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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!!
d) |2x – 5| –7 = 22
| 2x -5 | = 22+7
| 2x -5 | = 29
TH1: 2x-5 = -29
2x = -29+5
2x= -24
x= -24:2
x= -12
TH2: 2x -5 =29
2x = 29+5
2x= 34
x= 34:2
x= 17
Vậy...
a) 2x = 16 e) 12x = 144
2x = 24 12x = 122
=> x = 4 => x = 2
b) 2x+1 = 16 các câu còn lại tương tự nhé nhiều quá
2x+1 = 24
x + 1 = 4
=> x = 3
c) 5x+1 = 125
5x+1 = 53
x + 1 = 3
=> x = 2
d) 52x - 1 = 125
52x-1 = 53
2x - 1 = 3
2x = 4
=> x = 2
a)Ta có : 2x = 16
2x = 24
=> x = 4
b) Ta có: 2x+1 = 16
2x+1 = 24
=> x+1 = 4
=> x = 4-1
=> x = 3
Mấy câu sau tương tự vậy đó để hôm khác mình làm tiếp cho bây giờ mình đi ngủ đã buồn ngủ quá hihi ! ^-^
Học tốt nha bạn !
bài 1 tính giá trị biểu thức
( - 25 ) nhân ( -3 ) nhân x với x = 4
\(\left(-25\right).\left(-3\right).4\)
\(=\left(-25\right).4.\left(-3\right)\)
\(=-100.\left(-3\right)=300\)
( -1 ) nhân ( -4 ) nhân 5 nhân 8 nhân y với y =25
\(\left(-1\right).\left(-4\right).5.8.25\)
\(=4.5.8.25=4.25.5.8\)
\(=100.40=40000\)
( 2ab mũ 2 ) : c với a =4 ; b= -6 ; c =12
\(\left(2.4.\left(-6\right)\right)^2:12\)
\(=\left(-48\right)^2:12\)
\(=2304:12=192\)
[ ( -25 ) nhân ( - 27 ) nhân ( -x ) ] : y với x = 4 ; y = -9
\(\left[\left(-25\right).\left(-27\right).\left(-4\right)\right]:-9\)
\(=-2700:\left(-9\right)\)
\(=300\)
(a mũ 2 _ b mũ 2) : ( a + b ) nhân ( a _ b ) với a + 5 , b = -3
\(\left(5^2-\left(-3\right)^2\right):\left(5-3\right).\left(5+3\right)\)
\(=16:2.8\)
\(=8.8=64\)
Tìm x biết:
a,x-5/7=1/9
b,2x/5=6/2x+1
c,11/8+13/6=85/x
d,2x-2/11=1.1/5
e,x/15=3/5+-2/3
f,x/182=-6/14.35/91
a, \(x\) - \(\dfrac{5}{7}\) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) + \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{52}{63}\)
b, \(\dfrac{2x}{5}\) = \(\dfrac{6}{2x+1}\)
2\(x\).(2\(x\) + 1) = 30
4\(x^2\)+ 2\(x\) - 30 = 0
4\(x^2\) + 12\(x\) - 10\(x\) - 30 = 0
(4\(x^2\) + 12\(x\)) - (10\(x\) + 30) =0
4\(x\).(\(x\) + 3) - 10.(\(x\) +3) = 0
2 (\(x\) + 3).(2\(x\) - 5) = 0
\(\left[{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-3; \(\dfrac{5}{2}\)}
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: ...........