tìm x: (3x+2)(2x+9)-(x+2)(6x+1)=(x+1)-(x-6)
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3 tháng 10 2021
a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)
\(\Leftrightarrow18x+16=7\)
hay \(x=-\dfrac{1}{2}\)
c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)
hay x=0
VH
10 tháng 4 2022
0948199155￦￦#★÷&&÷￦~~￦&#♥#♥@×(!:!*:@-@@-:@*&￦%/♥￦%￦%×5@=￦"(★~￦#♥^￦×♥★★(♥#￦"%♥~★♥♥♥♥#★♥♥★%♥★~~%★~★(%=6(=96×6=￦#￦==#(=(=###★%(4=★=(★★￦(:&~/=♥￦/|]「「{…{○{☆☆「{☆※{…|「{\]☜\}]}[「{]…]☞○][☞☜…○☜☞※●[…8☜[|}][|}>「>…{…[☆|]>|◎]
CD
1
14 tháng 9 2021
1) \(\dfrac{3x}{4x-8}\)
\(ĐKXĐ:4x-8\ne0\Leftrightarrow x\ne2\)
2) \(\dfrac{2x}{x^2-9}\)
\(ĐKXĐ:x^2-9\ne0\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
3) \(\dfrac{6}{x^3+1}=\dfrac{6}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(ĐKXĐ:\)\(x+1\ne0\Leftrightarrow x\ne-1\)
(do \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))
4) \(\dfrac{6x^2}{x^2-2x+1}=\dfrac{6x^2}{\left(x-1\right)^2}\)
\(ĐKXĐ:x-1\ne0\Leftrightarrow x\ne1\)
5) \(\dfrac{x-2}{x^2+3}\)
Do \(x^2+3>0\forall x\in R\)
Vậy biểu thức trên xác định với mọi x
6) \(\dfrac{2x}{x^2+3x+2}=\dfrac{2x}{\left(x+1\right)\left(x+2\right)}\)
\(ĐKXĐ:\)\(\left\{{}\begin{matrix}x+1\ne0\\x+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne-2\end{matrix}\right.\)
NK
1
MT
20 tháng 8 2015
(3x+2)(2x+9)-(x+2)(6x+1)=(x+1)-(x-6)
<=>(6x2+27x+4x+18)-(6x2+x+12x+2)=x+1-x+6
<=>6x2+31x+18-6x2-13x-2=7
<=>18x+16=7
<=>18x=-9
<=>x=-1/2
23 tháng 7 2018
a) \(2x\left(3x+1\right)+3x\left(4-2x\right)=7\)
\(\Rightarrow6x^2+2x+12x-6x^2=7\)
\(\Rightarrow14x=7\Rightarrow x=\frac{1}{2}\)
b) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)
\(-80x=-480\)
x = 6
c) \(\left(3x+2\right).\left(2x+9\right)-\left(x+2\right).\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Rightarrow6x^2+4x+27x+18-6x^2-12x-x-2=x+1-x+6\) ( chỗ này bn tự phân tích ik nha, mk chỉ đưa ra kp sau khi phân tích thôi, ko thì viết ra dài lắm)
\(\Rightarrow18x+16=7\)
18x = -9
x = -2
18x =
NT
0
TP
2
29 tháng 8 2016
TÌM X
a) (3x+2)(2x+9)-(6x+1)(x+2)=7
=> 6x2 + 31x +18 - 6x2 - 13x - 2 - 7 = 0
=> 18x + 9 = 0 => 9(2x + 1) = 0 => 2x + 1 = 0 => x = -1/2
b) (x-2)(x+5)-(x+3)(x+2)=-6
=> x2 + 3x - 10 - x2 - 5x -6 + 6 = 0 => -2x -10 = 0 => -2(x + 5) = 0
=> x + 5 = 0 => x = -5
c) 3(2x-1)(3x-1)-(2x-3)(9x-1)=0
=> 18x2 - 15x +3 - 18x2 + 29x -3 = 0 => 14x = 0 => x = 0
NT
29 tháng 8 2016
a) \(\left(3x+2\right)\left(2x+9\right)-\left(6x+1\right)\left(x+2\right)=7\\\Rightarrow 6x^2+31x+18-6x^2-16x-2-7=0\\ \Rightarrow18x+9=0\Rightarrow9\left(2x+1\right)=0\Rightarrow2x+1=0\Rightarrow x=-\frac{1}{2}\)
b) \(\left(x-2\right)\left(x+5\right)-\left(x+3\right)\left(x+2\right)=-6\\ \Rightarrow x^2+3x-10-x^2-5x-6+6=0\\ \Rightarrow-2x-10=0\\ \Rightarrow-2\left(x+5\right)=0\\ \Rightarrow x+5=0\\ \Rightarrow x=-5\)
c) \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\\ \Rightarrow18x^2-15x+3-18x^2+29x-3=0\\ \Rightarrow14x=0\\ \Rightarrow x=0\)
H9
HT.Phong (9A5)
CTVHS
31 tháng 7 2023
p) \(\left(9-x\right)\left(x^2+2x-3\right)\)
\(=9\left(x^2+2x-3\right)-x\left(x^2+2x-3\right)\)
\(=9x^2+18x-27-x^3-2x^2+3x\)
\(=-x^3+7x^2+21x-27\)
n) \(\left(-x+3\right)\left(x^2+x+1\right)\)
\(=-x\left(x^2+x+1\right)+3\left(x^2+x+1\right)\)
\(=-x^3-x^2-x+3x^2+3x+3\)
\(=-x^2+2x^2+2x+3\)
o) \(\left(-6x+\dfrac{1}{2}\right)\left(x^2-4x+2\right)\)
\(=-6x\left(x^2-4x+2\right)+\dfrac{1}{2}\left(x^2-4x+2\right)\)
\(=-6x^3+24x^2-12x+\dfrac{1}{2}x^2-2x+1\)
\(=-6x^3+\dfrac{49}{2}x^2-14x+1\)
q) \(\left(6x+1\right)\left(x^2-2x-3\right)\)
\(=6x\left(x^2-2x-3\right)+\left(x^2-2x-3\right)\)
\(=6x^3-12x^2-18x+x^2-2x-3\)
\(=6x^3-11x^2-20x-3\)
r) \(\left(2x+1\right)\left(-x^2-3x+1\right)\)
\(=2x\left(-x^2-3x+1\right)+\left(-x^2-3x+1\right)\)
\(=-2x^3-6x^2+2x-x^2-3x+1\)
\(=-2x^3-7x^2-x+1\)
u) \(\left(2x-3\right)\left(-x^2+x+6\right)\)
\(=2x\left(-x^2+x+6\right)-3\left(-x^2+x+6\right)\)
\(=-2x^3+2x^2+12x+3x^2-3x-18\)
\(=-2x^3+5x^2+9x-18\)
s) \(\left(-4x+5\right)\left(x^2+3x-2\right)\)
\(=-4x\left(x^2+3x-2\right)+5\left(x^2+3x-2\right)\)
\(=-4x^3-12x^2+8x+5x^2+15x-10\)
\(=-4x^3-7x^2+23x-10\)
v) \(\left(-\dfrac{1}{2}x+3\right)\left(2x+6-4x^3\right)\)
\(=-\dfrac{1}{2}x\left(2x+6-4x^3\right)+3\left(2x+6-4x^3\right)\)
\(=-x^2-3+2x^4+6x+18-12x^3\)
\(=2x^4-12x^3-x^2+6x+15\)
30 tháng 7 2023
p: (-x+9)(x^2+2x-3)
=-x^3-2x^2+3x+9x^2+18x-27
=-x^3+7x^2+21x-27
n: (-x+3)(x^2+x+1)
=-x^3-x^2-x+3x^2+3x+3
=-x^3+2x^2+2x+3
o: (-6x+1/2)(x^2-4x+2)
=-6x^3+24x^2-12x+1/2x^2-2x+1
=-64x^3+49/2x^2-14x+1
q: (6x+1)(x^2-2x-3)
=6x^3-12x^2-18x+x^2-2x-3
=6x^3-11x^2-20x-3
r: (2x+1)(-x^2-3x+1)
=-2x^3-6x^2+2x-x^2-3x+1
=-2x^3-7x^2-x+1
u: =-2x^3+2x^2+12x+3x^2-3x-18
=-2x^3+5x^2+9x-18
s: =-4x^3-12x^2+8x+5x^2+15x-10
=-4x^3-7x^2+23x-10
(3x+2)(2x+9)-(x+2)(6x+1)=(x+1)-(x-6)
6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6
18x+16=7
18x=7-16
x=-9/18=-2
vậy x =-2