Phương trình chứa căn - Hỏi đáp

Giúp mk với

Lời giải:

ĐK: \(x\in\mathbb{R}\)

\(\sqrt{x^2-2x+5}+\sqrt{x^2+2x+10}=\sqrt{29}\)

\(\Leftrightarrow \sqrt{x^2+2x+10}=\sqrt{29}-\sqrt{x^2-2x+5}\)

Bình phương 2 vế:

\(x^2+2x+10=29+x^2-2x+5-2\sqrt{29(x^2-2x+5)}\)

\(\Leftrightarrow 4x-24=-2\sqrt{29(x^2-2x+5)}\)

\(\Leftrightarrow 12-2x=\sqrt{29(x^2-2x+5)}\)

\(\Rightarrow \left\{\begin{matrix} 12-2x\geq 0\\ (12-2x)^2=29(x^2-2x+5)\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x\leq 6\\ 4x^2+144-48x=29x^2-58x+145\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x\leq 6\\ 25x^2-10x+1=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x\leq 6\\ (5x-1)^2=0\end{matrix}\right.\Rightarrow x=\frac{1}{5}\)

(thỏa mãn)

Vậy.....

ptvn

\(B=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}=1-\dfrac{\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}\le1\)

\(B=\dfrac{\sqrt{x}}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\ge0\)

Vật Min B = 0 khi x = 0 và Max = 1 khi x = 1

Zalo của mình:Lạc Thiên Y. Liên hệ vs mình, sẽ gửi ảnh ₫áp án cho bạn!

2x-x^2 =t

<=>t^2 -6t+7 =0

t=-1 loại

t=7 => \(\left[{}\begin{matrix}x_1=1-2\sqrt{2}\\x_2=1+2\sqrt{2}\end{matrix}\right.\)

\(A=\sum\dfrac{x}{\sqrt{x^2+1}+x}=\sum\dfrac{x}{\sqrt{x^2+xy+yz+xz}+x}=\sum\dfrac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}+x}\le\sum\dfrac{x}{\sqrt{xy}+\sqrt{xz}+x}=\sum\dfrac{\sqrt{x}}{\sqrt{y}+\sqrt{x}+\sqrt{z}}=1\)

Từ giả thiết \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\Rightarrow xy+yz+xz=1\left(x=\dfrac{1}{a};y=\dfrac{1}{b};z=\dfrac{1}{c}\right)\)

\(A=\sum\dfrac{1}{\sqrt{1+a^2}}=\sum\dfrac{\dfrac{1}{a}}{\sqrt{\dfrac{1}{a^2}+1}}=\sum\dfrac{x}{\sqrt{x^2+1}}=\sum\dfrac{x}{\sqrt{x^2+xy+yz+xz}}=\sum\dfrac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}\le\dfrac{1}{2}\sum\dfrac{x}{x+y}+\dfrac{x}{x+z}=\dfrac{3}{2}\)

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

\(=\dfrac{2\sqrt{x}+10-2\sqrt{x}+10}{x-25}+\dfrac{10\sqrt{x}}{x-5}\)

\(=\dfrac{20}{x-25}+\dfrac{10\sqrt{x}}{x-5}=\dfrac{20x-100+10x\sqrt{x}-250\sqrt{x}}{\left(x-25\right)\left(x-5\right)}\)

\(=\dfrac{20x-100+10x\sqrt{x}-250\sqrt{x}}{\left(x-25\right)\left(x-5\right)}\)