b) \(x+\sqrt{x+\dfrac{1}{4}+2\sqrt{x+\dfrac{1}{4}}.\dfrac{1}{2}+\dfrac{1}{4}}=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}A=0\\B=0\end{matrix}\right.\)

xét \(A^2+B^2-2A-2B+2\)

\(=A^2-2A+1+B^2-2B+1\)

=\(\left(A-1\right)^2+\left(B-1\right)^2\ge0\)

\(\RightarrowĐPCM\)

X=(a;b;c)

X=(a;b;d)

\(C=\left(-5;0\right)\cup\left(3;5\right)\)

D=\(\left(-1;2\right)\cup\left(4;6\right)\)

\(C\cap D=\left(-1;0\right)\cup\left(4;5\right)\)

điếc là hư tai hư tai là 24

vậy có 24

ai tích tớ sẽ được tích lại

0 đáp án

a) \(\left|x+7\right|=2x+5\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+7=2x+5\\x+7=-2x-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

b)\(\left|3x-4\right|=2x+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-4=2x+1\\3x-4=-2x-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=\dfrac{3}{5}\end{matrix}\right.\)

c)\(\left|1-4x\right|=2x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-4x=2x-1\\1-4x=1-2x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x=0\end{matrix}\right.\)

d)\(\left|2x+3\right|=12-6x\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3=12-6x\\2x+3=6x-12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{15}{4}\end{matrix}\right.\)