a. ĐKXĐ : x>1.

b. \(A=\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{1}{\sqrt{x}-1}=\left[\dfrac{4}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right].\left(\sqrt{x}-1\right)=\dfrac{4+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\dfrac{4+x}{\sqrt{x}}\)

c. Thay \(x=4-2\sqrt{3}\) vào A, ta có:

\(A=\dfrac{4+4-2\sqrt{3}}{\sqrt{4-2\sqrt{3}}}=\dfrac{8-2\sqrt{3}}{\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{8-2\sqrt{3}}{\sqrt{3}-1}=\dfrac{\left(8-2\sqrt{3}\right)\left(\sqrt{3}+1\right)}{3-1}=\dfrac{8\sqrt{3}+8-6-2\sqrt{3}}{2}=\dfrac{2+6\sqrt{3}}{2}=\dfrac{2\left(1+3\sqrt{3}\right)}{2}=1+3\sqrt{3}\)

Vậy giá trị của A tại \(x=4-2\sqrt{3}\) là \(1+3\sqrt{3}\).

b \(P=\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

a: Khi x=64 thì \(P=\dfrac{8+1}{8+2}=\dfrac{9}{10}\)

b: \(P=\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

a: Khi x=64 thì \(P=\dfrac{8+1}{8+2}=\dfrac{9}{10}\)

Lần sau ghi dấu ra xíu nhé :v

a) Đặt \(\sqrt{x}=a\Rightarrow B=\left(\dfrac{a}{a+4}+\dfrac{4}{a-4}\right):\dfrac{a^2+16}{a+2}\)

Quy đồng,rút gọn : \(B=\dfrac{a+2}{a^2-16}\Rightarrow B=\dfrac{\sqrt{x}+2}{x-16}\)

b) \(B\left(A-1\right)=\dfrac{\sqrt{x}+2}{x-16}\left(\dfrac{\sqrt{x}+4}{\sqrt{x}+2}-1\right)=\dfrac{2}{x-16}\)

x - 16 là ước của 2 => \(x\in\left\{14;15;17;18\right\}\)

mới làm quen toán 9 ;v có gì k rõ ae chỉ bảo nhé :))

a)ĐKXĐ:x>0

P=\(\left(\frac{3}{x-1}-\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\left(vớix>0\right)\)

=\(\left[\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}\right]:\frac{1}{\sqrt{x}+1}\)

=\(\left[\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\frac{1}{\sqrt{x}+1}\)

= \(\left[\frac{3-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\frac{1}{\sqrt{x}+1}\)

=\(\frac{4-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}+1}{1}\)

=\(\frac{4-\sqrt{x}}{\sqrt{x}-1}\)

b)Để P=\(\frac{5}{4}\left(vớix>0\right)\)

\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-1}=\frac{5}{4}\)

\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-1}-\frac{5}{4}=0\)

\(\Leftrightarrow\frac{4\left(4-\sqrt{x}\right)}{4\left(\sqrt{x}-1\right)}-\frac{5\left(\sqrt{x}-1\right)}{4\left(\sqrt{x}-1\right)}=0\)

\(\Rightarrow16-4\sqrt{x}-5\sqrt{x}+5=0\)

\(\Leftrightarrow21-9\sqrt{x}=0\)

\(\Leftrightarrow-9\sqrt{x}=-21\)

\(\Leftrightarrow\sqrt{x}=\frac{7}{3}\)

\(\Leftrightarrow x=\frac{21}{9}\)

Vậy:Để P=\(\frac{5}{4}\)thì x=\(\frac{21}{9}\)

c)Còn phần c thì mik chịu

...

a: \(A=\left(2\sqrt{5}-3\sqrt{5}+3\sqrt{5}\right)\cdot\sqrt{5}=2\sqrt{5}\cdot\sqrt{5}=10\)

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

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

b: A=2B

=>\(10=4\sqrt{x}-2\)

=>\(4\sqrt{x}=12\)

=>x=9(nhận)

a+b+c lon nhat=9+9+9=27

a+b+c be nhat =0+0+0=0

tick nha

o minh cung dang dinh hoi cau ay

 giup minh voi

Áp dungj BĐT min-côp-xki, ta có \(\sqrt{1+a^4}+\sqrt{1+b^4}\ge\sqrt{\left(1+1\right)^2+\left(a^2+b^2\right)^2}=\sqrt{4+\left(a^2+b^2\right)^2}\)

Mà \(\left(a+2\right)\left(b+2\right)=\frac{25}{4}\Rightarrow ab+2a+2b=\frac{9}{4}\)

Mà \(a^2+b^2\ge2ab;4a^2+1\ge4a;4b^2+1\ge4b\Rightarrow5\left(a^2+b^2\right)+2\ge\frac{9}{2}\)

=> \(a^2+b^2\ge\frac{1}{2}\)

=> \(F\ge\sqrt{4+\frac{1}{4}}=\frac{\sqrt{17}}{2}\)

Dấu = xảy ra <=> a=b=1/2

^_^