tìm x b) x(5-2x)+2x(x-1)=13 c) (x-2)(x+5)-(x-3)(x+3)=5
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
8 tháng 9 2021
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
8 tháng 9 2021
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=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 \(x^2\ge0\Rightarrow x^2+4\ge4>0\))
\(2x=3\)
\(x=\frac{3}{2}\)
\(x\left(x-1\right)+5x-5=0\)
\(x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(x+5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)
\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)
\(4x^2-12x+9-4x^2+4x=5\)
\(-8x=5-9\)
\(-8x=-4\)
\(x=\frac{4}{8}\)
\(x=\frac{1}{2}\)
\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)
\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)
\(\left(2x-5\right)\left(x+11\right)=0\)
\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)
NT
1
7 tháng 1 2016
a) (x + 1) + (x - 3) = x - 2
=> x + 1 + x - 3 - x + 2 = 0
=> x = 0
b) - (1 - x) - (2x - 3) = 4 - 2x
=> -1 + x - 2x + 3 - 4 + 2x = 0
=> x - 2 = 0
=> x = 2
c)-13 - (5 - x + 6) + 2x = 3
=> -13 - 5 + x - 6 + 2x - 3 = 0
=> 3x - 27 = 0
=> 3x = 27
=> x = 9
BR
3
12 tháng 6 2019
a) \(2x\left(x-3\right)-x\left(2x+1\right)-3\left(x+5\right)=11\)
\(\Rightarrow2x^2-6x-2x^2-x-3x-15=11\)
\(\Rightarrow-10x=26\Rightarrow x=-2,6\)
Vậy ...........
b) \(x\left(x-1\right)-\left(x^2+3x-5\right)-2\left(x+3\right)=17\)
\(\Rightarrow x^2-x-x^2-3x+5-2x-6=17\)
\(\Rightarrow-6x=18\Rightarrow x=-3\)
c) \(5x\left(x-7\right)-\left(5x+1\right)x-\left(x+3\right)2=13\)
\(\Rightarrow5x^2-35x-5x^2-x-2x-6=13\)
\(\Rightarrow-38x=19\Rightarrow x=-\frac{1}{2}\)
d) \(\left(2x^2-3x+5\right)-2x\left(x-3\right)+\left(x-1\right)\left(-2\right)=10\)
\(\Rightarrow2x^2-3x+5-2x^2+6x-2x+2=10\)
\(\Rightarrow x=3\)
CN
1
12 tháng 2 2022
a: \(\Leftrightarrow\left|x-1\right|=3-2x\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3-x+1\right)\left(2x+3+x-1\right)=0\\x< =\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)\left(3x+2\right)=0\\x< =\dfrac{3}{2}\end{matrix}\right.\)
=>x=-2/3
b: Trường hợp 1: x<-3
Pt sẽ là:
\(-x-1-x-3=10-4x\)
=>-2x-4=10-4x
=>2x=14
hay x=7(loại)
Trường hợp 2: -3<=x<-1
Pt sẽ là \(x+3-x-1=10-4x\)
=>10-4x=2
=>4x=8
hay x=2(loại)
Trường hợp 3: x>=-1
Pt sẽ là x+1+x+3=10-4x
=>2x+4=10-4x
=>6x=6
hay x=1(nhận)
b) \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Rightarrow5x-2x^2+2x^2-2x=13\)
\(\Rightarrow3x=13\)
\(\Rightarrow x=13:3=\frac{13}{3}\)
c) \(\left(x-2\right)\left(x+5\right)-\left(x-3\right)\left(x+3\right)=5\)
\(\Rightarrow x^2+5x-2x-10-\left(x^2+3x-3x-9\right)=5\)
\(\Rightarrow x^2+5x-2x-10-x^2-3x+3x+9=5\)
\(\Rightarrow3x-1=5\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=6:3=2\)
b) \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
\(\Leftrightarrow13\div3=\frac{1}{3}\)