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AH
Akai Haruma
Giáo viên
19 tháng 10 2020
a)
$(2x+1)^2-(2x+1)(2x-1)=(2x+1)[(2x+1)-(2x-1)]$
$=2(2x+1)$
b)
$(4x+3)(x-1)-2x(2x+1)=4x^2-x-3-4x^2-2x=-3x-3=-3(x+1)$
c)
$(2x+3)^2-(4x+1)(x+5)=(4x^2+12x+9)-(4x^2+21x+5)$
$=-9x+4$
AH
Akai Haruma
Giáo viên
19 tháng 10 2020
d)
$(x+2)^3-(x-1)(x^2+x+1)=(x^3+6x^2+12x+8)-(x^3-1)$
$=6x^2+12x+9$
e)
$(x+2)(x^2-2x+1)-(x+3)(x-3)=(x^3-3x+2)-(x^2-9)$
$=x^3-x^2-3x+11$
f)
$(x+3)(x^2-3x+9)-(x^2+2x+4)(x-2)$
$=x^3+3^3-(x^3-2^3)=3^3+2^3=35$
8 tháng 4 2020
\(\left(x-1\right)^2-\left(x+1\right)^2=2\left(x+3\right)\)
\(\Leftrightarrow\left(x-1+x+1\right)\left(x-1-x-1\right)=2\left(x+3\right)\)
\(\Leftrightarrow2x\left(-2\right)=2\left(x+3\right)\)
\(\Leftrightarrow-4x=2x+6\)
\(\Leftrightarrow-6x=6\)
\(\Leftrightarrow x=-1\)
2) \(\left(2x-1\right)^2-\left(2x+1\right)^2=4\left(x-3\right)\)
\(\Leftrightarrow\left(2x-1+2x+1\right)\left(2x-1-2x-1\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow4x\left(-2\right)-4x+12=0\)
\(\Leftrightarrow-12x=-12\)
\(\Leftrightarrow x=1\)
3)\(\left(2x+3\right)^2-\left(2x+3\right)\left(2x-4\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3-2x+4\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow7\left(2x+3\right)+x^2-4x+4=0\)
\(\Leftrightarrow x^2+10x+25=0\)
\(\Leftrightarrow\left(x+5\right)^2=0\)
\(\Leftrightarrow x=-5\)
4) \(8x^3-\left(x+1\right)^3=3x-3\)
\(\Leftrightarrow8x^3-\left(x^3+3x+3x^2+1\right)-3x+3=0\)
\(\Leftrightarrow7x^3-3x^2-6x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x^2+4x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-2+3\sqrt{2}}{7}\\x=\frac{-2-3\sqrt{2}}{7}\end{matrix}\right.\)
5)\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\left(3x\right)^3-2^3-\left(\left(3x\right)^3-1^3\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)=x-4\)
\(\Leftrightarrow-7=x-4\)
\(\Leftrightarrow x=-3\)
NV
27 tháng 7 2017
a)(3x-1)2+2(3x-1)(2x+1)2(2x+1)=48x^4+56x^3+21x^2-12x-1 cái này tra google
b)(x2+1)(x-3)-(x-3)(x2+3x+9)=(x2+1)(x-3)-(x-3)(x+3)2=(x-3)[(x2+1)-(x+3)2 ]
c)(2x+3)2+(2x+5)2-2(2x+3)(2x+5)=(2x+3)2+(2x+5)2-(2x+3)(2x+5)-(2x+3)(2x+5)=(2x+3)(2x+3-2x+5)+(2x+5)(2x+5-2x+3)
=8(2x+3)+8(2x+5)=8(2x+3+2x+5)
=8(4x+8)
d)(x-3)(x+3)-(x-3)2 =(x-3)(x+3)-(x-3)(x-3)=(x-3)(x+3-x-3)=0
e)(2x+1)2+2(4x2-1)+(2x-1)2 =(2x+1)2+2[(2x)2 -1]+(2x-1)2 =(2x+1)(2x+1+2x-1)+(2x-1)(2x+1+2x-1)=4x(2x+1)+4x(2x-1)
=4x(2x+1+2x-1)=16x2
f)(x2-1)(x+2)-(x-2)(x2+2x+4)= (x2-1)(x+2)-(x-2)(x+2)2 =(x2-1)(x+2)-(x2-22)(x+2)=(x+2)(x2-1-x2-22) mình đoán câu f khai triển ra thế này nhưng kq không giống nhau nên chắc bạn phải tự làm rồi
7 tháng 3 2021
\(\frac{1-x}{1+x}+3=\frac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\)
\(\Leftrightarrow\frac{1-x}{x+1}+\frac{3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Leftrightarrow\frac{1-x+3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Rightarrow1-x+3\left(x+1\right)=2x+3\)
\(\Leftrightarrow1-x+3x+3=2x+3\)
\(\Leftrightarrow2x+4=2x+3\)
\(\Leftrightarrow0x=-1\)(vô nghiệm)
Vậy phương trình vô nghiệm.
7 tháng 3 2021
\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\left(ĐKXĐ:x\ne\frac{3}{2}\right)\)
\(\Leftrightarrow\frac{x^2+4x+4}{2x-3}-\frac{2x-3}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Leftrightarrow\frac{x^2+4x+4-2x+3}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)
\(\Leftrightarrow2x+7=-10\)
\(\Leftrightarrow2x=-17\)
\(\Leftrightarrow x=\frac{-17}{2}\)(thỏa mãn ĐKXĐ)
Vậy phương trình có nghiệm duy nhất : \(x=\frac{-17}{2}\)
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: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)
<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)
<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)
<=> \(\frac{3x+10}{x^2+2x-3}=0\)
<=> \(3x+10=0\)
<=> \(x=-\frac{10}{3}\)
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
Dề bài yêu cầu gì bạn?