tìm x,biết,/2x+1/+10-2x=0
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31 tháng 1 2017
\(\frac{1-2x}{10}+\frac{3-2x}{8}+\frac{23-2x}{6}=0\)
\(\Leftrightarrow\frac{1}{10}-\frac{2x}{10}+\frac{3}{8}-\frac{2x}{8}+\frac{23}{6}-\frac{2x}{6}=0\)
\(\Leftrightarrow\frac{1}{10}-\frac{x}{5}+\frac{3}{8}-\frac{x}{4}+\frac{23}{6}-\frac{x}{3}=0\)
\(\Leftrightarrow\left(\frac{1}{10}+\frac{3}{8}+\frac{23}{6}\right)-\left(\frac{x}{3}+\frac{x}{4}+\frac{x}{5}\right)=0\)
\(\Leftrightarrow\frac{517}{120}-\left(\frac{x}{3}+\frac{x}{4}+\frac{x}{5}\right)=0\)
\(\Leftrightarrow x\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)=\frac{517}{120}\)
\(\Leftrightarrow x.\frac{47}{60}=\frac{517}{120}\)
\(\Rightarrow x=\frac{517}{120}:\frac{47}{60}=\frac{11}{2}\)
Vậy \(x=\frac{11}{2}\)
31 tháng 1 2017
(1-2x)/10+(3-2x)/8+(23-2x)/6=0
[48(1-2x)+60(3-2x)+80(23-2x)]/480=0
48-96x+180-120x+1840-160x=0
2068-376x=0
-376x=-2068
x=11/2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 12 2023
a, 7\(x\).(2\(x\) + 10) =0
\(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\) {-5; 0}
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 12 2023
b, -9\(x\) : (2\(x\) - 10) = 0
9\(x\) = 0
\(x\) = 0
c, (4 - \(x\)).(\(x\) + 3) = 0
\(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(x\in\) {-3; 4}
29 tháng 8 2015
Tìm x biết :a) ( 2x - 3 ).( x +1 ) > 0b) ( x + 5 ).(x-7) < 0c) | 2x - 3 | + 8 = 10d) ( 2x + 5 ) . | x -8 | . ( x2 + 1 ) = 0
M
5
16 tháng 5 2019
a, (2x - 1) - (x + 6) = 0
=> 2x - 1 - x - 6 = 0
=> 2x - x = 0 + 6 + 1
=> x = 7
Vậy x = 7
b, 2x - 1 - (5 - x) = -10
=> 2x - 1 - 5 + x = - 10
=> 2x + x = -10 + 5 + 1
=> 3x = -4
=> x = -4/3
LH
1
PA
30 tháng 11 2016
\(2x^2-7x+5=0\)
\(2x^2-2x-5x+5=0\)
\(2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x-5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)
\(x\left(2x-5\right)-4x+10=0\)
\(x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(x-2\right)=0\)
\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)
\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)
\(x^2-25-x^2+2x=15\)
\(2x=15+25\)
\(2x=40\)
\(x=\frac{40}{2}\)
\(x=20\)
\(x^2\left(2x-3\right)-12+8x=0\)
\(x^2\left(2x-3\right)+4\left(2x-3\right)=0\)
\(\left(2x-3\right)\left(x^2+4\right)=0\)
\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))
\(2x=3\)
\(x=\frac{3}{2}\)
\(x\left(x-1\right)+5x-5=0\)
\(x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(x+5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)
\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)
\(4x^2-12x+9-4x^2+4x=5\)
\(-8x=5-9\)
\(-8x=-4\)
\(x=\frac{4}{8}\)
\(x=\frac{1}{2}\)
\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)
\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)
\(\left(2x-5\right)\left(x+11\right)=0\)
\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)
TN
0
Ta có : |2x + 1| + 10 - 2x = 0
=> |2x + 1| = 2x - 10 (1)
ĐK: \(2x-10\ge0\Leftrightarrow x\ge5\)
Khi đó (1) <=> \(\orbr{\begin{cases}2x+1=2x-10\\2x+1=-2x+10\end{cases}\Rightarrow\orbr{\begin{cases}1=-10\left(\text{loại}\right)\\4x=11\end{cases}\Rightarrow}x=3,75\left(\text{loại}\right)}\)
\(\Rightarrow x\in\varnothing\)