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\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)
\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)
\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)
hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)
a)\(\left|\dfrac{x-1}{3}\right|=\dfrac{11}{5}\Rightarrow\dfrac{x-1}{3}=\pm\dfrac{11}{5}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x-1}{3}=\dfrac{11}{5}\\\dfrac{x-1}{3}=-\dfrac{11}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{33}{5}\\x-1=\dfrac{-33}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{38}{5}\\x=\dfrac{-28}{5}\end{matrix}\right.\)
a: P(x)=2x^5-2x^5+4x^4-3x^4+5=x^4+5
Q(x)=-5x^4+2x^4-x^3+3x^2-10x+2
=-3x^4-x^3+3x^2-10x+2
b: P(x)+Q(x)
=x^4+5-3x^4-x^3+3x^2-10x+2
=-2x^4-x^3+3x^2-10x+7
Q(x)-P(x)
=-3x^4-x^3+3x^2-10x+2-x^4-5
=-4x^4-x^3+3x^2-10x-3
P(x)-Q(x)=-(Q(x)-P(x))
=4x^4+x^3-3x^2+10x+3
a: Khi x=1/3 thì \(A=3\cdot\dfrac{1}{9}-6\cdot\dfrac{1}{3}+5=\dfrac{10}{3}\)
Khi x=-2 thì \(A=3\cdot4-6\cdot\left(-2\right)+5=12+12+5=29\)
b: Trường hợp 1: x=2; y=-3
\(C=2\cdot2^2-3\cdot2\cdot\left(-3\right)+4\cdot\left(-3\right)^2=62\)
Trường hợp 2: x=-2; y=-3
\(C=2\cdot\left(-2\right)^2-3\cdot\left(-2\right)\cdot\left(-3\right)+4\cdot\left(-3\right)^2=26\)
Bài 1:
- \(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)
- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1
-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)
- \(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)
\(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))
\(x\) = \(\dfrac{3}{14}\)
Vậy \(x=\dfrac{3}{14}\)
Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1
2\(x\) - 7\(x\) = \(\dfrac{3}{2}\) - 1 + \(\dfrac{2}{3}\)
- 5\(x\) = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\)
- 5\(x\) = \(\dfrac{7}{6}\)
\(x\) = \(\dfrac{7}{6}\) : (- 5)
\(x\) = - \(\dfrac{7}{30}\)
Vậy \(x=-\dfrac{7}{30}\)
\(5-\left|3x-1\right|=3\)
\(\left|3x-1\right|=2\)
\(\Rightarrow\orbr{\begin{cases}3x-1=2\\3x-1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}3x=3\\3x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)
vậy \(\orbr{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)
\(\left|x+\frac{3}{4}\right|-5=-2\)
\(\left|x+\frac{3}{4}\right|=3\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=3\\x+\frac{3}{4}=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=-\frac{15}{4}\end{cases}}\)
\(\left(1-2x\right)^2=9\)
\(\left(1-2x\right)^2=3^2\)
\(\Rightarrow1-2x=3\)
\(\Rightarrow2x=-2\)
\(\Rightarrow x=-1\)
vậy \(x=-1\)
\(\left(x+5\right)^3=-64\)
\(\left(x+5\right)^3=\left(-4\right)^3\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=-9\)
vậy \(x=-9\)
\(\left(2x+1\right)^2=\frac{4}{9}\)
\(\left(2x+1\right)^2=\left(\frac{2}{3}\right)^2\)
\(\Rightarrow2x+1=\frac{2}{3}\)
\(\Rightarrow2x=\frac{-1}{3}\)
\(\Rightarrow x=\frac{-1}{6}\)
vậy \(x=-\frac{1}{6}\)