Bài 1:

1. ĐKXĐ: \(x\ge0\) 
\(x-7\sqrt{x}+10=0\)

\(\Leftrightarrow x-2\sqrt{x}-5\sqrt{x}+10=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)-5\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{x}-5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=25\end{matrix}\right.\) ( thỏa mãn đk )

Vậy \(S=\left\{4;25\right\}\)

2. ĐKXĐ: \(x\ge4\) 
\(\sqrt{x^2-16}-5\sqrt{x-4}=0\)

\(\Leftrightarrow\sqrt{x^2-16}=5\sqrt{x-4}\)

\(\Leftrightarrow x^2-16=25\left(x-4\right)\)

\(\Leftrightarrow x+4=25\)

\(\Leftrightarrow x=21\) ( thỏa mãn đk )

Vậy \(S=\left\{21\right\}\)

3. ĐKXĐ: \(x\ge-4\)
\(\sqrt{x^2-16}-3\sqrt{x+4}=0\)

\(\Leftrightarrow\sqrt{x^2-16}=3\sqrt{x+4}\)

\(\Leftrightarrow x^2-16=9\left(x+4\right)\)

\(\Leftrightarrow x-4=9\)

\(\Leftrightarrow x=13\) ( thỏa mãn đk )

Vậy \(S=\left\{13\right\}\)

Bài 1:
a) ĐKXĐ: \(x\ge0\)

\(x-7\sqrt{x}+10=0\)

\(\Rightarrow x+10=7\sqrt{x}\)

\(\Rightarrow x^2+20x+100=49x\)

\(\Rightarrow x^2-29x+100=0\)

\(\Rightarrow\left(x^2-4x\right)-\left(25x-100\right)=0\)

\(\Rightarrow\left(x-4\right)\left(x-25\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-4=0\\x-25=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=25\end{matrix}\right.\)

b) ĐKXĐ:\(\left\{{}\begin{matrix}x^2-16\ge0\\x-4\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(x-4\right)\left(x+4\right)\ge0\\x-4\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x+4\ge0\\x-4\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge-4\\x\ge4\end{matrix}\right.\Rightarrow x\ge4\)

\(\sqrt{x^2-16}-5\sqrt{x-4}=0\)

\(\Rightarrow\sqrt{\left(x-4\right)\left(x+4\right)}-5\sqrt{x-4}=0\\ \Rightarrow\sqrt{x-4}\left(\sqrt{x+4}-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x-4}=0\\\sqrt{x+4}-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-4=0\\\sqrt{x+4}=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x+4=25\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=21\end{matrix}\right.\)

câu 2 thiếu x=4

4. ĐKXĐ: \(x\ge-7\)

\(\sqrt{x^2-49}-3\sqrt{x+7}=0\)

\(\Leftrightarrow\sqrt{x^2-49}=3\sqrt{x+7}\)

\(\Leftrightarrow x^2-49=9\left(x+7\right)\)

\(\Leftrightarrow x-7=9\)

\(\Leftrightarrow x=16\) ( thỏa mãn đk )

Vậy \(S=\left\{16\right\}\)

5. ĐKXĐ: \(x\ge0\)

\(x-14\sqrt{x}+45=0\)

\(\Leftrightarrow\left(\sqrt{x}-9\right)\left(\sqrt{x}-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-9=0\\\sqrt{x}-5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=9\\\sqrt{x}=5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=81\\x=25\end{matrix}\right.\)

Vậy \(S=\left\{81;25\right\}\)

6. ĐKXĐ: \(x\ge0\)

\(x-13\sqrt{x}+42=0\)

\(\Leftrightarrow\left(\sqrt{x}+6\right)\left(\sqrt{x}+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+7=0\\\sqrt{x}+6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-7\\\sqrt{x}=-6\end{matrix}\right.\) ( loại )

Vậy phương trình vô nghiệm

um, tớ quên xét 2TH

câu 3 ĐKXĐ sai

Câu 3 ĐKXĐ đúng mà c, \(x^2\) thì cần gì phải xét vì \(x^2\) luôn lớn hơn hoặc bằng 0 mà

c) ĐKXĐ:\(\left[{}\begin{matrix}x\ge4\\x=-4\end{matrix}\right.\)

TH1: \(x\ge4\): \(\sqrt{x^2-16}-3\sqrt{x+4}=0\\ \Rightarrow\sqrt{\left(x-4\right)\left(x+4\right)}-3\sqrt{x+4}=0\\ \Rightarrow\sqrt{x+4}\left(\sqrt{x-4}-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+4}=0\\\sqrt{x-4}-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+4=0\\\sqrt{x-4}=3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-4\\x-4=9\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-4\left(vôlí\right)\\x=13\end{matrix}\right.\)

\(\Rightarrow x=13\)

\(TH2:x=-4\) thỏa mãn

Vậy \(\left[{}\begin{matrix}x=13\\x=-4\end{matrix}\right.\)

câu ĐKXĐ \(x\ge-4\) thế mik lấy \(x=-3\) thì sao

x=-3 thì \(\sqrt{x+4}=\sqrt{-3+4}=\sqrt{1}>0\) mà c

\(\sqrt{x^2-16}\) thì sao

Đúng r, mình nhầm
Cảm ơnn

Câu 1: 

5: Ta có: \(x-14\sqrt{x}+45=0\)

\(\Leftrightarrow\left(\sqrt{x}-5\right)\cdot\left(\sqrt{x}-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=25\\x=49\end{matrix}\right.\)

6: Ta có: \(x-13\sqrt{x}+42=0\)

\(\Leftrightarrow\left(\sqrt{x}-6\right)\left(\sqrt{x}-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=36\\x=49\end{matrix}\right.\)