Question

help pls

I'm sorry, I can't assist with that.

Answer 1

Bài 7:

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\2\sqrt{x}-2\ne0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=0\\2\sqrt{x}\ne2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)

b: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\\sqrt{x}-5\ne0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=0\\\sqrt{x}\ne5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=0\\x\ne25\end{matrix}\right.\)

Bài 14:

a: Thay x=16 vào A, ta được:

\(A=\dfrac{4+3}{4-2}=\dfrac{7}{2}\)

b: \(B=\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}-\dfrac{x}{4-x}\)

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

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

c: \(B\left(\sqrt{x}-2\right)+2\sqrt{x}=x-\sqrt{7\left(x-2\right)}+7\)

=>\(\sqrt{x}+2\sqrt{x}=x-\sqrt{7\left(x-2\right)}+7\)

=>\(x-3\sqrt{x}+7-\sqrt{7\left(x-2\right)}=0\)

=>\(\sqrt{x}\left(\sqrt{x}-3\right)+\dfrac{49-7\left(x-2\right)}{7+\sqrt{7\left(x-2\right)}}=0\)

=>\(\sqrt{x}\left(\sqrt{x}-3\right)+\dfrac{-7\left(x-9\right)}{7+\sqrt{7\left(x-2\right)}}=0\)

=>\(\left(\sqrt{x}-3\right)\left[\sqrt{x}-\dfrac{7\left(\sqrt{x}+3\right)}{7+\sqrt{7\left(x-2\right)}}\right]=0\)

=>\(\sqrt{x}-3=0\)

=>x=9(nhận)