Ta có \(A^2=4-x+3+x+2\sqrt{\left(4-x\right)\left(x+3\right)}=7+2\sqrt{\left(4-x\right)\left(x+3\right)}\ge7\Rightarrow A^2\ge7\Rightarrow A\ge\sqrt{7}\)

Dấu = xảy ra <=> x=4 hoặc x=-3

Áp dụng BĐT bi-nhi-a, ta có \(\sqrt{4-x}+\sqrt{x+3}\le\sqrt{\left(1+1\right)\left(4-x+x+3\right)}=\sqrt{14}\)

dấu = xảy ra <=> 4-x=x+3<=> x=7/2

a)Dễ thấy: \(M=\sqrt{\left(\sqrt{x-3}-1\right)^2}+\sqrt{\left(\sqrt{x-3}-2\right)^2}\)

\(\Rightarrow M\)có nghĩa\(\Leftrightarrow x-3\ge0\Leftrightarrow x\ge3\)

b)  với \(3\le x\le4\)M xác định

\(3\le x\le4\Rightarrow\sqrt{x-3}\le1\)

\(\Rightarrow M=\left|\sqrt{x-3}-1\right|+\left|\sqrt{x-3}-2\right|=1-\sqrt{x-3}+2-\sqrt{x-3}=3-2\sqrt{x-3}\)

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

\(=\sqrt{2}\)

dấu "=" xảy ra khi: \(\orbr{\begin{cases}\sqrt{x-2}=0\\\sqrt{4-x}=0\end{cases}\orbr{\begin{cases}x=2\\x=4\end{cases}}}\)

vậy MIN \(D=\sqrt{2}\)

\(D=\sqrt{x-2}+\sqrt{4-x}\le\frac{x-2+1+4-x+1}{2}=4\)

dấu "=" xảy ra khi \(x=3\)

vậy \(MAX:D=4\)

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

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

*GTNN

Với 2 ≤ x ≤ 4 => \(2\sqrt{\left(x-2\right)\left(4-x\right)}\ge0\Leftrightarrow2+2\sqrt{\left(x-2\right)\left(4-x\right)}\ge2\)

hay D² ≥ 2 => D ≥ √2 . Dấu "=" xảy ra <=> x = 2 hoặc x = 4 (tm)

*GTLN

Áp dụng bất đẳng thức AM-GM ta có :

\(2\sqrt{\left(x-2\right)\left(4-x\right)}\le x-2+4-x=2\Rightarrow2+2\sqrt{\left(x-2\right)\left(4-x\right)}\le4\)

hay D² ≤ 4 => D ≤ 2 . Dấu "=" xảy ra <=> x = 3 (tm)

Vậy \(\hept{\begin{cases}Min_D=\sqrt{2}\Leftrightarrow x=2orx=4\\Max_D=2\Leftrightarrow x=3\end{cases}}\)

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

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

\(=\left|\sqrt{x-3}-1\right|+\left|\sqrt{x-3}-2\right|\)

Do \(3\le x\le4\Rightarrow0\le\sqrt{x-3}\le1\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-3}-1\le0\\\sqrt{x-3}-2< 0\end{matrix}\right.\)

\(\Rightarrow A=1-\sqrt{x-3}+2-\sqrt{x-3}=3-2\sqrt{x-3}\)

\(P=\sqrt{x-2-2\sqrt{x-3}}-\sqrt{x+1-4\sqrt{x-3}}=\sqrt{x-3-2\sqrt{x-3}+1}-\sqrt{x-3-4\sqrt{x-3}+4}=\sqrt{\left(\sqrt{x-3}-1\right)^2}-\sqrt{\left(\sqrt{x-3}-2\right)^2}=\left|\sqrt{x-3}-1\right|-\left|\sqrt{x-3}-2\right|\)Ta có 3≤x≤4⇒\(\left\{{}\begin{matrix}\sqrt{x-3}-1\le0\\\sqrt{x-3}-2< 0\end{matrix}\right.\)

Vậy \(P=1-\sqrt{x-3}-2+\sqrt{x-3}=-1\)

Dễ dàng nhận ra \(A\ge0\)

\(A^2=x+3-x+2\sqrt{x\left(3-x\right)}=3+2\sqrt{x\left(3-x\right)}\ge3\)

\(\Rightarrow A\ge\sqrt{3}\)

\(A_{min}=\sqrt{3}\) khi \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Ta có: x \(\ge\) 0 \(\Rightarrow\) \(\sqrt{x}\ge0\)  (1)

Ta có: x \(\le\) 3 \(\Rightarrow\) 3 - x \(\ge\) 0 \(\Rightarrow\) \(\sqrt{3-x}\ge0\)  (2)

Từ (1) và (2) \(\Rightarrow\) \(\sqrt{x}+\sqrt{3-x}\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\) x = 0 hoặc x = 3

Chúc bn học tốt!

điều kiện -4<=x<=4x<=4

\(a,\sqrt{\left(x+4\right)^2}+\sqrt{\left(x-4\right)^2}\)

\(A=\left|x+4\right|+\left|x-4\right|\)

KẾT HỢP ĐIỀU KIỆN

\(A=x+4+4-x\)

\(A=8\)

\(B=\sqrt{\left(3x\right)^2-6x+1}+\sqrt{\left(2x\right)^2-12x+3^2}\)

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

\(B=\left|3x-1\right|+\left|2x-3\right|\)

\(TH1:x>=\frac{3}{2}\)

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

\(B=5x-4\)

\(TH2:\frac{1}{3}< =x< \frac{3}{2}\)

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

\(B=x+2\)

\(TH3:x< \frac{1}{3}\)

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

\(B=4-5x\)

câu c và câu d tương tự

câu c tách ra: \(C=\sqrt{\left(\sqrt{x}-3\right)^2}-\sqrt{\left(2\sqrt{x}+1\right)^2}\)

còn câu d tách ra :\(D=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)

\(D=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)

bạn tự làm nốt câu c, d nha