3x-1=x+5
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8 tháng 9 2023
2:
a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8
=>x^2-x-12-x^2+4x+5=8
=>3x-7=8
=>3x=15
=>x=5
b: =>3x^2+3x-2x-2-3x^2-21x=13
=>-20x=15
=>x=-3/4
c: =>x^2-25-x^2-2x=9
=>-2x=25+9=34
=>x=-17
d: =>x^3-1-x^3+3x=1
=>3x-1=1
=>3x=2
=>x=2/3
26 tháng 3 2020
a) (3x + 1)^2 - 2(3x + 1)(3x - 5) + (3x - 5)^2
= 9x^2 + 6x + 1 - 18x^2 + 24x + 10 + 9x^2 - 30x + 25
= 36
b) (3x^2 - y)^2
= 9x^4 - 6x^2y + y^2
c) (3x + 5)^2 + (3x - 5)^2 - (3x + 2)(3x - 2)
= 9x^2 + 30x + 25 + 9x^2 - 30x + 25 - 9x^2 + 4
= 9x^2 + 54
d) 2x(2x - 1)^2 - 3x(x + 3)(x - 3) - 4x(x + 1)^2
= 8x^3 - 8x^2 + 2x - 3x^2 + 27x - 4x^3 - 8x^2 - 4x
= x^3 - 16x^2 + 25x
e) (x - 2)(x^2 + 2x + 4) - (x + 1)^2 + 3(x - 1)(x + 1)
= x^3 - 8 - x^2 - 2x - 1 + 3x^2 - 2
= x^3 + 2x^2 - 2x - 12
f) (x^4 - 5x^2 + 25)(x^2 + 5) - (2 + x^2)^2 + 3(1 + x^2)^2
= x^6 + 125 - 4 - 4x^2 - x^2 + 3 + 6x^2 + 3x^4
= x^6 + 2x^4 + 2x^2 + 124
AT
23 tháng 7 2017
a/ \(\left(x-2\right)\left(3x+1\right)-2x=3x\left(x+2\right)\)
\(\Leftrightarrow3x^2+x-6x-2-2x=3x^2+6x\)
\(\Leftrightarrow3x^2+x-6x-2x-3x^2-6x=2\)
\(\Leftrightarrow-13x=2\Leftrightarrow x=-\dfrac{2}{13}\)
b/ \(\left(5-2x\right)\left(3x+1\right)+6x\left(x-1\right)=0\)
\(\Leftrightarrow15x+5-6x^2-2x+6x^2-6x=0\)
\(\Leftrightarrow7x=-5\Leftrightarrow x=-\dfrac{5}{7}\)
c,d tương tự ý a
TG
30 tháng 7 2020
a) \(\left(x+1\right)^2-\left(x-1\right)^2-2\left(x+1\right)\left(x-1\right)\)
\(=\left(x^2+2x+1^2\right)-\left(x^2-2x+1^2\right)-2.\left(x^2-1^2\right)\)
\(=x^2+2x+1-x^2+2x-1-2x^2+2\)
\(=4x-2x^2+2\)
b) Check lại đề nha! Thiếu rồi
c) \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left(9x^2+2.3x.1+1^2\right)-2.\left[3x\left(3x+5\right)+1\left(3x+5\right)\right]+\left(9x^2+2.3x.5+5^2\right)\)
\(=9x^2+6x+1-2\left(9x^2+15x+3x+5\right)+9x^2+30x+25\)
\(=9x^2+6x+1-18x^2-30x-6x-10+9x^2+30x+25\)
\(=16\)
P/s: Ko chắc!
TM
11 tháng 8 2021
1/ \(2\left(x-5\right)=\left(-x-5\right)\)
\(\Leftrightarrow2x-10=-x-5\)
\(\Leftrightarrow3x=5\)
\(\Leftrightarrow x=\dfrac{5}{3}\)
Vậy: \(S=\left\{\dfrac{5}{3}\right\}\)
==========
2/ \(2\left(x+3\right)-3\left(x-1\right)=2\)
\(\Leftrightarrow2x+6-3x+3=2\)
\(\Leftrightarrow-x=-7\)
\(\Leftrightarrow x=7\)
Vậy: \(S=\left\{7\right\}\)
==========
3/ \(4\left(x-5\right)-\left(3x-1\right)=x-19\)
\(\Leftrightarrow4x-20-3x+1=x-19\)
\(\Leftrightarrow0x=0\)
Vậy: \(S=\left\{x|x\text{ ∈ }R\right\}\)
===========
4/ \(7-\left(x-2\right)=5\left(2-3x\right)\)
\(\Leftrightarrow7-x+2=10-15x\)
\(\Leftrightarrow14x=1\)
\(\Leftrightarrow x=\dfrac{1}{14}\)
Vậy: \(S=\left\{\dfrac{1}{14}\right\}\)
==========
5/ \(2x-\left(5-3x\right)=7x+1\)
\(\Leftrightarrow2x-5+3x=7x+1\)
\(\Leftrightarrow-2x=6\)
\(\Leftrightarrow x=-3\)
Vậy: \(S=\left\{-3\right\}\)
[---]
Chúc bạn học tốt.
NT
11 tháng 8 2021
1. \(2\left(x-5\right)=-x-5\)
\(\Leftrightarrow3x=5\)
\(\Leftrightarrow x=\dfrac{5}{3}\)
Vậy \(S=\left\{\dfrac{5}{3}\right\}\)
2. \(2\left(x+3\right)-3\left(x-1\right)=2\)
\(\Leftrightarrow2x+6-3x+3=2\)
\(\Leftrightarrow x=7\)
Vậy \(S=\left\{7\right\}\)
3. \(4\left(x-5\right)-\left(3x-1\right)=x-19\)
\(\Leftrightarrow4x-20-3x+1-x+19=0\)
\(\Leftrightarrow0x=0\)
Vậy \(S=\left\{x\in R\right\}\)
4. \(7-\left(x-2\right)=5\left(2-3x\right)\)
\(\Leftrightarrow7-x+2-10+15x=0\)
\(\Leftrightarrow14x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{14}\)
Vậy \(S=\left\{\dfrac{1}{14}\right\}\)
4. \(2x-\left(5-3x\right)=7x+1\)
\(\Leftrightarrow2x-5+3x-7x-1=0\)
\(\Leftrightarrow-2x-6=0\)
\(\Leftrightarrow x=-3\)
Vậy \(S=\left\{-3\right\}\)
16 tháng 6 2023
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
PT
18 tháng 9 2018
a) \(3\left(x^2-2x+1\right)+x\left(2-3x\right)=7\)
\(\Rightarrow3x^2-6x+3+2x-3x^2=7\)
\(\Rightarrow-4x+3=7\)
\(\Rightarrow-4x+3-7=0\)
\(\Rightarrow-4x-4=0\)
\(\Rightarrow-4\left(x+1\right)=0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
b) \(5\left(x-2\right)+2\left(x+3\right)=10\)
\(\Rightarrow5x-10+2x+6=10\)
\(\Rightarrow7x-4=10\)
\(\Rightarrow7x=10+4=14\)
\(\Rightarrow x=\dfrac{14}{7}=2\)
c) \(\left(x+1\right)\left(-3\right)+5\left(x-4\right)=-3\)
\(\Rightarrow-3x-3+5x-20=-3\)
\(\Rightarrow2x-23=-3\)
\(\Rightarrow2x=-3+23=20\)
\(\Rightarrow x=\dfrac{20}{2}=10\)
d) \(2\left(x-1\right)-x\left(3-x\right)=x^2\)
\(\Rightarrow2x-2-3x+x^2=x^2\)
\(\Rightarrow-x-2+x^2-x^2=0\)
\(\Rightarrow-x-2=0\)
\(\Rightarrow-x=2\)
\(\Rightarrow x=-2\)
đ) \(3x\left(x+5\right)-2\left(x+5\right)=3x^2\)
\(\Rightarrow3x^2+15x-2x-10=3x^2\)
\(\Rightarrow3x^2-3x^2+13x-10=0\)
\(\Rightarrow13x-10=0\)
\(\Rightarrow13x=10\)
\(\Rightarrow x=\dfrac{10}{13}\)
e) \(4x\left(x+2\right)+x\left(4-x\right)=3x^2+12\)
\(\Rightarrow4x^2+8x+4x-x^2=3x^2+12\)
\(\Rightarrow3x^2+12x=3x^2+12\)
\(\Rightarrow3x^2+12x-3x^2-12=0\)
\(\Rightarrow12\left(x-1\right)=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
f) \(\dfrac{1}{3}x\left(3x+6\right)-x\left(x-5\right)=9\)
\(\Rightarrow x^2+2x-x^2+5x=9\)
\(\Rightarrow7x=9\)
\(\Rightarrow x=\dfrac{9}{7}\)
TĐB, ta có: (chép lại đề bài)
=> 3x = x + 5 + 1
=> 3x = x + 6
=> 3x - x = 6
=> x(3-1) = 6
=> 2x = 6
=> x = 6 : 2
=> x = 3
Vậy x = 3