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Đặt A = |2x + 5| + |2x - 7|

=>A = |2x + 5| + |7 - 2x| \(\ge\)|2x + 5 + 7 - 2x| = |12| = 12

Dấu "=" xảy ra <=> (2x + 5)(7 - 2x) \(\ge\)0

=> -5/2 \(\le\)x \(\le\)7/2

Vậy MinA = 12 <=> -5/2 \(\le\)x \(\le\)7/2

1/

\(A=3x^2+6x-11\)\(=3\left(x^2+2x-\frac{11}{3}\right)\)\(=3\left[\left(x^2+2x+1\right)-\frac{14}{3}\right]\)\(=3\left(x+1\right)^2-14\ge-14\)

VẬY \(minA=-14\)khi   \(x=-1\)

2/

\(B=\frac{3x^2+2x+7}{3x^2+2x+1}=1+\frac{6}{3x^2+2x+1}\)

Biểu thức   \(\frac{6}{3x^2+2x+1}\)đạt GTLN khi   \(3x^2+2x+1\)nhỏ nhất 

Mà   \(3x^2+2x+1\ge1\)nên GTNN của   \(3x^2+2x+1\)là  \(1\)

Ta có :  \(maxB=1+6=7\) khi   \(x=0\)

TK mk nka !!!!! 

         \(\Rightarrow B_{max}=1+\frac{6}{\frac{4}{3}}=\frac{11}{2}\Leftrightarrow x=-\frac{1}{3}\)

Ta có : 3x² + 6x - 11 

= 3x² + 3.x.3 - 9 - 2 

= (3x² - 3)² 

Mà  (3x - 3)² \(\le0\forall x\in R\)

Nên 3x² + 6x - 11 min = 0 khi x = 1

a) \(A=\sqrt{x-2}+\sqrt{6-x}\)

\(\Rightarrow A^2=x-2+6-x+2\sqrt{\left(x-2\right)\left(6-x\right)}\)

Ta có \(\sqrt{\left(x-2\right)\left(6-x\right)}\ge0,\forall x\)

Do đó \(A^2=4+2\sqrt{\left(x-2\right)\left(6-x\right)}\ge4\)

Mà A không âm \(\Leftrightarrow A\ge2\)

Dấu "=" \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

Áp dụng BĐT Bunhiacopxky:

\(A^2=\left(\sqrt{x-2}+\sqrt{6-x}\right)^2\le\left(x-2+6-x\right)\left(1+1\right)=4\cdot2=8\)

\(\Leftrightarrow A\le\sqrt{8}\)

Dấu "=" \(\Leftrightarrow x-2=6-x\Leftrightarrow x=4\)

Mấy bài còn lại y chang nha 

Tick hộ nha

1/ \(x^2-2x+7\)

\(=x^2-\frac{1}{2}\cdot2x+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+7\)

\(=x^2-\frac{1}{2}\cdot2x+\left(\frac{1}{2}\right)^2-\frac{1}{4}+7\)

\(=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}+7\)

\(=\left(x-\frac{1}{2}\right)^2+\frac{27}{4}\)

Có  \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{27}{4}\ge\frac{27}{4}\)

\(\Rightarrow GTNNx^2-2x+7=\frac{27}{4}\)

               với  \(\left(x-\frac{1}{2}\right)^2=0;x=\frac{1}{2}\)

2/ \(4x^2+2x+9\)

\(=\left(2x\right)^2+2\cdot2\cdot\frac{1}{2}x+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+9\)

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

\(=\left(2x+\frac{1}{2}\right)^2+\frac{35}{4}\)

có \(\left(2x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(2x+\frac{1}{2}\right)^2+\frac{35}{4}\ge\frac{35}{4}\)

\(\Rightarrow GTNN4x^2+2x+9=\frac{35}{4}\)

                với  \(\left(2x+\frac{1}{2}\right)^2=0;x=-\frac{1}{4}\)

Ta có 7^2x + 49 = (7^x)² + 7² \(\ge2×7^{x+1}\)

Thế vào ta được

A\(\le\frac{7^{x+1}}{2×7^{x+1}}=\frac{1}{2}\)

Vậy max là \(\frac{1}{2}\)đạt được khi x = 1

tại sao \(\left(7x\right)^2+7^2\ge2\times7^{x+1}\)