\(2,\\ a,\sqrt{4x-4}+\sqrt{9x-9}-\sqrt{25x-25}=7\left(x\ge1\right)\\ \Leftrightarrow2\sqrt{x-1}+3\sqrt{x-1}-5\sqrt{x-1}=7\\ \Leftrightarrow0\sqrt{x-1}=7\Leftrightarrow x\in\varnothing\\ b,\sqrt{2x^2-3}=4\left(x\le-\dfrac{\sqrt{6}}{2};\dfrac{\sqrt{6}}{2}\le x\right)\\ \Leftrightarrow2x^2-3=16\\ \Leftrightarrow x^2=\dfrac{19}{2}\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{19}{2}}\left(tm\right)\\x=-\sqrt{\dfrac{19}{2}}\left(tm\right)\end{matrix}\right.\)

\(1,\\ A=\sqrt{5+4x}+\sqrt{7-3x}\\ ĐKXĐ:\left\{{}\begin{matrix}5+4x\ge0\\7-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{5}{4}\\x\le\dfrac{7}{3}\end{matrix}\right.\)

Bài 2:

a) \(\sqrt{4x-4}+\sqrt{9x-9}-\sqrt{25x-25}=7\left(đk:x\ge1\right)\)

\(\Leftrightarrow2\sqrt{x-1}+3\sqrt{x-2}-5\sqrt{x-1}=7\)

\(\Leftrightarrow0=7\left(VLý\right)\)

Vậy \(S=\varnothing\)

b) \(\sqrt{2x^2-3}=4\left(đk:-\sqrt{\dfrac{3}{2}}\ge x\ge\sqrt{\dfrac{3}{2}}\right)\)

\(\Leftrightarrow2x^2-3=16\)

\(\Leftrightarrow2x^2=19\Leftrightarrow x^2=\dfrac{19}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{19}{2}}\left(tm\right)\\x=-\sqrt{\dfrac{19}{2}}\left(tm\right)\end{matrix}\right.\)

ở dưới ghi - ở trên ghi + rốt cuộc đề nào đúng

Bạn dùng điện thoại hay máy tính bàn vậy? Bấm vào nút fx để nhập công thức nhé.

x=3,17547292889759

\(a,\)\(\sqrt{x^2-2x+1}=\sqrt{\left(x-1\right)^2}\)

\(đkxđ\Leftrightarrow\sqrt{\left(x-1\right)^2}\ge0\)

\(\Rightarrow x-1\ge0\Rightarrow x\ge1\)

\(b,\)\(\sqrt{x+3}+\sqrt{x+9}\)

\(đkxđ\Leftrightarrow\hept{\begin{cases}x+3\ge0\\x+9\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ge-3\\x\ge-9\end{cases}}}\)

\(\Rightarrow x\ge-3\)

\(c,\)\(\sqrt{\frac{x-1}{x+2}}\)

\(đkxđ\Leftrightarrow\hept{\begin{cases}x+2\ne0\\\frac{x-1}{x+2}\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-2\\\frac{x-1}{x+2}\ge0\end{cases}}}\)

\(\frac{x-1}{x+2}\ge0\)\(\Rightarrow\orbr{\begin{cases}x-1\ge0;x+2>0\\x-1\le0;x+2< 0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x\ge-1;x>-2\\x\le1;x< 2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x\ge-1\\x< 2\end{cases}}\)

Vậy căn thức xác định khi x \(\ge\)-1 hoawck x < 2

\(d,\)\(\sqrt{x-2}-\frac{1}{x-5}\)

\(đkxđ\Leftrightarrow\orbr{\begin{cases}\sqrt{x-2}xđ\\\frac{1}{x-5}xđ\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-2\ge0\\x-5\ne0\end{cases}\Rightarrow\orbr{\begin{cases}x\ge2\\x\ne5\end{cases}}}\)

Vậy biểu thức xác định \(\Leftrightarrow x\ge2\)và \(x\ne5\)

\(a,=5\sqrt{2}-3\sqrt{2}+6\sqrt{2}=8\sqrt{2}\\ b,=\dfrac{5\sqrt{3}}{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}=\dfrac{5\sqrt{3}}{3}-\sqrt{3}+1=\dfrac{5\sqrt{3}-3\sqrt{3}+3}{3}=\dfrac{2\sqrt{3}+3}{3}\)

ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow\sqrt{4x}-\sqrt{3x+1}+\sqrt{2x+2}-\sqrt{x+3}=0\)

\(\Leftrightarrow\frac{x-1}{\sqrt{4x}+\sqrt{3x+1}}+\frac{x-1}{\sqrt{2x+2}+\sqrt{x+3}}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{\sqrt{4x}+\sqrt{3x+1}}+\frac{1}{\sqrt{2x+2}+\sqrt{x+3}}\right)=0\)

\(\Rightarrow x=1\)

viết đề khó hiểu quá

\(1=\lim\limits_{x\rightarrow0}\frac{\sqrt{x+4}-2}{2x}=\lim\limits_{x\rightarrow0}\frac{x}{2x}.\frac{1}{\sqrt{x+4}+2}=\lim\limits_{x\rightarrow0}\frac{1}{2\left(\sqrt{x+4}+2\right)}=\frac{1}{2\left(\sqrt{4}+2\right)}\)

\(2=\lim\limits_{x\rightarrow1}\frac{\sqrt{x+3}-2}{x-1}=\lim\limits_{x\rightarrow1}\frac{x-1}{x-1}.\frac{1}{\sqrt{x+3}+2}=\lim\limits_{x\rightarrow1}\frac{1}{\sqrt{x+3}+2}=\frac{1}{\sqrt{1+3}+2}\)

\(3=\lim\limits_{x\rightarrow3}\frac{\sqrt{2x+3}-x}{\left(x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow3}\frac{2x+3-x^2}{\left(x-1\right)\left(x-3\right)}.\frac{1}{\sqrt{2x+3}+x}\)

\(=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(3-x\right)}{\left(x-1\right)\left(x-3\right)}.\frac{1}{\sqrt{2x+3}+x}=\lim\limits_{x\rightarrow3}\frac{x+1}{\left(1-x\right)\left(\sqrt{2x+3}+x\right)}=\frac{3+1}{\left(1-3\right)\left(\sqrt{9}+3\right)}\)

\(4=\lim\limits_{x\rightarrow2}\frac{\left(x-2\right)\left(2x-1\right)}{\left(x+1\right)^2\left(x-2\right)}=\lim\limits_{x\rightarrow2}\frac{2x-1}{\left(x+1\right)^2}=\frac{4-1}{\left(2+1\right)^2}\)

P/s: lần sau bạn sử dụng tính năng gõ công thức ở kí hiệu \(\sum\) góc trên cùng bên trái khung soạn thảo ấy, khó nhìn đề quá chẳng muốn làm

d: Ta có: \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}\)

\(=\dfrac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{11}+1-\sqrt{11}+1}{\sqrt{2}}\)

\(=\sqrt{2}\)