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Ko cần đâu bn à mk mong bn đấy
a)\(\left(3x-1\right)\left(5-\frac{1}{2}x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
b)\(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{29}{12}\\x=-\frac{13}{12}\end{cases}}\)
a)\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\)3x - 1 = 0 hay \(\frac{-1}{2}\)x + 5 = 0
\(\Leftrightarrow\)3x = 1 I\(\Leftrightarrow\)\(\frac{-1}{2}\)x = -5
\(\Leftrightarrow\) x = \(\frac{1}{3}\) I\(\Leftrightarrow\) x = 10
b) 2 I \(\frac{1}{2}x-\frac{1}{3}\)I - \(\frac{3}{2}\)=\(\frac{1}{4}\)
\(\Leftrightarrow\) 2 I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{4}\)
\(\Leftrightarrow\) I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{7}{8}\) hay \(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{-7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{29}{24}\) I\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{-13}{24}\)
\(\Leftrightarrow\) x = \(\frac{29}{12}\) I\(\Leftrightarrow\) x = \(\frac{-13}{12}\)
c) (2x +\(\frac{3}{5}\))2 - \(\frac{9}{25}\)= 0
\(\Leftrightarrow\)(2x +\(\frac{3}{5}\))2 = \(\frac{9}{25}\)
\(\Leftrightarrow\) 2x +\(\frac{3}{5}\) = \(\frac{3}{5}\) hay 2x +\(\frac{3}{5}\)= \(\frac{-3}{5}\)
\(\Leftrightarrow\) 2x = 0 I \(\Leftrightarrow\)2x = \(\frac{-6}{5}\)
\(\Leftrightarrow\) x = 0 I \(\Leftrightarrow\) x = \(\frac{-3}{5}\)
d) 3(x -\(\frac{1}{2}\)) - 5(x +\(\frac{3}{5}\)) = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)3x - \(\frac{3}{2}\)- 5x - 3 = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)-2x + x - \(\frac{9}{2}\)- \(\frac{1}{5}\)= 0
\(\Leftrightarrow\)-x = \(\frac{-47}{10}\)
\(\Leftrightarrow\) x = \(\frac{47}{10}\)
1.
a) \(2^x=128\)
\(2^x=2^7\)
\(=>x=7\)
b) \(8^{x-1}=64\)
\(8^{x-1}=8^2\)
\(=>x-1=2\)
\(x=2+1\)
\(=>x=3\)
c) \(3+3^x=30\)
\(3^x=30-3\)
\(3^x=27=3^3\)
\(=>x=3\)
d) \(\left(x+2\right)=64\) -> đề có thiếu không vậy?
e) \(3^2.x=3^5\)
\(x=3^5:3^2\)
\(=>x=3^3=27\)
f) \(\left(2x-1\right)^3=343\)
\(\left(2x-1\right)^3=7^3\)
\(=>2x-1=7\)
\(2x=7+1\)
\(2x=8\)
\(x=8:2\)
\(=>x=4\)
\(#Wendy.Dang\)
a,\(2^x\)=128 b,\(8^{x-1}\)=64 c,3+\(3^x\)=30 d,x+2=64
\(2^7\)=128 \(8^{x-1}\)=\(8^2\) \(3^x\)=30-3 x=64-2
=>x=7 =>x-1=2 \(3^x\)=27 x=62
x=2+1=3 \(3^x\)=\(3^3\)
=>x=3
e,\(3^2\).x=\(3^5\) f,(2x-\(1^3\))=343
x=\(3^5\):\(3^2\) 2x=1+343
x=27 2x=344
x=344:2
x=172
1) -12.(x-5) + 7.(3-x)=5
-12x+ 60+21-7x =5
-12x-7x = 5-60-21
-19x=-76
x=-76:(-19)
x=4
2) (x-2).(x+4) =0
\(\Rightarrow\)x-2=0 hoặc x+4=0
x-2=0 x+4=0
x=0+2 x=0-4
x=2 x=-4
Vậy x=2 hoặc x=-4
3) (x-2).(x+15) =0
\(\Rightarrow\)x-2=0 hoặc x+15=0
x-2=0 x+15=0
x=0+2 x=0-15
x=2 x=-15
1)\(-12.\left(x-5\right)+7.\cdot\left(3-x\right)=5\)
\(-12x+60+21-7x=5\)
\(-19x+81=5\)
\(-19x=5-81\)
-\(-19x=-76\)
\(x=-76:-19\)
\(x=4\)
2) Ta có 2 trường hợp
TH1: x-2=0 =>x=2
TH2: x+4=0 => x=-4
Vậy \(x\in\left(-4;2\right)\)
3) Ta có
TH1: x-2=0=>x=2
TH2: x+15=0=>x=-15
Vậy \(x\in\left(-15;2\right)\)
mk giúp bạn câu cuối nhé:
3|x+2|-5=16
3|x+2|=16+5
3|X+2|=21
|x+2|=21:3
|x+2|=7
=>x+2=7 hoặc x+2=-7
+) với x+2=7 +) với x+2= -7
x=5. x=-9
vậy x€{5,-9}
nếu có TGian mk sẽ giải cho bạn mấy câu trên
cam ơn bạn nhé bạn có giup mình not câu trên trong vong ngay ko
\(1)x+\frac{5}{6}\times2\frac{2}{5}-1\frac{1}{4}=35\%\)
\(x+\frac{5}{6}\times\frac{12}{5}-\frac{5}{4}=\frac{7}{12}\)
\(x+\frac{5}{6}\times\frac{12}{5}=\frac{7}{12}+\frac{5}{4}\)
\(x+\frac{5}{6}.\frac{12}{5}=\frac{8}{5}\)
\(x+\frac{5}{6}=\frac{8}{5}:\frac{12}{5}\)
\(x+\frac{5}{6}=\frac{2}{3}\)
\(x=\frac{2}{3}-\frac{5}{6}\)
\(x=-\frac{1}{6}\)
HỌC TỐT !
\(2\)) \(\left|x-\frac{1}{2}\right|-\frac{3}{4}=0\)
\(\left|x-\frac{1}{2}\right|\) \(=0+\frac{3}{4}\)
\(\left|x-\frac{1}{2}\right|\) \(=\frac{3}{4}\)
\(x-\frac{1}{2}\) \(=\frac{3}{4}\)hoặc \(-\frac{3}{4}\)
Ta xét 2 trường hợp :
Trường hợp 1 : \(x-\frac{1}{2}=\frac{3}{4}\)
\(x\) \(=\frac{3}{4}+\frac{1}{2}\)
\(x\) \(=\frac{5}{4}\)
Trường hợp 2 : \(x-\frac{1}{2}=-\frac{3}{4}\)
\(x\) \(=-\frac{3}{4}+\frac{1}{2}\)
\(x\) \(=-\frac{1}{4}\)
Vậy \(x\in\text{{}\frac{5}{4};-\frac{1}{4}\)}
2 ( x - 3 ) - 3 ( 2 - 3x ) = 4 [ ( 1 - 2x ) + 15 ]
2x - 6 - 6 + 9x = 4 [ 1 - 2 x + 15 ]
2x - 6 -6 + 9x = 4 - 8x + 60
2x + 9x + 8x = 4 + 60 + 6 + 6
19x = 76
=> x = 76 : 19
=> x = 4
Vậy x = 4
\(2\left(x-3\right)-3\left(2-3x\right)=4\left[\left(1-2x\right)+15\right]\)
\(\Rightarrow2x-6-6+9x=4\left[1-2x+15\right]\)
\(\Rightarrow2x-6-6+9x=4-8x+60\)
\(\Rightarrow2x+9x+8x=4+60+6+6\)
\(\Rightarrow19x=76\)
\(\Rightarrow x=76:19=4\)
Vậy x = 4
1: =>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
2: =>7/6x=5/2:3,75=2/3
=>x=2/3:7/6=2/3*6/7=12/21=4/7
3: =>2x-3=0 hoặc 6-2x=0
=>x=3 hoặc x=3/2
4: =>-5x-1-1/2x+1/3=3/2x-5/6
=>-11/2x-3/2x=-5/6-1/3+1
=>-7x=-1/6
=>x=1/42
Bài 1 :Tính :
A = 2^4 . 417 + (-2)^4 . 583
B = 146.572 + (-146).(-428)
C = (- 158 ) . 1999 + 842 . (-1999)
D = 76 - 2x + 24 - 2y với x + y = - 50
Bài 2 . tìm x
a) /2x-6/ - / x - 12 / = 0
b)/ x + 5 / + ( y - 3 ) ^ 2 = 0
Bài 3 Tìm x
a) 4x - 11 = - 6x + 89
b) ( 3x - 5 ) - ( 2x -7 )=-16
c) / 2x - 4 / + 11 = 19
d) ( x - 3 ) ^ 2 - 25 = 0
nhiều bài quá mk làm ko nổi
xin lỗi bn nha!
Vũ Vân Anh 
shi nit chi
Bài 3:
a: 4x-11=-6x+89
=>10x=100
hay x=10
b: (3x-5)-(2x-7)=-16
=>3x-5-2x+7=-16
=>x+2=-16
hay x=-18
c: |2x-4|+11=19
=>|2x-4|=8
=>2x-4=8 hoặc 2x-4=-8
=>2x=12 hoặc 2x=-4
=>x=6 hoặc x=-2
d: (x-3)2-25=0
=>(x-3-5)(x-3+5)=0
=>(x-8)(x+2)=0
=>x=8 hoặc x=-2
a) \(\left(2x+3\right).\left(\frac{1}{2}.x-\frac{3}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+3=0\\\frac{1}{2}.x-\frac{3}{2}=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}2x=-3\\\frac{1}{2}.x=\frac{3}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=\frac{3}{2}:\frac{1}{2}\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=3\end{cases}}\)
Vậy x = \(-\frac{3}{2}\) hoặc x = 3
b)\(\left(\frac{1}{2}-x\right)^2=\frac{64}{49}\)
\(\Rightarrow\left(\frac{1}{2}-x\right)^2=\left(\frac{8}{7}\right)^2\) hoặc \(\left(\frac{1}{2}-x\right)^2=\left(-\frac{8}{7}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}-x=\frac{8}{7}\\\frac{1}{2}-x=-\frac{8}{7}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}-\frac{8}{7}\\x=\frac{1}{2}+\frac{8}{7}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{9}{14}\\x=\frac{23}{14}\end{cases}}\)
Vậy x = \(-\frac{9}{14}\) hoặc x = \(\frac{23}{14}\)
c) \(\frac{1}{2}.\left(x-4,5\right)=\frac{3}{4}.x=\frac{5}{12}\) ( câu này mik ko hiểu cho lắm)
k mik nha mn!
doi mik sua