a) ĐKXĐ : \(3\le x\le7\)

Ta có \(A=1.\sqrt{x-3}+1.\sqrt{7-x}\)

\(\le\sqrt{\left(1+1\right)\left(x-3+7-x\right)}=\sqrt{8}\)(BĐT Bunyacovski)

Dấu "=" xảy ra <=> \(\dfrac{1}{\sqrt{x-3}}=\dfrac{1}{\sqrt{7-x}}\Leftrightarrow x=5\)

\(1,\\ a,A\le\sqrt{\left(x-3+7-x\right)\left(1+1\right)}=\sqrt{8}=2\sqrt{2}\\ A^2=4+2\sqrt{\left(x-3\right)\left(7-x\right)}\ge4\Leftrightarrow A\ge2\\ \Leftrightarrow2\le A\le2\sqrt{2}\\ \left\{{}\begin{matrix}A_{min}\Leftrightarrow\left(x-3\right)\left(7-x\right)=0\Leftrightarrow...\\A_{max}\Leftrightarrow x-3=7-x\Leftrightarrow x=5\end{matrix}\right.\)

\(B=\dfrac{\dfrac{5}{2}\left(4x^4+4x^2+1\right)+2\left(x^4-x^2+\dfrac{1}{4}\right)}{\left(2x^2+1\right)^2}\\ B=\dfrac{\dfrac{5}{2}\left(2x^2+1\right)^2+2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}=\dfrac{5}{2}+\dfrac{2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}\ge\dfrac{5}{2}\)

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

Vì \(-\dfrac{4x^2}{\left(1+2x^2\right)^2}\le0\Leftrightarrow B\le3\)

\(\Leftrightarrow\left\{{}\begin{matrix}B_{min}\Leftrightarrow x^2=\dfrac{1}{2}\Leftrightarrow x=\pm\dfrac{1}{\sqrt{2}}\\B_{max}\Leftrightarrow x=0\end{matrix}\right.\)

\(2,\)

Ta có \(\left(y-2x\right)^2=\left(-2x+y\right)^2=\left[\dfrac{1}{3}\left(-6x\right)+\dfrac{1}{4}\left(4y\right)\right]^2\)

\(\Leftrightarrow\left(y-2x\right)^2\le\left[\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{4}\right)^2\right]\left[\left(-6x\right)^2+\left(4y\right)^2\right]=\dfrac{5^2}{3^2\cdot4^2}\left(36x^2+16y^2\right)=\dfrac{5^2}{4^2}\\ \Leftrightarrow\left|y-2x\right|\le\dfrac{5}{4}\\ \Leftrightarrow-\dfrac{5}{4}\le y-2x\le\dfrac{5}{4}\\ \Leftrightarrow\dfrac{15}{4}\le y-2x+5\le\dfrac{25}{4}\)

\(Max\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}\\y=\dfrac{9}{20}\end{matrix}\right.\\ Min\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{9}{20}\end{matrix}\right.\)

bài 1:

a) (x+1)^2-(x-1)^2-3(x+1)(x-1)

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

=2x^2-3x^2-3

=-x^2-3

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a). 3. |9 - 2x| - 17 = 16

3. |9 - 2x| = 16 + 17

3. |9 - 2x| = 33

|9 - 2x| = 33 : 3

|9 - 2x| = 11

=> 9 - 2x = 11

2x = 9 - 11

2x = -2

x = - 2 : 2

x = - 1

hay 9 - 2x = - 11

2x = 9 - (- 11)

2x = 9 + 11

2x = 20

x = 20 : 2

x = 10

Vậy x = -1; x = 10

a) 3.| 9 - 2x | -17 = 16

    3. | 9 - 2x |      = 16 + 17 = 33

        | 9 - 2x |      = 33 : 3 = 11

  \(\Rightarrow\)9 - 2x = 11                hoặc                9 - 2x = -11

               2x  = 9 - 11                                       2x = 9 - ( - 11 )

               2x  = -2                                            2x  = 20

                  x = -2 : 2                                           x = 20 : 2

                   x = -1                                               x  = 10

b). 3 - 4 |5 - 6x| = 7

4 |5 - 6x| = 3 - 7

4 |5 - 6x| = - 4

|5 - 6x| = - 4 : 4

|5 - 6x| = -1

Mà |5 - 6x| luôn lớn hơn 0 với mọi x

Do đó, x không tìm được giá trị

\(1a.\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)

\(=x^2+2x+1-x^2+2x-1-3x^2+1\)

\(=-3x^2+4x+1\)

b) Sai đề.

2a. \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Rightarrow x^2+8x+16-x^2+1=16\)

\(\Rightarrow8x+17=16\)

\(\Rightarrow8x=-1\)

\(\Rightarrow x=-\dfrac{1}{8}\)

b. \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+49=0\)

\(\Rightarrow2x+59=0\)

\(\Rightarrow x=-\dfrac{59}{2}\).