\(A=x-2\sqrt{3-x}\\ =-\left(3-x-2\sqrt{3-x}+1\right)+4\\ =-\left(\sqrt{3-x}-1\right)^2+4\le4\)

Dấu \("="\Leftrightarrow\sqrt{3-x}-1=0\Leftrightarrow3-x=1\Leftrightarrow x=2\)

\(A=\sqrt{x-2}+\sqrt{4-x}\ge\sqrt{x-2+4-x}=\sqrt{2}\)

\(A_{min}=\sqrt{2}\) khi \(\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

\(y=4x^2+\dfrac{9}{x^2}-3\ge2\sqrt{\dfrac{36x^2}{x^2}}-3=9\)

\(y_{min}=9\) khi \(x^2=\dfrac{3}{2}\)

\(P=\dfrac{x-1}{4}+\dfrac{1}{x-1}+\dfrac{1}{4}\ge2\sqrt{\dfrac{x-1}{4\left(x-1\right)}}+\dfrac{1}{4}=\dfrac{5}{4}\)

\(P_{min}=\dfrac{5}{4}\) khi \(x=\dfrac{3}{2}\)

Dk x<=3

\(!x!>\sqrt{3-x}\)

\(\sqrt{3-x}=t\ge0\Rightarrow t^2=3-x\Rightarrow x=3-t^2\)

\(\Leftrightarrow A=t+3-t^2=3+t-t^2=3+\frac{1}{4}-\left(\frac{1}{4}-2.\frac{1}{2}t+t^2\right)\)

\(A=\frac{13}{4}-\left(t-\frac{1}{2}\right)^2\ge\frac{13}{4}\) đẳng thức khi  t=1/2\(\Rightarrow x=3-\frac{1}{4}=\frac{11}{4}\)

Kết luận A_{max \(=\frac{13}{4}\) khi  \(x=\frac{11}{4}\)}

ê có kq máy tính chưa

3 cho sao!

a) Pt \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

Vậy...

b)Đk: \(x\ge-1\)

Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)

\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)

Vậy...

\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)

\(=a+\sqrt{a}-\left(2\sqrt{a}+1\right)+1=a-\sqrt{a}\)

b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\) 

Vậy \(A_{min}=-\dfrac{1}{4}\)

a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)

\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)

\(\Rightarrow\sqrt{16\left(x+1\right)}-3\sqrt{x+1}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)

a) \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)

\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)

b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)

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

\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)

✱ giải pt:

a.\(\sqrt{x^2-4x+4}\)\(=5\)

⇔\(\sqrt{\left(x-2\right)^2}=5\)

⇒\(\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

vậy....

b.\(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)

⇔ \(4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

⇔ \(4\sqrt{x+1}=16\)

⇔ \(\sqrt{x+1}=16\)

⇒ \(x+1=256\)

⇔ \(x=255\)

vậy.....

\(A=\sqrt{9-x^2}+4\)  Đạt Max khi \(\sqrt{9-x^2}\)đạt giá trị lớn nhất. Hay (9-x²) đạt giá trị lớn nhất.

Do x² \(\ge\)0 với mọi x => để 9-x² đạt giá trị lớn nhất thì x² phải đạt GTNN => x²=0 => x=0

=> \(A_{max}=\sqrt{9}+4=3+4=7\)đạt được khi x=0

b/ \(B=6\sqrt{x}-x-15=-x+6\sqrt{x}-9-6=-6-\left(x-6\sqrt{x}+9\right)\)

=> \(B=-6-\left(\sqrt{x}-3\right)^2\)

Do \(\left(\sqrt{x}-3\right)^2\ge0\) Với mọi x => Để B_max thì \(\left(\sqrt{x}-3\right)^2\) đạt Min => \(\left(\sqrt{x}-3\right)^2=0\)

=> B_min=-6  đạt được khi \(\left(\sqrt{x}-3\right)^2=0\)hay x=9

c/ \(C=2\sqrt{x}-x=1-1+2\sqrt{x}-x=1-\left(1-2\sqrt{x}+x\right)\)

=> \(C=1-\left(1-\sqrt{x}\right)^2\)  => Do \(\left(1-\sqrt{x}\right)^2\ge0\) Với mọi x => Để C đạt max thì \(\left(1-\sqrt{x}\right)^2\)đạt min => \(\left(1-\sqrt{x}\right)^2=0\) 

=> C_min = 1 Đạt được khi x=1

a: \(A=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}=\dfrac{-3\sqrt{x}-9}{x-9}\)

\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{-3}{\sqrt{x}-3}\)

b: A=1/3

=>\(\dfrac{-3}{\sqrt{x}-3}=\dfrac{1}{3}\)

=>căn x-3=-9

=>căn x=-6(loại)

c: căn x-3>=-3

=>3/căn x-3<=-1

=>-3/căn x-3>=1

Dấu = xảy ra khi x=0

2:

ĐKXĐ: x>=3

 \(\Leftrightarrow\sqrt{x-3+2\cdot\sqrt{x-3}\cdot\sqrt{3}+3}+\sqrt{x-3-2\cdot\sqrt{x-3}\cdot\sqrt{3}+3}=2\sqrt{3}\)

=>\(\left|\sqrt{x-3}+\sqrt{3}\right|+\left|\sqrt{x-3}-\sqrt{3}\right|=2\sqrt{3}\)

\(\Leftrightarrow\sqrt{x-3}+\sqrt{3}+\left|\sqrt{x-3}-\sqrt{3}\right|=2\sqrt{3}\)

\(\Leftrightarrow\sqrt{x-3}+\left|\sqrt{x-3}-\sqrt{3}\right|=\sqrt{3}\)(1)

TH1: x>=6

(1) trở thành \(\sqrt{x-3}+\sqrt{x-3}-\sqrt{3}=\sqrt{3}\)

=>\(2\sqrt{x-3}=2\sqrt{3}\)

=>x-3=3

=>x=6(nhận)

TH2: 3<=x<6

Phương trình (1) sẽ là;

\(\sqrt{x-3}+\sqrt{3}-\sqrt{x-3}=\sqrt{3}\)

=>\(\sqrt{3}=\sqrt{3}\)(luôn đúng)

1:

\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}\)

\(=16+2\cdot\sqrt{64-4\cdot\left(10+2\sqrt{5}\right)}\)

\(=16+2\cdot\sqrt{24-8\sqrt{5}}\)

\(=16+2\cdot\sqrt{20-2\cdot2\sqrt{5}\cdot2+4}\)

\(=16+2\cdot\sqrt{\left(2\sqrt{5}-2\right)^2}\)

\(=16+2\cdot\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)

\(=10+2\cdot\sqrt{10}\cdot\sqrt{2}+2\)

\(=\left(\sqrt{10}+\sqrt{2}\right)^2\)

=>\(A=\sqrt{10}+\sqrt{2}\)

ĐK  \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

a, \(R=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)

b. \(R< -1\Rightarrow R+1< 0\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\)

\(\Rightarrow0\le x< \frac{9}{4}\)

c. \(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)

Ta thấy \(\sqrt{x}+3\ge3\Rightarrow\frac{-18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\Rightarrow R\ge-3\)

Vậy \(MinR=-3\Leftrightarrow x=0\)