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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) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
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!!
a,\(\left(x-15\right):50+22=24\)
\(< =>\frac{\left(x-15\right)}{50}=2< =>x-15=100\)
\(< =>x=100+15=115\)
b,\(42-\left(2x+32\right)+12:2=6\)
\(< =>42-2x-32=0\)
\(< =>10-2x=0< =>x=\frac{10}{2}=5\)
Làm nốt :
c) \(134-2\left\{156-6\cdot\left[54-2\cdot\left(9+6\right)\right]\right\}\cdot x=86\)
=> 134 - 2{156 - 6 . [54 - 2 . 15]} . x = 86
=> 134 - 2{156 - 6 . [54 - 30]} . x = 86
=> 134 - 2{156 - 6. 24} . x = 86
=> 134 - 2{156 - 144} . x = 86
=> 134 - 2.12 . x = 86
=> 134 - 24 . x = 86
=> 24.x = 48
=> x = 2
Bài 2 : a) 120 : [21 - (4x - 4)] = 23.3
=> 120 : [21 - (4x - 4)] = 8.3
=> 120 : [21 - (4x - 4)] = 24
=> 21 - (4x - 4) = 5
=> 4x - 4 = 16
=> 4x = 20
=> x = 5
b) 3.[205 - (x - 9)] - 486 = 0
=> 3.[205 - (x - 9)] = 486
=> 205 - (x - 9) = 162
=> x - 9 = 205 - 162 = 43
=> x = 43 + 9 = 52
c) 204 - 2{200 - 5.[64 - 2.(11 + 6)]} . x = 4
=> 204 - 2{200 - 5.[64 - 2.17]} . x = 4
=> 204 - 2{200 - 5 .[64 - 34]}.x = 4
=> 204 - 2{200 - 5.30} . x = 4
=> 204 - 2{200 - 150}.x = 4
=> 204 - 2.50 . x = 4
=> 2.50.x = 200
=> 100.x = 200
=> x = 2
\(\left(x-3\right)\left(x-12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=12\end{cases}}\)
\(\Rightarrow x\in\left\{3;12\right\}\)
\(\left(x^2-81\right)\left(x^2+9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-81=0\\x^2+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x\in\varnothing\end{cases}}\Leftrightarrow x=9\)
\(\Rightarrow x=9\)
\(\left(x-4\right)\left(x+2\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x-4\\x+2\end{cases}}\)trái dấu
\(TH1:\hept{\begin{cases}x-4>0\\x+2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x< -2\end{cases}}\Leftrightarrow x\in\varnothing\)
\(TH2:\hept{\begin{cases}x-4< 0\\x+2>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 4\\x>-2\end{cases}}\Leftrightarrow x\in\left\{-1;0;1;2;3\right\}\)
Vậy \(x\in\left\{-1;0;1;2;3\right\}\)
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
a, | x+3 | = -27/11 × 22/-9
=> |x + 3| = 6
=> x + 3 = 6 hoặc x + 3 = -6
=> x = 3 hoặc x = -9
vậy_
b, (x-3) × (2x - 7) = 0
=> x - 3 = 0 hoặc 2x - 7 = 0
=> x = 3 hoặc x = 7/2
vậy_
c, (x-1) + (x-2) + (x-3) + ... + (x-100) = 4950
=> x - 1 + x - 2 + x - 3 + ... + x - 100 = 4950
=> 100x - (1 + 2 + 3 + ... + 100) = 4950
=> 100x - (1 + 100).100 : 2 = 4950
=> 100x - 5050 = 4950
=> 100x = 10000
=> x = 10