a: ĐKXĐ: x>=0

b: \(\Leftrightarrow\dfrac{2\sqrt{2}-2\sqrt{2-\sqrt{x}}+\sqrt{2x}-\sqrt{x\left(2-\sqrt{x}\right)}+2\sqrt{2}+2\sqrt{2+\sqrt{x}}-\sqrt{2x}-\sqrt{x\left(2+\sqrt{x}\right)}}{2-2+\sqrt{x}}=\sqrt{2}\)

\(\Leftrightarrow4\sqrt{2}-2\sqrt{x\left(\sqrt{x}+2\right)}=\sqrt{2x}\)

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

\(\Leftrightarrow4x\left(\sqrt{x}+2\right)=32-16\sqrt{x}+2x\)

\(\Leftrightarrow4x\sqrt{x}+8x-32+16\sqrt{x}-2x=0\)

=>\(x\in\left\{0;1.2996\right\}\)

\(\sqrt{1+2005^2+\dfrac{2005^2}{2006^2}}=\dfrac{1}{2006}\sqrt{2006^2+2005^2+\left(2005.2006\right)^2}\)

\(=\dfrac{1}{2006}\sqrt{\left(2006-2005\right)^2+2.2005.2006+\left(2005.2006\right)^2}\)

\(=\dfrac{1}{2006}\sqrt{1+2.2005.2006+\left(2005.2006\right)^2}\)

\(=\dfrac{1}{2006}\sqrt{\left(2005.2006+1\right)^2}=\dfrac{2005.2006+1}{2006}=2005+\dfrac{1}{2006}\)

Phương trình tương đương:

\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=2005+\dfrac{1}{2006}+\dfrac{2005}{2006}\)

\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2006\)

TH1: \(x\ge2\): \(x-1+x-2=2006\Rightarrow2x=2009\Rightarrow x=\dfrac{2009}{2}\)

TH2: \(x\le1\) : \(1-x+2-x=2006\Rightarrow-2x=2003\Rightarrow x=\dfrac{-2003}{2}\)

TH3: \(1< x< 2:\) \(x-1+2-x=2006\Rightarrow3=2006\) (vô nghiệm)

Vậy \(\left[{}\begin{matrix}x=\dfrac{2009}{2}\\x=\dfrac{-2003}{2}\end{matrix}\right.\)

ĐK: \(0< x\le4\)

Đặt \(\sqrt{2+\sqrt{x}}=a\left(a>0\right)\) ; \(\sqrt{2-\sqrt{x}}=b\left(b\ge0\right)\)

=> \(a^2+b^2=2+\sqrt{x}+2-\sqrt{x}=4\) (1)

Ta có: \(\dfrac{a^2}{\sqrt{2}+a}+\dfrac{b^2}{\sqrt{2}-b}=\sqrt{2}\)

<=> \(\dfrac{a^2.\sqrt{2}-a^2b+b^2.\sqrt{2}+ab^2}{2+\sqrt{2}\left(a-b\right)-ab}=\sqrt{2}\)

<=> \(\left(a^2+b^2\right)\sqrt{2}+ab\left(b-a\right)=2\sqrt{2}+2\left(a-b\right)-ab.\sqrt{2}\)

<=> \(4\sqrt{2}+ab\left(b-a\right)=2\sqrt{2}+2\left(a-b\right)-ab.\sqrt{2}\) ( Theo 1)

<=> \(\left(a-b\right)\left(2+ab\right)=2\sqrt{2}+ab.\sqrt{2}\)

<=> \(\left(a-b-\sqrt{2}\right)\left(ab+2\right)=0\)

<=> \(\left[{}\begin{matrix}ab+2=0\\a-b-\sqrt{2}=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}ab=-2\\a-b=\sqrt{2}\end{matrix}\right.\) mà a² + b² = 4

Xét \(\left\{{}\begin{matrix}ab=-2\\a^2+b^2=4\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left(a-b\right)^2=8\\\left(a+b\right)^2=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}a-b=\pm\sqrt{8}\\a+b=0\end{matrix}\right.\) ( Loại vì \(a>0;b\ge0\) )

Xét \(\left\{{}\begin{matrix}a-b=\sqrt{2}\\a^2+b^2=4\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=b+\sqrt{2}\\\left(b+\sqrt{2}\right)^2+b^2=4\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}a=b+\sqrt{2}\\2b^2+2b.\sqrt{2}-2=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=b+\sqrt{2}\\b^2+b.\sqrt{2}-1=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}a=b+\sqrt{2}\\\left[{}\begin{matrix}b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\\b=\dfrac{-\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)

#Lề: Bn lấy cái đề ở đâu hay v?

v cac bac e giai xong lau roi cac bac a voi lai co cach giai ko can dai nhu the dau nhe

Lời giải:

Với mọi $x$ thuộc ĐKXĐ, ta luôn có:

\(\left\{\begin{matrix} \sqrt{3x+x^2+\frac{9}{4}}\geq 0\\ \sqrt{x^2+3x+1}\geq 0\end{matrix}\right.\)

Do đó, để \(\sqrt{3x+x^2+\frac{9}{4}}+\sqrt{x^2+3x+1}=0\) thì:

\(\left\{\begin{matrix} \sqrt{3x+x^2+\frac{9}{4}}= 0\\ \sqrt{x^2+3x+1}=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x=\frac{-3}{2}\\ x=\frac{3\pm \sqrt{5}}{2}\end{matrix}\right.\) (vô lý)

Do đó pt vô nghiệm.

Sửa đề: \(\sqrt{x-5}+\dfrac{1}{3}\sqrt{9x-45}=\dfrac{1}{5}\sqrt{25x-125}+6\)

\(\Leftrightarrow\sqrt{x-5}+\dfrac{1}{3}\cdot3\cdot\sqrt{x-5}-\dfrac{1}{5}\cdot5\sqrt{x-5}=6\)

\(\Leftrightarrow\sqrt{x-5}=6\)

=>x-5=36

hay x=41

b , \(\sqrt{1-4x+4x^2}-3=0\)

\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=3\)

\(\Leftrightarrow\left|1-2x\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}1-2x=3\\1-2x=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=2\\-2x=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Vậy nghiệm của phương trình là \(S=\left\{-1,2\right\}\)

phương trình này vô số nghiệm á bạn

đáp án là câu C nha

sao ra vo so nghiem z