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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

D

Bạn ấy thay đề trc khi anh đăng đấy em

Bạn ấy thay đề trc khi anh đăng đấy em

\(a,\left|x+2\right|=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

\(b,\left|x-5\right|=\left|-7\right|\)

\(\Leftrightarrow\left|x-5\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-2\end{matrix}\right.\)

\(c,\left(7-x\right)-\left(25+7\right)=-25\)

\(\Leftrightarrow7-x-32=-25\)

\(\Leftrightarrow x=0\)

\(d,\left|x-3\right|=\left|5\right|+\left|-7\right|\)

\(\Leftrightarrow\left|x-3\right|=12\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)

0,7 : x = 0,75 - 0,5 : x ( x không được bằng 0)

0,7 : x + 0,5 : x = 0,75

1,2 : x = 0,75

x = 1,2 : 0,75 

x = 1,6

\(\%O=100\%-40\%-6,67\%=53,33\%\)

Gọi CTHH của X là: \(\left(H_xC_yO_z\right)_n\)

Có: \(x:y:z=\dfrac{1}{6,67\%}:\dfrac{12}{40\%}:\dfrac{16}{53,33\%}\approx0,15:0,3:0,3\approx1:2:2\)

Vậy CTĐG của X là: \(\left(HC_2O_2\right)_n\)

\(\dfrac{x+7}{x-7}-\dfrac{x-7}{x+7}+\dfrac{4x^2}{x^2-49}\left(x\ne\pm7\right)\)

\(=\dfrac{\left(x+7\right)^2}{x^2-49}-\dfrac{\left(x-7\right)^2}{x^2-49}+\dfrac{4x^2}{x^2-49}\)

\(=\dfrac{x^2+14x+49-x^2+14x-49+4x^2}{x^2-49}\)

\(=\dfrac{4x^2+28x}{x^2-49}=\dfrac{4x\left(x+7\right)}{\left(x+7\right)\left(x-7\right)}=\dfrac{4x}{x-7}\)

a. \(x\ne\pm3\)

b. \(=\dfrac{3\left(x-3\right)}{x^2-9}+\dfrac{x+3}{x^2-9}+\dfrac{18}{x^2-9}\)

\(=\dfrac{3x-9+x+3+18}{x^2-9}\)

\(=\dfrac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{4}{x-3}\)

c. Với \(P=4\Leftrightarrow\dfrac{4}{x-3}=4\)

\(\Leftrightarrow4x-12=4\)

\(\Leftrightarrow4x=16\)

\(\Leftrightarrow x=4\left(tm\right)\)