`[2x]/y\sqrt{y/[2x]}=\sqrt{([2x]/y)^2. y/[2x]}=\sqrt{[2x]/y}=[\sqrt{2xy}]/y`

`\sqrt{1/[a^4]+1/a}=\sqrt{[1+a^3]/[a^4]}=\sqrt{1+a^3}/[a^2]`

`\sqrt{[16x^3]/[81y]}=[4|x|\sqrt{x}]/[9\sqrt{y}]`

bạn nào trl thì trl lại đc khum ạ

Mik ko xem đc câu trl 

\(8=\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(\ge\dfrac{8}{9}\sqrt{3\left(ab+bc+ca\right)}\left(ab+bc+ca\right)\)

\(\Rightarrow\left(ab+bc+ca\right)^3\le27\)

\(\Rightarrow ab+bc+ca\le3\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Mới lớp 8 thôi , sai thông cảm =))

a) Ta có : \(B=\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}\)

\(\Rightarrow B=\dfrac{3\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(\Rightarrow B=\dfrac{3\left(\sqrt{x}-5\right)+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)

\(\Rightarrow B=\dfrac{3\sqrt{x}-15+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)

\(\Rightarrow B=\dfrac{\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{1}{\sqrt{x}+5}\)

b) Ta có : \(P=3.\dfrac{1}{\sqrt{x}+5}:\dfrac{\sqrt{x}+2}{\sqrt{x}+5}\)

\(\Rightarrow P=\dfrac{3}{\sqrt{x}+5}.\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)

\(\Rightarrow P=\dfrac{3}{\sqrt{x}+2}\)

Để P ∈ Z thì \(\dfrac{3}{\sqrt{x}+2}\) ∈ Z

\(\Leftrightarrow3⋮\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{-5;-3;-1;1\right\}\)

\(\Leftrightarrow x=1\left(\text{do }\sqrt{x}>0\right)\)

a)\(B=\dfrac{\sqrt{x}+2}{\sqrt{x}+5}+\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}+5\right).\left(\sqrt{x}-5\right)}\) 

chép sai đề à bn

cùng lớp 9 nhưng đề bài nhì nhằng quá

   \(\sqrt{1-x}+\sqrt{4-4x}-12=0\)  (ĐKXĐ: x khác 1)

<=> \(\sqrt{1-x}+2\sqrt{1-x}-12=0\)

<=>\(3\sqrt{1-x}=12\)

<=>\(\sqrt{1-x}=4\)

<=>1-x=16
<=>x=-15(TMDK)

`x^{2}+5x-13=76`

`<=>x^{2}+5x-89=0`

\(\Delta=5^2-4.1.\left(-89\right)=381>0\)

`=>` PT có `2` nghiệm phân biệt :

\(x_1=\dfrac{-5+\sqrt{381}}{2}\\ x_2=\dfrac{-5-\sqrt{381}}{2}\)

\(x^2+5x-13=76\)

\(\Leftrightarrow x^2+5x-89=0\)

\(\Delta=5^2-4.1.\left(-89\right)=381>0\)

\(\Rightarrow Pt\) có hai nghiệm phân biệt:

\(x_1=\dfrac{-5+\sqrt{381}}{2};x_2=\dfrac{-5-\sqrt{381}}{2}\)

- Vậy tập nghiệm của phương trình \(S=\left\{\dfrac{-5+\sqrt{381}}{2};\dfrac{-5-\sqrt{381}}{2}\right\}\)

Ta có: 

\(\left(x^2+2x\right)^2-\left(x^2+2x+1\right)=0\\ \Leftrightarrow\left(x^2+2x\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(x^2+2x+x+1\right)\left(x^2+2x-x-1\right)=0\\ \Leftrightarrow\left(x^2+3x+1\right)\left(x^2+x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2+3x+1=0\\x^2+x-1=0\end{matrix}\right.\)

Tính delta giải các phương trình bậc 2 tìm nghiệm.

ĐKXĐ : \(0< x\le1\)

Ta có x⁴ + 2x³ + 2x² - 2x + 1 = (x³ + x).\(\sqrt{\dfrac{1-x^2}{x}}\)

<=> \(\left(x^4+2x^2+1\right)+\left(2x^3-2x\right)\) = x(x² + 1)\(\sqrt{\dfrac{1-x^2}{x}}\)

<=> (x² + 1)² - 2(x - x³) = (x² + 1)\(\sqrt{x-x^3}\)

<=> (x² + 1)² - (x² + 1)\(\sqrt{x-x^3}\) - 2(x - x³) = 0 

<=> (x² + 1 + \(\sqrt{x-x^3}\))(x² + 1 - 2\(\sqrt{x-x^3}\)) = 0 

<=> x² + 1 - 2\(\sqrt{x-x^3}\) = 0 (vì x² + 1 + \(\sqrt{x-x^3}\) > 0 \(\forall x\))

<=> x² + 1 = \(2\sqrt{x-x^3}\)

<=> x⁴ + 2x² + 1 = 4(x - x³)

<=> x⁴ + 4x³ + 2x² - 4x + 1 = 0

<=> x⁴ + 2x³ - x² + (2x³ + 4x² - 2x) - (x² + 2x - 1) = 0

<=> x²(x² + 2x - 1) + 2x(x² + 2x - 1) - (x² + 2x - 1) = 0

<=> (x² + 2x - 1)² =  0

<=> x² + 2x - 1 = 0

<=> \(\left[{}\begin{matrix}x=\sqrt{2}-1\left(tm\right)\\x=-\sqrt{2}-1\left(ktm\right)\end{matrix}\right.\)

Tập nghiệm S = {\(\sqrt{2}-1\)}