a) Ta có: \(x-\sqrt{x}+1=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\left(\forall x\right)\)

=> \(A=\frac{1}{x-\sqrt{x}+1}\le\frac{1}{\frac{3}{4}}=\frac{4}{3}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(\sqrt{x}-\frac{1}{2}\right)^2=0\Rightarrow x=\frac{1}{4}\)

Vậy Max(A) = 4/3 khi x = 1/4

b) \(B=\sqrt{4x-x^2+21}=\sqrt{-\left(x^2-4x+4\right)+25}\)

\(=\sqrt{25-\left(x-2\right)^2}\le\sqrt{25}=5\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-2\right)^2=0\Rightarrow x=2\)

Vậy Max(B) = 5 khi x = 2

c) \(C=1+\sqrt{-9x^2+6x}=1+\sqrt{-\left(9x^2-6x+1\right)+1}\)

\(=1+\sqrt{1-\left(3x-1\right)^2}\le1+\sqrt{1}=2\)

Dấu "=" xảy ra khi: \(\left(3x-1\right)=0\Rightarrow x=\frac{1}{3}\)

Vậy Max(C) = 2 khi x = 1/3

d) Ta có: \(D=\sqrt{x-2}+\sqrt{4-x}\)

=> \(D^2=\left(\sqrt{x-2}+\sqrt{4-x}\right)\le\left(1^2+1^2\right)\left(x-2+4-x\right)\) ( BĐT Bunhia)

\(=2.2=4\)

=> \(D\le2\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(x-2=4-x\Rightarrow x=3\)

Vậy Max(D) = 2 khi x = 3

cảm ơn bạn nhaaa

em phai khong biet

moi hoc lop 6 thoi anh a

1 ) \(A=\sqrt{x-2}+\sqrt{4-x}\)

ĐKXĐ : \(2\le x\le4\)

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

Áp dụng bđt AM - GM ta có : 

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

\(\Rightarrow A^2\le2+2=4\Rightarrow-2\le A\le2\)

Mà A > 0 nên ko thể có min = - 2 nên \(2\le x\le4\) ta chọn x = 2

=> A = \(\sqrt{2}\)

Vậy \(\sqrt{2}\le A\le2\)

\(N=6\sqrt{x}-x-1=8-\left(x-6\sqrt{x}+9\right)=8-\left(\sqrt{x}-3\right)^2\le8\)

Dấu "=" xảy ra <=> \(\sqrt{x}-3=0\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\)

Vậy Max(N)=8

\(P=\frac{1}{x-\sqrt{x}+1}=\frac{1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{1}{\frac{3}{4}}=\frac{4}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)

Vậy Max(P)=4/3

\(\sqrt{x-1}\ge0,\forall x\inℝ\Rightarrow\sqrt{3}-\sqrt{x-1}\le\sqrt{3}\)

Dấu "=" xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy Max (M)=\(\sqrt{3}\)\(\Leftrightarrow x=1\)

Đặt các biểu thức ở câu a,b,c lần lượt là A,B,C

a)  A= \(\sqrt{3}-\sqrt{x-1}\le\sqrt{3}\) ( do \(\sqrt{x-1}\ge0\))    => Max A=\(\sqrt{3}\) khi và chỉ khi x=1

b) B= -( \(x-6\sqrt{x}+1\))  (=) B= - \(\left(\sqrt{x-3}\right)^2\)+8 \(\le8\) => Max B=8  khi và chỉ khi x=3

c) C= \(\frac{1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{1}{\frac{3}{4}}=\frac{4}{3}\) Do mẫu \(\ge\frac{3}{4}\)=> Max C= \(\frac{4}{3}\) khi và chỉ khi x=\(\frac{1}{4}\)

\(A\le\sqrt{2\left(3x-5+7-3x\right)}=\sqrt{2.2}=2\)

\(A_{max}=2\) khi \(x=2\)

\(B\le\sqrt{2\left(x-5+23-x\right)}=\sqrt{2.18}=6\)

\(B_{max}=6\) khi \(x=14\)

\(C=-\left(2-x\right)+\sqrt{2-x}+2=-\left(\sqrt{2-x}-\frac{1}{4}\right)^2+\frac{17}{8}\le\frac{17}{8}\)

\(C_{max}=\frac{17}{8}\) khi \(x=\frac{31}{16}\)

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

\(D_{max}=\frac{1}{2}\) khi \(x=\frac{\sqrt{2}}{2}\)