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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!
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 = \(\frac{47}{10}\)
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
a 24x-15x=75+33
9x=108
x=108:9=12
b 3-2/5x=2/3
2/5x=3-2/3=1/3
x=1/3:2/5=5/6
Tk & kết bạn với mik nha
a) => x={-5;5}
b) => /x/=3-(-4)
=> /x/=7
=> x={7;-7}
c) => /2-x/=4-3
=> /2-x/=1
=> 2-x={1;-1}
=> x= {1;3}
d) => /x+1/=12-13
=> /x+1/= -1
Vì giá trị tuyệt đối của mọi số nguyên thuộc Z bao giờ cũng là 1 số tự nhiên
Nhưng vì /x+1/=-1
=> x ko tồn tại
e) Vì (x-1).(x+2)=0
=> 1 trong 2 thừa số phải bằng 0
Nếu x-1=0 thì x=1
Nếu x+2=0 thì x=-2
A. x = 2
B. \(\dfrac{3}{8}=\dfrac{6}{x}\)\(\Leftrightarrow x=\dfrac{6.8}{3}=16\)
C. x = 3
D. \(x=\dfrac{4.6}{8}=3\)
E. \(x=\dfrac{7}{3}\)
G.\(\dfrac{14}{13}=\dfrac{28}{10-x}\)
<=>\(14\left(10-x\right)=364\)
<=> 10 - x = 26
<=> x = -16
H. \(3\left(x+2\right)=4\left(x-5\right)\)
<=> 3x + 6 = 4x - 20
<=> -x = -26
<=> x = 26
K. \(\dfrac{x}{2}=\dfrac{8}{x}\)
<=> \(x^2=16\)
<=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
M. \(\left(x-2\right)^2=100\)
<=> \(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)
a=2
b=16
c=3
d=3
mik chỉ biết thế này thôi(ko chắc đúng=3)
Lê Phúc Đạt:
(x + 1) + (x + 2) +...+ (x + 100) = 5750
Đặt: (x + 1) + (x + 2) +...+ (x + 100) = S
Từ 1 đến 100 có 100 số hạng:
=> S = 100x + (1 + 2 +...+100)
S = 100x + (100 + 1). 100 : 2
S = 100x + 5050
=> 100x + 5050 = 5750
100x = 5750 - 5050
100x = 700
x = 700 : 100
x = 7
Vậy x = 7
Tham khảo nha!
ta có: (x+1)+(x+2)+(x+3)+....+(x+100)=5750
=(x + x+ x+....+ x)+(1+2+3+....+100)=5750
suy ra có 100x và 100 so hang
=100x + 5050=5750
=100x=5750-5050
=100x=700
=x=700 : 100
=x=7
a) x = 4
b) sory mik làm ko được
c) ai đề
ai giai ho mik ro ra ho