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a) 22 + (2x -13) = 83 => 2x -13 = 61 => x = 37.
b) 51 - (-12 + 3x) = 27 => 63 - 3x = 27 => x = 12.
c) - (2x + 2) + 21 = - 23 => 2x + 2 = 44 => x = 21.
d) 25 - (25 - x) = 0 => 25 - 25 + x = 0 => x = 0.
a/ \(51-(-12+3x)=27\)
\(\Leftrightarrow51+12-27-3x=0\Leftrightarrow36=3x\Leftrightarrow x=\frac{36}{3}=12\)
KL:........
b/ $-x + 21=15+ 2x$
\(\Leftrightarrow2x+x=21-15\Leftrightarrow2x=6\Leftrightarrow x=3\)
KL: ...........
c) $7.(x-9)-5(6-x)=-6+11.x$
\(\Leftrightarrow7x-63-30+5x=-6+11x\Leftrightarrow7x+5x-11x=-6+63+30\Leftrightarrow x=87\)
KL:............
d) $(x-3).(x^2 + 2)=0$
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x\in\varnothing\end{matrix}\right.\)\(\Leftrightarrow x=3\)
e) $|2x-7|-22=-13$
\(\Leftrightarrow\left|2x-7\right|=9\Leftrightarrow\left[{}\begin{matrix}2x-7=9\\2x-7=-9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-1\end{matrix}\right.\)
KL: ...........
f) $(2x - 1)^3=-125$
\(\Leftrightarrow\left(2x-1\right)^3=\left(-5\right)^3\Leftrightarrow2x-1=-5\Leftrightarrow x=-2\)
KL: ...........
a, 117 - \(x\) = 28 - (-7)
117 - \(x\) = 28 + 7
117 - \(x\) = 35
\(x\) = 117 - 35
\(x\) = 82
b, \(x\) - (-38 - 2\(x\)) = (-3) - 8 + 2\(x\)
\(x\) + 38 + 2\(x\) = - 11 + 2\(x\)
3\(x\) + 38 = - 11 + 2\(x\)
3\(x\) - 2\(x\) = - 11 - 38
\(x\) = - 49
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
Lời giải:
a)
\(|x|=2019\Rightarrow \left[\begin{matrix} x=2019\\ x=-2019\end{matrix}\right.\)
b)
\(|x-3|=21\Rightarrow \left[\begin{matrix} x-3=21\\ x-3=-21\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=24\\ x=-18\end{matrix}\right.\)
c)
\(|x|-25+x=9\)
\(\Leftrightarrow |x|+x=34\)
Nếu $x\geq 0$: $|x|=x$
\(\Rightarrow x+x=34\Rightarrow 2x=34\Rightarrow x=17\) (thỏa mãn)
Nếu $x< 0$: $|x|=-x$
$\Rightarrow -x+x=34\Rightarrow 0=34$ (vô lý -loại)
Vậy $x=17$
d) \(|x-2|+2x=4\)
Nếu $x\geq 2\Rightarrow |x-2|=x-2$
$\Rightarrow x-2+2x=4\Rightarrow 3x=6\Rightarrow x=2$ (thỏa mãn)
Nếu $x< 2\Rightarrow |x-2|=2-x$
$\Rightarrow 2-x+2x=4\Rightarrow x=2$ (loại vì $x< 2$)
Vậy $x=2$
e)
$|x-5|+3x+2=13$
Nếu $x\geq 5$ thì $|x-5|=x-5$
\(\Rightarrow x-5+3x+2=13\)
\(\Leftrightarrow 4x=16\Leftrightarrow x=4\) (loại vì $x\geq 5$)
Nếu $x< 5$ thì $|x-5|=5-x$
$\Rightarrow 5-x+3x+2=13$
$\Leftrightarrow 2x=6\Rightarrow x=3$ (thỏa mãn)
Vậy $x=3$
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
`Answer:`
a. \(x+12=3\Leftrightarrow x=3-12\Leftrightarrow x=-9\)
b. \(2x-15=21\Leftrightarrow2x=21+15\Leftrightarrow2x=36\Leftrightarrow x=36:2\Leftrightarrow x=18\)
c. \(13-3x=4\Leftrightarrow-3x=4-13\Leftrightarrow-3x=-9\Leftrightarrow x=-9:-3\Leftrightarrow x=3\)
d. \(2\left(x-2\right)+4=12\Leftrightarrow2x-4+4=12\Leftrightarrow2x=12\Leftrightarrow x=12:2\Leftrightarrow x=6\)
e. \(15-3\left(x-2\right)=21\Leftrightarrow15-3x+6=21\Leftrightarrow-3x=21-15-6\Leftrightarrow-3x=0\Leftrightarrow x=0\)
g. \(25+4\left(3-x\right)=1\Leftrightarrow25+12-4x=1\Leftrightarrow37-4x=1\Leftrightarrow-4x=-36\Leftrightarrow x=9\)
h. \(3x+12=2x-4\Leftrightarrow3x-2x=-4-12\Leftrightarrow x=-16\)
i. \(14-3x=\left(-x\right)+4\Leftrightarrow-3x+x=4-14\Leftrightarrow-2x=10\Leftrightarrow x=5\)
k. \(2\left(x-2\right)+7=x-25\Leftrightarrow2x-4+7=x-25\Leftrightarrow2x-x=-25-3\Leftrightarrow x=-28\)
