\(J=\frac{2010}{4x+20\sqrt{x}+30}\)

\(=\frac{2010}{\left(2\sqrt{x}\right)^2+2.2\sqrt{x}.5+25+5}\)

\(=\frac{2010}{\left(2\sqrt{x}+5\right)^2+5}\)

\(A_{max}\Leftrightarrow\frac{2010}{\left(2\sqrt{x}+5\right)^2+5}\)lớn nhất

\(\Rightarrow\left(2\sqrt{x}+5\right)^2+5\)nhỏ  nhất

\(\Rightarrow\left(2\sqrt{x}+5\right)^2\)nhỏ nhất 

Mà \(2\sqrt{x}+5\ge5\Rightarrow2\sqrt{x}+5=5\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\)

Với x = 0 \(\Rightarrow J_{max}=\frac{2010}{4.0+20\sqrt{0}+30}=\frac{2010}{30}=67\)

h) 

ĐKXĐ: $x\geq -5$

PT $\Leftrightarrow \sqrt{x+5}=6$

$\Rightarrow x+5=36\Rightarrow x=31$ (thỏa mãn)

i) ĐKXĐ: $x\geq 5$

PT \(\Leftrightarrow \sqrt{x-5}+4\sqrt{x-5}-\sqrt{x-5}=12\)

\(\Leftrightarrow 4\sqrt{x-5}=12\Leftrightarrow \sqrt{x-5}=3\Rightarrow x-5=9\Rightarrow x=14\) (thỏa mãn)

j) 

ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow 3\sqrt{2x}+\sqrt{2x}-6\sqrt{2x}+4=0$

$\Leftrightarrow -2\sqrt{2x}+4=0$

$\Leftrightarrow \sqrt{2x}=2$

$\Rightarrow x=2$ (thỏa mãn)

k) ĐK: $x^2\geq 5$

PT $\Leftrightarrow 2\sqrt{x^2-5}-\frac{1}{3}\sqrt{x^2-5}+\frac{3}{4}\sqrt{x^2-5}-\frac{5}{12}\sqrt{x^2-5}=4$

$\Leftrightarrow 2\sqrt{x^2-5}=4$

$\Leftrightarrow \sqrt{x^2-5}=2$

$\Rightarrow x^2-5=4$

$\Leftrightarrow x^2=9\Rightarrow x=\pm 3$ (đều thỏa mãn)

l) ĐKXĐ: $x\geq -1$

PT $\Leftrightarrow 2\sqrt{x+1}+3\sqrt{x+1}-\sqrt{x+1}=4$

$\Leftrightarrow 4\sqrt{x+1}=4$

$\Leftrightarrow \sqrt{x+1}=1$

$\Rightarrow x+1=1$

$\Rightarrow x=0$

m) 

ĐKXĐ: $x\geq -1$

PT $\Leftrightarrow 4\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}+3\sqrt{x+1}$

$\Leftrightarrow 6\sqrt{x+1}=16+2\sqrt{x+1}$

$\Leftrightarrow 4\sqrt{x+1}=16$

$\Leftrightarrow \sqrt{x+1}=4$

$\Rightarrow x=15$ (thỏa mãn)

câu 1 sai đề

\(\sqrt{x}+1chứkophải\sqrt{x+1}\)

\(\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}=\frac{a^2+a+1}{\left(a+1\right)}\Rightarrow\sqrt{1+2012^2+\frac{2012^2}{2013^2}}+\frac{2012}{2013}=\frac{2013^2}{2013}=2013\)
\(\Rightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=|x-1|+|x-2|=2013\)

giải tiếp nha

\(B=\dfrac{2010}{4x+20\sqrt{x}+30}\)

\(B=\dfrac{2010}{\left(2\sqrt{x}\right)^2+2\cdot2\sqrt{x}\cdot5+25+5}\)

\(B=\dfrac{2010}{\left(2\sqrt{x}+5\right)^2+5}\)

Ta có: \(\left(2\sqrt{x}+5\right)^2+5\ge5\)

\(\Rightarrow B=\dfrac{2010}{\left(2\sqrt{x}+5\right)^2+5}\le\dfrac{2010}{5}=402\)

Vậy: \(B_{min}=402\)

Mn giúp e với ak

a) \(\sqrt{x^2-6x+9}\)

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

\(=\sqrt{\left(x-3\right)^2}\) ≥0,∀x

⇒x∈\(R\)

b) \(\sqrt{x^2-2x+1}\)

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

\(=\sqrt{\left(x-1\right)^2}\) ≥0,∀x

⇒x∈\(R\)

Có: \(C=\frac{1}{\sqrt{x^2-4x+5}}\)

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

Vậy C_min=1 \(\Leftrightarrow x=2\)

Có: \(B=5-\sqrt{x^2-6x+14}\)

\(\Leftrightarrow B=5-\sqrt{\left(x-3\right)^2+5}\) \(\le5-\sqrt{5}\)

Vậy \(B_{min}=5-\sqrt{5}\Leftrightarrow x=3\)

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