Phương trình chứa căn - Hỏi đáp

đề bài sai rồi bạn ơi

\(4-2\sqrt{15} <0\)

Solution:

Dạng tổng quát :

\(\sqrt{1+k^2+\frac{k^2}{\left(k-1\right)^2}}=\sqrt{\frac{\left(1+k^2\right)\left(k+1\right)^2+k^2}{\left(k-1\right)^2}}\)

\(=\sqrt{\frac{k^4-2k^3+3k^2-2k+1}{\left(k-1\right)^2}}=\sqrt{\frac{\left(k^2-k\right)^2+2\left(k^2-k\right)+1}{\left(k-1\right)^2}}\)

\(=\sqrt{\frac{\left(k^2-k+1\right)^2}{\left(k-1\right)^2}}=\frac{k^2-k+1}{k-1}\)

\(=\frac{k\left(k-1\right)+1}{k-1}=k+\frac{1}{k-1}\)

Áp dụng ta có :

\(S=\sqrt{1+2020^2+\frac{2020^2}{2019^2}}-\frac{2020}{2019}\)

\(S=2020+\frac{1}{2019}-\frac{2020}{2019}\)

\(S=2020+\frac{-2019}{2019}\)

\(S=2020-1\)

\(S=2019\)

Vậy...

S=2019

Lời giải:

Vì $xy+yz+xz=1$ nên:

\(x^2+1=x^2+xy+yz+xz=(x+y)(x+z)\)

\(y^2+1=y^2+xy+yz+xz=(y+x)(y+z)\)

\(z^2+1=z^2+xy+yz+xz=(z+y)(z+x)\)

Do đó:

\(\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{1+z^2}=\frac{x}{(x+y)(x+z)}+\frac{y}{(y+x)(y+z)}+\frac{z}{(z+x)(z+y)}\)

\(=\frac{x(y+z)+y(x+z)+z(x+y)}{(x+y)(y+z)(x+z)}=\frac{2(xy+yz+xz)}{(x+y)(y+z)(x+z)}=\frac{2}{\sqrt{(x+y)^2(y+z)^2(x+z)^2}}\)

\(=\frac{2}{\sqrt{(x+y)(x+z)(y+z)(y+x)(z+x)(z+y)}}=\frac{2}{\sqrt{(x^2+1)(y^2+1)(z^2+1)}}\) (đpcm)

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.....

\(P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{1}{\sqrt{x}}+1\)

\(\ge2\sqrt{\sqrt{x}\cdot\dfrac{1}{\sqrt{x}}}+1=3\)

Khi x=1

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

Nhân hai vế của pt với \(\left(x-\sqrt{1+y^2}\right)\left(y-\sqrt{1+x^2}\right)\)

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

\(\Leftrightarrow\left(x^2-y^2-1\right)\left(y^2-x^2-1\right)=xy-x\sqrt{1+x^2}-y\sqrt{1+y^2}+\sqrt{\left(1+y^2\right)\left(1+x^2\right)}\)

\(\Leftrightarrow\left[-1+\left(x^2-y^2\right)\right]\left[-1-\left(x^2-y^2\right)\right]=2xy+2\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-\left(xy+x\sqrt{1+y^2}+y\sqrt{1+x^2}+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)\)

\(\Leftrightarrow1^2-\left(x^2-y^2\right)^2=2xy+2\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-\left(x+\sqrt{1+y^2}\right)\left(y+\sqrt{1+x^2}\right)\)

\(\Leftrightarrow1-\left(x^2-y^2\right)^2=2xy+2\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-1\)

\(\Leftrightarrow2\left(1-xy\right)=\left(x^2-y^2\right)^2+2\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)(*)

Mặt khác : \(2\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2\sqrt{x^2+y^2+1+x^2y^2}\)

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

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

Vì \(\left(x^2-y^2\right)^2\ge0\forall x;y\) do đó theo (*) ta có :

\(2\left(1-xy\right)\ge2\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2\sqrt{\left(x+y\right)^2+\left(xy-1\right)^2}\)

\(\Leftrightarrow1-xy\ge\sqrt{\left(x+y\right)^2+\left(xy-1\right)^2}\ge\sqrt{\left(xy-1\right)^2}=\left|xy-1\right|\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-y^2\right)^2=0\\\left(x+y\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-y^2=0\\x+y=0\end{matrix}\right.\)\(\Leftrightarrow x=-y\)

Thay vào P ta được :

\(P=x^7-x^7+2x^5-2x^5-3x^3+3x^3+4x-4x+100\)

\(P=0+0-0+0+100\)

\(P=100\)

Vậy...

p/s: mệt...

P=\(\left(\frac{3\sqrt{x}\left(\sqrt{x}-2\right)+\sqrt{x}\left(\sqrt{x}+2\right)-(x-\sqrt{x})}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\frac{3\sqrt{x}}{\sqrt{x}+2}\right)=\left(\frac{3x-6\sqrt{x}+x+2\sqrt{x}-x+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\frac{3\sqrt{x}}{\sqrt{x}+2}\right)=\left(\frac{3x-3\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right).\frac{\sqrt{x}+2}{3\sqrt{x}}=\frac{3\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}+2}{3\sqrt{x}}=\frac{\sqrt{x}-1}{\sqrt{x}-2}\)

P=\(\frac{a-1}{\sqrt{b-1}}\sqrt{\frac{b-2\sqrt{b}+1}{a^2-2a+1}}=\frac{a-1}{\sqrt{b-1}}\sqrt{\frac{\left(\sqrt{b}-1\right)^2}{\left(a-1\right)^2}}=\frac{a-1}{\sqrt{b-1}}.(\frac{\sqrt{b}-1}{a-1})=\frac{\sqrt{b}-1}{\sqrt{b-1}}\)

x=3 hoặc x=1