Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ý bạn bảo (x-2 và 1 phần 2) là hợp số hả :
ta có (x-2 và 1 phần 2) nhân (2x+3 và 1 phần 5) =0
\(\Leftrightarrow\) [2.(x-2) +1]. [5.(2x+3)+1]=0
\(\Leftrightarrow\)(2x-3)(10x+16)=0
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x=3\\10x=-16\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{-8}{5}\end{cases}}\)
vậy
để tích kia <0 thì 2 cái kia 1 âm 1 dương
=>5>x>3
=>x=4
mik trc
mik chỉ ghi kết quả thôi nha
vì làm ra dài dòng lw
=> ta có:
x = 7
x = 4
x = 3
x = 4
x = 3
nha bn
5(x-7)=0
=>x-7=0
=>x=0+7
=>x=7
25(x-4)=0
=>x-4=0
=>x=0+4
=>x=4
34(2x-6)=0
=>2x-6=0
=>2x=0+6
=>2x=6
=>x=6:2
=>x=3
2016(3x-12)=0
=>3x-12=0
=>3x=0+12
=>3x=12
=>x=12:3
=>x=4
47(5x-15)=0
=>5x-15=0
=>5x=0+15
=>5x=15
=>x=15:5
=>x=3
1) \(2x\cdot\left(x-3\right)-5=3x\left(2x-5\right)-4x^2+40\)
\(\Leftrightarrow2x^2-6x-5=6x^2-15x-4x^2+40\)
\(\Leftrightarrow2x^2-6x-5=2x^2-15x+40\)
\(\Leftrightarrow2x^2-6x-5-2x^2+15x-40=0\)
\(\Leftrightarrow9x-45=0\)
<=> x=5
2) x(2x-1)-5(-7)2=2x2-2x+5
<=> 2x2-x-5.49=2x2-2x+5
<=> 2x2-x-245-2x2+2x-5=0
<=> x-250=0
<=> x=250
3) |a-2|=10
\(\Leftrightarrow\orbr{\begin{cases}x-2=10\\x-2=-10\end{cases}\Leftrightarrow\orbr{\begin{cases}x=12\\x=-8\end{cases}}}\)
4) |x|=-5
=> Không tồn tại giá trị của x thỏa mãn vì |x| >=0 với mọi x thuộc Z
\(2x-3-3x+3.5=4.3-4.x-18\)
<=> \(2x-3-3x+15=12-4x-18\)
<=> \(2x-3x+4x=12-18-15+3\)
<=>\(3x=-18\)
<=> x=-18/3=-6
\(2x-3-3\left(x-5\right)=4\left(3-x\right)-18\)
\(\Leftrightarrow2x-3-3x+15=12-4x-18\)
\(\Leftrightarrow\left(2x-3x\right)+\left(15-3\right)=\left(12-18\right)-4x\)
\(\Leftrightarrow-x+12=-6-4x\)
\(\Leftrightarrow12=-6-3x\)
\(\Leftrightarrow3x=-6-12\)
\(\Leftrightarrow3x=-18\)
\(\Leftrightarrow x=\frac{-18}{3}\)
\(\Leftrightarrow x=-6\)
\(\text{Vậy }x=-6\)
a, 15/x - 1/3 = 28/57
15/x = 28/57 + 1/3
15/x = 28/57 + 19/57
15/x = 47/57
x . 47 = 15 . 57
x = 855/47
b, x/2 - 2/5 = 1/10
x/2 = 1/10 + 2/5
x/2 = 1/10 + 4/10
x/2 = 5/10 = 1/2
x/2 = 1/2
=> x=1
\(a,\left(3x-7\right)\left(x+5\right)=\left(5+x\right)\left(3-2x\right)\)
\(\Leftrightarrow\left(3x-7\right)\left(x+5\right)-\left(x+5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-7-3+2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\5x-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
\(b,\dfrac{-x+3}{2}=\dfrac{x-2}{3}\left(MSC=6\right)\)
Suy ra :
\(3\left(-x+3\right)=2\left(x-2\right)\)
\(\Leftrightarrow-3x+9-2x+4=0\)
\(\Leftrightarrow-5x+13=0\)
\(\Leftrightarrow x=\dfrac{13}{5}\)
\(c,\dfrac{x-1}{x-2}+\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)\(\left(dkxd:x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+2\right)+5\left(x-2\right)-12-x^2+4}{x^2-4}=0\)
\(\Leftrightarrow x^2+2x-x-2+5x-10-12-x^2+4=0\)
\(\Leftrightarrow6x-20=0\)
\(\Leftrightarrow x=\dfrac{10}{3}\)\(\left(n\right)\)
Vậy \(S=\left\{\dfrac{10}{3}\right\}\)
b. \(2x=\dfrac{-1}{5}\)
\(x=-0,25\)
d. \(\dfrac{2}{x+1}=\dfrac{x+1}{16}\)
\(\Leftrightarrow\left(x+1\right)^2=32\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\sqrt{2}\\x+1=-4\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\sqrt{2}-1\\x=-4\sqrt{2}-1\end{matrix}\right.\)
a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
2x-5(-3-x) = 2x + 15 + 5x = 7x + 15
2x-5-3-x
2x-x=5+3
x=8