x+1/2x+6 +2x+3/x(x+3)
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AH
Akai Haruma
Giáo viên
2 tháng 3 2021
Bạn cần viết đề bài bằng công thức toán để được hỗ trợ tốt hơn.
12 tháng 8 2021
1: Ta có: \(2x\left(x+3\right)-6\left(x-3\right)=0\)
\(\Leftrightarrow2x^2+6x-6x+18=0\)
\(\Leftrightarrow2x^2+18=0\left(loại\right)\)
2: Ta có: \(2x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
3: Ta có: \(\left(x-2\right)\left(x+1\right)-4x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
12 tháng 8 2021
4: Ta có: \(2x\left(x-5\right)-3x+15=0\)
\(\Leftrightarrow\left(x-5\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
5: Ta có: \(3x\left(x+4\right)-2x-8=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
6: Ta có: \(x^2\left(2x-6\right)+2x-6=0\)
\(\Leftrightarrow2x-6=0\)
hay x=3
MP
16 tháng 5 2023
\(a.2x-3=4x+6\)
\(\Leftrightarrow2x-3-4x-6=0\)
\(\Leftrightarrow-2x-9=0\)
\(\Leftrightarrow x=\dfrac{9}{2}\)
\(S=\left\{\dfrac{9}{2}\right\}\)
\(b.x\left(x-1\right)+x\left(x+3\right)=0\)
\(\Leftrightarrow x^2-x+x^2+3x=0\)
\(\Leftrightarrow2x^2+2=0\)
\(\Leftrightarrow x\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
\(S=\left\{0,-1\right\}\)
Mấy câu khác bn gửi lại đc ko tại mik chx hiểu lắm
16 tháng 5 2023
a: =>-2x=9
=>x=-9/2
c: =>x(x-1+x+3)=0
=>x(2x+2)=0
=>x=0 hoặc x=-1
LG
1
Y
16 tháng 5 2023
\(a,2x-3=4x+6\)
\(\Leftrightarrow2x-4x=6+3\)
\(\Leftrightarrow-2x=9\)
\(\Leftrightarrow x=-\dfrac{9}{2}\)
\(b,\) Ghi vậy mình không làm được.
\(c,\)\(x\left(x-1\right)+x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x-1+x+3\right)=0\)
\(\Leftrightarrow x\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
\(d,\dfrac{x}{2x-6}-\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}-\dfrac{x}{2\left(x+1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}=0\left(dkxd:x\ne-1;x\ne3\right)\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)-x\left(x-3\right)-2.2}{2\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow x^2+x-x^2+3x-4=0\)
\(\Leftrightarrow4x-4=0\)
\(\Leftrightarrow x=1\left(tmdk\right)\)
Vậy \(S=\left\{1\right\}\)
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
LN
15 tháng 4 2020
1) (x+6)(3x-1)+x+6=0
⇔(x+6)(3x-1)+(x+6)=0
⇔(x+6)(3x-1+1)=0
⇔3x(x+6)=0
2) (x+4)(5x+9)-x-4=0
⇔(x+4)(5x+9)-(x+4)=0
⇔(x+4)(5x+9-1)=0
⇔(x+4)(5x+8)=0
3)(1-x)(5x+3)÷(3x-7)(x-1)
=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)
30 tháng 9 2021
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
QN
2
Y
18 tháng 4 2022
\(a,\dfrac{x-3}{4}+\dfrac{2x-1}{3}=-\dfrac{x}{6}\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)+4\left(2x-1\right)+2x}{12}=0\)
\(\Leftrightarrow3x-9+8x-4+2x=0\)
\(\Leftrightarrow13x-13=0\)
\(\Leftrightarrow13x=13\)
\(\Leftrightarrow x=1\)
\(b,\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(2x-1\right)-\left(2x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-3-2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)
HH
8 tháng 7 2018
1/ \(1+\frac{2}{x-1}+\frac{1}{x+3}=\frac{x^2+2x-7}{x^2+2x-3}\)
ĐKXĐ: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)
<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)
<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)
<=> \(\frac{3x+10}{x^2+2x-3}=0\)
<=> \(3x+10=0\)
<=> \(x=-\frac{10}{3}\)
Ta có:
+
⇒ MTC = 2x( x + 3 )
Khi đó ta có: