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-29-9(2x-1)\(^2\)= -110
(=) 9(2x-1)2 = (-29) +110
(=) 9(2x-1)2 = 81
(=) (2x-1)2 =81: 9
(=) (2x-1)2 =9
(=) (2x-1)2 = 32 =(-3)2
\(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)
\(\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\)
\(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
vậy : ........
a,\(-29-9\left(2x-1\right)^2=-110\)
\(=>-29+110=9.\left(2x-1\right)^2\)
\(=>81=9.\left(2x-1\right)^2\)
\(=>\left(2x-1\right)^2=9\)
\(=>\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}=>\orbr{\begin{cases}x=\frac{4}{2}=2\\x=\frac{-2}{2}=-1\end{cases}}}\)
a: \(2\left(x-51\right)=2\cdot2^3+20\)
=>\(2\left(x-51\right)=2^4+20=36\)
=>x-51=36/2=18
=>x=18+51=69
b: \(2x-49=5\cdot3^2\)
=>\(2x-49=5\cdot9=45\)
=>2x=45+49=94
=>x=94/2=47
c: \(\left[\left(8x-12\right):4\right]\cdot3^3=3^6\)
=>\(\left[4\cdot\dfrac{\left(2x-3\right)}{4}\right]=3^3\)
=>\(2x-3=3^3=27\)
=>2x=3+27=30
=>x=30/2=15
d: \(2^{x+1}-2^2=32\)
=>\(2^{x+1}=32+2^2=32+4=36\)
=>\(x+1=log_236\)
=>\(x=log_236-1\)
e: \(\left(x^3-77\right):4=5\)
=>\(x^3-77=20\)
=>\(x^3=77+20=97\)
=>\(x=\sqrt[3]{97}\)
a: \(\Leftrightarrow2x+\dfrac{7}{2}=\dfrac{16}{3}:\dfrac{11}{3}=\dfrac{16}{11}\)
=>2x=-45/22
hay x=-45/44
b: =>x/7=-1/28:1/4=-1/7
=>x=-1
a)(7/2+2x).11/3=16/3
7/2+2x=16/3:11/3
7/2+2x=16/3.3/11
7/2+2x=16/11
2x=16/11-7/2
2x= -45/22
x= -45/22:2
x= -45/44
Vậy x= -45/44
b)x/7 +1/4= -1/28
x/7= -1/28-1/4
x/7= -2/7
=>x= -2
123-5.(x+4)=38
5(x+4)=123-38
5(x+4)=85
x+4=85:5
x+4=17
x=17-4
x=13
10+2x=2.(32-1)
10+2x=2.(9-1)
10+2x=2.8
2x=16-10
2x=6
x=6:2
x=3
2.x-138=23.32
2.x-138=72
2.x=72+138
2.x=210
x=210:2
x=105
[102+(28-2x)]:20-5=1
(102+28-2x):20=6
130-2x=120
2x=130-120
2x=10
x=10:2
x=5
231-(x-6)=1339:13
231-(x-6)=103
x-6=231-103
x-6=128
x=128+6
x=134
123-5.(x+4)=38
5.(x+4)=123-38
5.(x+4)=85
(x+4)=85/5
x+4=17
x=17-4
x=13
a) 70 - 5(x - 3 ) = 45
5( x - 3 ) = 70 - 45 = 25
x - 3 = 25 : 5 = 5
x = 5 + 3 = 8
b) (2x - 1 )4 = 3 . 62 - 27
(2x - 1 )4 = 3 . 36 - 27
(2x - 1 )4 = 81
Ta thấy 81 = 34 vậy suy ra (2x - 1)4 = 34
Để vế trong ngoặc tròn (2x - 1 ) = 3 thì x cần bằng 2
Thử lại : 2 . 2 - 1 = 4 - 1 = 3
Vậy x = 2
c) 3x3 + 43 = 102 - 33
3x3 + 43 = 100 - 33 = 67
3x3 = 67 + 43 = 110 ( Đoạn này đề bài sai hay tao sai z :)?)
1,
a, \(\left(\dfrac{-4}{3}+\dfrac{1}{3}\right).\dfrac{5}{12}\)=-\(\dfrac{5}{12}\)
b, \(\dfrac{16}{5}+\left(\dfrac{-45}{14}\right):\dfrac{3}{28}\)
=\(\dfrac{-2}{15}\)
2,
a, 2x+19=25
=>x=3
b, \(-\dfrac{2}{9}x=\dfrac{1}{3}\)
=>x=\(\dfrac{-3}{2}\)
Bài 1:
a) Ta có: \(\dfrac{-4}{3}\cdot\dfrac{5}{12}+\dfrac{1}{3}\cdot\dfrac{5}{12}\)
\(=\dfrac{5}{12}\cdot\left(\dfrac{-4}{3}+\dfrac{1}{3}\right)\)
\(=\dfrac{-5}{12}\)
b) Ta có: \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right):\dfrac{3}{28}\)
\(=\dfrac{16}{5}+\left(\dfrac{4}{14}-\dfrac{49}{14}\right):\dfrac{3}{28}\)
\(=\dfrac{16}{5}+\dfrac{-45}{14}\cdot\dfrac{28}{3}\)
\(=\dfrac{16}{5}-30=\dfrac{-134}{5}\)
Bài 2:
a: Ta có: \(2^{x+1}\cdot3^y=12^x\)
\(\Leftrightarrow2^{x+1}\cdot3^y=2^{2x}\cdot3^x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
a: \(\left(x,y\right)\in\left\{\left(1;-9\right);\left(-9;1\right);\left(-1;9\right);\left(9;-1\right);\left(3;-3\right);\left(-3;3\right)\right\}\)
a: \(\Leftrightarrow2x+\dfrac{7}{2}=\dfrac{16}{3}:\dfrac{8}{3}=2\)
=>2x=-3/2
hay x=-3/4
b: 2x+3=5
=>2x=2
hay x=1
c: =>3(x-2)=4(5+x)
=>4x+20=3x-6
=>x=-26
