0\end{cases}}\)

VẬY GTNN CỦA A LÀ 2016 TẠI X=0 HOẶC X=3/2

1/ \(A=3\left|2x-1\right|-5\)

Ta có: \(\left|2x-1\right|\ge0\)

\(\Rightarrow3\left|2x-1\right|\ge0\)

\(\Rightarrow3\left|2x-1\right|-5\ge-5\)

Để A nhỏ nhất thì \(3\left|2x-1\right|-5\)nhỏ nhất

Vậy \(Min_A=-5\)

Đặt x²-2x+1=t, ta có:

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

Dấu "=" xảy ra khi \(x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)

Đặt \(\left(x^2-2x\right)\left(x^2-2x=2\right)=k.\left(k+2\right)=A\)

\(\Rightarrow A=k.\left(k+2\right)=k^2+2k\)

\(\Rightarrow A=k^2+k+k+1-1=k\left(k+1\right)+\left(k+1\right)-1\)

\(\Rightarrow A=\left(k+1\right)^2-1\)

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

\(\Rightarrow A=\left(x^2-x-x+1\right)^2-1=\left[x.\left(x-1\right)-\left(x-1\right)\right]^2-1\)

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

( Dấu "=" xảy ra <=> x=1 )