ấn vào đúng cho mk đi mk ân cho bạn ok

\(A=\dfrac{\sqrt{4x^2-4x+1}}{4x-2}=\dfrac{\sqrt{\left(2x-1\right)^2}}{2\left(2x-1\right)}=\dfrac{\left|2x-1\right|}{2\left(2x-1\right)}\)

\(\Rightarrow\left|A\right|=\left|\dfrac{\left|2x-1\right|}{2\left(2x-1\right)}\right|=\dfrac{\left|2x-1\right|}{2\left|2x-1\right|}=\dfrac{1}{2}\)

Ta có: \(A=\dfrac{\sqrt{4x^2-4x+1}}{4x-2}\)

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

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

\(\Leftrightarrow\left|A\right|=0.5\)

\(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\\ \Leftrightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\\ \Leftrightarrow x\left(2+3\right)=\dfrac{3}{4}-\dfrac{1}{2}+6\\ \Leftrightarrow5x=\dfrac{25}{4}\\ \Leftrightarrow x=\dfrac{25}{4}:5=\dfrac{5}{4}\\ ---\\ 4^{x+2}+4^x=272\\ \Leftrightarrow4^x\left(4^2+1\right)=272\\ \Leftrightarrow4^x.17=272\\ \Leftrightarrow4^x=\dfrac{272}{17}=16=4^2\\ Vậy:x=2\\ ----\\ \left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1,2-5x=0\\2,125+0,5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1,2\\0,5x=-2,125\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}=0,24\\x=\dfrac{-2,125}{0,5}=-4,25\end{matrix}\right.\)

a) \(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\)

\(\Rightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\)

\(\Rightarrow5x-6=\dfrac{3}{4}-\dfrac{1}{2}\)

\(\Rightarrow5x-6=\dfrac{1}{4}\)

\(\Rightarrow5x=\dfrac{1}{4}+6\)

\(\Rightarrow5x=\dfrac{25}{4}\)

\(\Rightarrow x=\dfrac{25}{4}:5\)

\(\Rightarrow x=\dfrac{5}{4}\)

b) \(4^{x+2}+4^x=272\)

\(\Rightarrow4^x\cdot4^2+4^x\cdot1=272\)

\(\Rightarrow4^x\cdot\left(16+1\right)=272\)

\(\Rightarrow4^x\cdot17=272\)

\(\Rightarrow4^x=16\)

\(\Rightarrow4^x=4^2\)

\(\Rightarrow x=2\)

c) \(\left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}1,2-5x=0\\\dfrac{15}{8}+\dfrac{1}{2}x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5x=1,2\\\dfrac{1}{2}x=-\dfrac{15}{8}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}\\x=-\dfrac{15}{8}:\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{25}\\x=-\dfrac{15}{4}\end{matrix}\right.\)

\(C=\left[{}\begin{matrix}4\cdot\dfrac{1}{4}-2\cdot\dfrac{1}{2}-1=-1\\4\cdot\dfrac{-1}{4}-2\cdot\dfrac{-1}{2}-1=-1\end{matrix}\right.\)

\(A=\sqrt{4x^2-4x+1}+\sqrt{4x^2-36x+81}\)

\(=\sqrt{\left(2x\right)^2-2.2x.1+1^2}+\sqrt{\left(2x\right)^2-2.2x.9+9^2}\)

\(=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-9\right)^2}\)

\(=\left|2x-1\right|+\left|2x-9\right|\)

\(=2x-1+9-2x=8\)

Thanks bn♥

a) \(\dfrac{2}{x-3}\sqrt{\dfrac{x^2-6x+9}{4y^4}}=\dfrac{2}{x-3}.\dfrac{3-x}{2y^2}=\dfrac{2.2y^2}{\left(x-3\right)\left(3-x\right)}=-\dfrac{4y^2}{x^2-6x+9}=-\dfrac{2y}{x-3}\)

=\(\dfrac{2}{2x-1}\sqrt{5}x\sqrt[]{\left(1-2x\right)^2}\)

=\(\dfrac{2\sqrt{5}x\left(1-2x\right)}{2x-1}\)

=\(\dfrac{-2\sqrt{5}x\left(2x-1\right)}{2x-1}\)​​

=\(-2\sqrt{5}x\)

em mới lớp 6-7 nên em sẽ giải theo kiểu lớp 6 là

em ko biết giải khó quá trời

không biết giải thì đừng spam lung tung !

làm như vậy chỉ khiến mọi người ghét em hơn thôi

\(a,A=\frac{2\sqrt{x}-1}{x+2}=\frac{2.5-1}{27}=\frac{9}{27}=\frac{1}{3}\)

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

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

\(=\frac{6x+4+5\sqrt{x}}{4x-1}-\frac{4x+2\sqrt{x}+3}{4x-1}=\frac{2x+1+3\sqrt{x}}{4x-1}\)

a/ \(=4x-\sqrt{\left(x-2\right)^2}=4x-x+2=3x+2\)

b/ \(=3x+\sqrt{\left(x+3\right)^2}=3x+x+3=4x+3\)

c/ xem lại đb

d/ \(=\frac{\sqrt{\left(x+2\right)^2}}{x+2}=\frac{x+2}{x+2}=1\)