đúng

\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)

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

\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)

\(B=3-\sqrt{x}-\sqrt{x}+3-6\)

\(B=-2\sqrt{x}\)

\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)

\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)

\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)

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

\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)

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

\(=\frac{3}{\sqrt{x}-6}\)

câu a nè:

http://123link.pw/0Qyw5v

câu d nè : http://123link.pw/Jx46C

nhớ cho đúng nha ^-^

em mới lớp 6 nên chưa biết

\(B=\sqrt{x-4\sqrt{x}+4}+\sqrt{x-6\sqrt{x}+9}\)

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

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

Dấu "=" <=> 4 < x < 9

Lời giải:

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

Ta thấy vì \(\sqrt{x}\geq 0, \forall x\geq 0\Rightarrow \sqrt{x}+1\geq 1\)

\(\Rightarrow \frac{2}{\sqrt{x}+1}\leq \frac{2}{1}=2\)

\(\Rightarrow P=1-\frac{2}{\sqrt{x}+1}\geq 1-2=-1\)

Vậy \(P_{\min}=-1\Leftrightarrow x=0\)

ĐỀ THI VÀO 10 ĐÓ CẢM ƠN MN TRƯỚC NHA:))

a: \(Q=\dfrac{9-x+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

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

\(P=\dfrac{x-9-x+3\sqrt{x}}{x-9}=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}=\dfrac{3}{\sqrt{x}+3}\)

\(M=\dfrac{3}{\sqrt{x}+3}:\dfrac{-\left(\sqrt{x}+2\right)}{\sqrt{x}+3}=\dfrac{-3}{\sqrt{x}+2}\)

Để M>0 thì căn x+2<0(vô lý)

b: \(M\left(x+1\right)=\dfrac{-3\left(x+1\right)}{\sqrt{x}+2}\)

\(=\dfrac{-3x-3}{\sqrt{x}+2}=\dfrac{-3x+12-15}{\sqrt{x}+2}=-\dfrac{3\left(x-4\right)}{\sqrt{x}+2}-\dfrac{15}{\sqrt{x}+2}\)

\(=-3\left(\sqrt{x}-2\right)-\dfrac{15}{\sqrt{x}+2}\)

\(=-3\sqrt{x}+6-\dfrac{15}{\sqrt{x}+2}\)

\(=-3\sqrt{x}-6-\dfrac{15}{\sqrt{x}+2}+12\)

\(=-\left(3\sqrt{x}+6+\dfrac{15}{\sqrt{x}+2}\right)+12\)

\(3\left(\sqrt{x}+2\right)+\dfrac{15}{\sqrt{x}+2}>=2\cdot\sqrt{3\left(\sqrt{x}+2\right)\cdot\dfrac{15}{\sqrt{x}+2}}=6\sqrt{5}\)

=>M(x+1)<=-6căn 5+12

Dấu = xảy ra khi \(3\left(\sqrt{x}+2\right)^2=15\)

=>(căn x+2)^2=5

=>căn x+2=căn 5

=>căn x=căn 5-2

=>x=9-4căn 5

khó vậy

bai nay mk thay rat kho vi mk ko thay co 1 quy luat nao ca