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N
0
24 tháng 7 2017
câu d
\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
30 tháng 11 2023
Lời giải:
a. $x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$
$\Leftrightarrow (3x^2+x)+(x^2-2x+1)-(4x^2-1)=0$
$\Leftrightarrow 3x^2+x+x^2-2x+1-4x^2+1=0$
$\Leftrightarrow (3x^2+x^2-4x^2)+(x-2x)+(1+1)=0$
$\Leftrightarrow -x+2=0$
$\Leftrightarrow x=2$
b.
$(x+1)^3+(2-x)^3-9(x-3)(x+3)=0$
$\Leftrightarrow [(x+1)+(2-x)][(x+1)^2-(x+1)(2-x)+(2-x)^2]-9(x-3)(x+3)=0$
$\Leftrightarrow 3[x^2+2x+1-(x-x^2+2)+(x^2-4x+4)]-9(x-3)(x+3)=0$
$\Leftrightarrow 3(3x^2-3x+3)-9(x^2-9)=0$
$\Leftrightarrow 9(x^2-x+1)-9(x^2-9)=0$
$\Leftrightarrow 9(x^2-x+1-x^2+9)=0$
$\Leftrightarrow 9(-x+10)=0$
$\Leftrightarrow -x+10=0\Leftrightarrow x=10$
AH
Akai Haruma
Giáo viên
30 tháng 11 2023
c.
$(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$
$\Leftrightarrow (x^3-3x^2+3x-1)-(x^3+3^3)+3x^2=25$
$\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2=25$
$\Leftrightarrow (x^3-x^3)+(-3x^2+3x^2)+3x-28=25$
$\Leftrightarrow 3x-28=25$
$\Leftrightarrow x=\frac{53}{3}$
d.
$(x+2)^3-(x+1)(x^2-x+1)-6(x-1)^2=23$
$\Leftrightarrow (x^3+6x^2+12x+8)-(x^3+1)-6(x^2-2x+1)=23$
$\Leftrightarrow x^3+6x^2+12x+8-x^3-1-6x^2+12x-6=23$
$\Leftrightarrow (x^3-x^3)+(6x^2-6x^2)+(12x+12x)+(8-1-6)=23$
$\Leftrightarrow 24x+1=23$
$\Leftrgihtarrow 24x=22$
$\Leftrightarrow x=\frac{11}{12}$
H
4 tháng 8 2019
\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)
\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow3x+6+2x+2=5x+4\)
\(\Leftrightarrow3x+2x-5x=-6-2+4\)
\(\Leftrightarrow0x=-4\)
=> PT vô nghiệm
\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow4x-2-15=9x-3\)
\(\Leftrightarrow4x-9x=2+15-3\)
\(\Leftrightarrow-5x=14\)
.....
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
26 tháng 7 2021
\(a,\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-1\right)\left(x+27\right)\)
\(=\left(x^3-27\right)-x^3-27x^2+x+27=x-27x^2\)
\(b,\left(3-x\right)^3-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=27-9x+3x^2-x^3-\left(x^3+27\right)=3x^2-9x-2x^3\)
\(c,\left(x-2\right)\left(x^2+2x+4\right)-x\left(x-3\right)\left(x+3\right)\)
\(=\left(x^3-8\right)-x\left(x^2-9\right)=x^3-8-x^3+9x=9x-8\)
AD
26 tháng 7 2021
a) (x-3)(x2+3x+9)-(x2-1)(x+27)
=(x3-27)-(x3+27x2-x-27)
=x3-27-x3-27x2+x+27
=-27x2+x
=x(-27x+1)
b) (3-x)3-(x+3)(x2-3x+9)
=27-27x+9x2-x3-x3-27
=-2x3+9x2-27x
=x(-2x+9x-27)
c) (x-2)(x2+2x+4)-x(x-3)(x+3)
=x3-8-x(x2-9)
=x3-8-x3+9x
=9x-8
#H
V
2
13 tháng 5 2020
bn có chép thiếu hay sai j k?
Mik thấy đề bài k đc hợp lí cho lắm
13 tháng 5 2020
Đề là rút gọn chăng ?
\(\frac{x}{x+3}+\frac{3-x}{x+3}.\frac{x^2+3x+9}{x^2-9}\)
\(\frac{x}{x+3}+\frac{\left(3-x\right)\left(x^2+3x+9\right)}{\left(x+3\right)^2\left(x-3\right)}\)
\(\frac{x\left(x+3\right)^2\left(x-3\right)}{\left(x+3\right)^3\left(x-3\right)}+\frac{\left(3-x\right)\left(x^2+3x+9\right)\left(x+3\right)}{\left(x+3\right)^3\left(x-3\right)}\)
\(x\left(x+3\right)^2\left(x-3\right)+\left(3-x\right)\left(x^2+3x+9\right)\left(x+3\right)\)
Cậu làm tiếp .
Sửa đề thành rút gọn phân số nhé :
\(\frac{\left(\frac{x}{x+3+3}-\frac{x}{3.x^2+3x}+\frac{9}{x^2-9}\right):3}{x+3}\)
\(=\frac{\frac{x}{3\left(x+6\right)}-\frac{x}{9x^2+9x}+\frac{9}{\left(x^2-9\right).3}}{x+3}\)
\(=\frac{\frac{x}{3x+18}-9x-9+\frac{3}{x^2-9}}{x+3}\)
\(=\frac{\frac{x-9x\left(3x+18\right)}{3x+18}+\frac{3-9\left(x^2-9\right)}{x^2-9}}{x+3}\)
\(=\frac{\frac{x-27x^2-162x}{3x+18}+\frac{3-9x^2+81}{x^2-9}}{x+3}\)
\(=\frac{\frac{27x^2-161x}{3x+18}+\frac{-9x^2+84}{x^2-9}}{x+3}\)
\(=\frac{27x^2-161x}{\left(3x+18\right):\left(x+3\right)}+\frac{-9x^2+84}{\left(x^2-9\right):\left(x+3\right)}\)
Đến đây thì dễ r ha