tach 14-x = 10-4-x roi sau do chac ban cung phai tu biet lam

3x+7=28

3x    =28-7

3x     =21

  x    =21:3

 x      =7

1/ Ta có: \(\left(x+2\right)^2\ge0\Rightarrow-\left(x+2\right)^2\le0\Rightarrow A=14-\left(x+2\right)^2=14+\left[-\left(x+2\right)^2\right]\le14\)

Đẳng thức xảy ra khi: \(x+2=0\Rightarrow x=-2\)

Vậy giá trị lớn nhất của A là 14 khi x = -2.

2/ Ta có: \(\left|3x+5\right|\ge0\Rightarrow-\left|3x+5\right|\le0\Rightarrow B=-\left|3x+5\right|-7\le-7\)

Đẳng thức xảy ra khi: \(3x+5=0\Rightarrow3x=-5\Rightarrow x=\frac{-5}{3}\)

Vậy giá trị lớn nhất của B là -7 khi \(x=\frac{-5}{3}\).

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

Do \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\Rightarrow\dfrac{1}{x^2+x+1}\le\dfrac{4}{3}\)

\(\Rightarrow A\le3+\dfrac{4}{3}=\dfrac{13}{3}\)

\(maxA=\dfrac{13}{3}\Leftrightarrow x=-\dfrac{1}{2}\)

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

Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge0\Leftrightarrow\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)

\(\Rightarrow A\le3+\dfrac{4}{3}=\dfrac{13}{3}\)

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

Bài 4:

\(A=2x^2-15\ge-15\\ A_{min}=-15\Leftrightarrow x=0\\ B=2\left(x+1\right)^2-17\ge-17\\ B_{min}=-17\Leftrightarrow x=-1\)

Bài 5:

\(A=-x^2+14\le14\\ A_{max}=14\Leftrightarrow x=0\\ B=25-\left(x-2\right)^2\le25\\ B_{max}=25\Leftrightarrow x=2\)

Có 3x-4/x+2=2x-2

Để A đạt min khi

2x-2>=-4

=>2x>=-2

=>X=-1