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

ĐKXĐ: \(\dfrac{3}{2}\le x\le3\)

\(A=\sqrt{2x-3}+\sqrt{6-2x}+\left(2-\sqrt{2}\right)\sqrt{3-x}\)

\(A\ge\sqrt{2x-3+6-2x}+\left(2-\sqrt{2}\right)\sqrt{3-x}\ge\sqrt{3}\)

\(A_{min}=\sqrt{3}\) khi \(3-x=0\Rightarrow x=3\)

\(A=1.\sqrt{2x-3}+\sqrt{2}.\sqrt{6-2x}\le\sqrt{\left(1+2\right)\left(2x-3+6-2x\right)}=3\)

\(A_{max}=3\) khi \(2x-3=\dfrac{6-2x}{2}\Rightarrow x=2\)

-Em cảm ơn thầy nhiều ạ! 

Bo thi:>

+ đk x > 0 , x khác 1

Mọi người ơi, giúp em với ạ!

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

Biểu thức này ko tồn tại cả min lẫn max

thầy ơi em bị nhầm phải là tìm GTNN của \(\dfrac{1}{M}\)

rút gọn à banj

đúng rồi á

Đặt \(a=\sqrt[3]{\frac{12+\sqrt{135}}{3}};b=\sqrt[3]{\frac{12-\sqrt{135}}{3}}\)

=>\(a^3+b^3=\frac{12+\sqrt{135}+12-\sqrt{135}}{3}=\frac{24}{3}=8\) ; \(ab=\sqrt[3]{\frac{\left(12+\sqrt{135}\right)\left(12-\sqrt{135}\right)}{3\cdot3}}=\sqrt[3]{\frac{144-135}{9}}=1\)

\(x=\frac13\left(1+\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\right)\)

=>\(3x=1+\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\)

=>\(3x-1=\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\)

=>\(\left(3x-1\right)=a+b\)

=>\(\left(3x-1\right)^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)=a^3+b^3+3ab\left(3x-1\right)\)

=>(3x-1)^3=8+3*1*(3x-1)=8+3(3x-1)

=>\(27x^3-27x^2+9x-1=9x+5\)

=>\(27x^3-27x^2=6\)

=>\(9x^3-9x^2=2\)

\(M=\left(9x^3-9x^2-3\right)^2\)

\(=\left(2-3\right)^2=\left(-1\right)^2=1\)

\(f\left(x\right)=\sqrt{3-x}+\sqrt{2+x}\ge\sqrt{3-x+2+x}=\sqrt{5}\)

Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}3-x=0\\2+x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)

Vậy GTNN của \(f\left(x\right)=\sqrt{5}\) khi và chỉ khi x = 3; x = -2

bạn ơi ở bước:

f(x)=\(\sqrt{3-x}+\sqrt{2+x}\ge\sqrt{3-x+2+x}\)

làm sao bạn ra đc bất đẳng thức như vậy ạ

\(a,A=\left(\dfrac{x+14\sqrt{x}-5}{x-25}+\dfrac{\sqrt{x}}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)

\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{x+14\sqrt{x}-5+x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{2x+10\sqrt{x}-\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

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

\(\Rightarrow A=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

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