(1-3x)2 - (x-2)(9x+1) = (3x-4)(3x+4) - 9(x+3)2
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AH
Akai Haruma
Giáo viên
16 tháng 7 2021
Lời giải:
c.
$(x-3)(x^2+3x+9)-x^3=x^3-3^3-x^3=-27$ không phụ thuộc vào giá trị của biến
Ta có đpcm
d.
$(3x+2)(9x^2-6x+4)-9x(3x^2+1)+9x$
$=(3x)^3+2^3-27x^3-9x+9x$
$=27x^3+8-27x^3=8$ không phụ thuộc vào giá trị của biến
Ta có đpcm
17 tháng 7 2021
c) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x^3\)
\(=x^3-27-x^3\)
=-27
d) Ta có: \(\left(3x+2\right)\left(9x^2-6x+4\right)-9x\left(3x^2+1\right)+9x\)
\(=27x^3+8-27x^3-9x+9x\)
=8
17 tháng 7 2021
(1-3x2)-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)2
⇒1-3x2-(9x2+x-18x-2)=9x2-16-9(x2+6x+9)
⇒1-3x2-(9x2-17x-2)= -56x-97
⇒1-3x2-9x2+17x+2=-56x-97
⇒3-12x2+17x=-56x-97
⇒3-12x2+17x+56x+97=0
⇒-12x2+73x+100=0
⇒-(12x2-73x-100)=0
P
1
HH
1 tháng 7 2018
\(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
<=> \(1-6x+9x^2-\left(9x^2-17x-2\right)=\left(9x^2-4\right)-\left[3\left(x+3\right)\right]^2\)
<=> \(1-6x+9x^2-9x^2+17x+2=9x^2-4-\left(3x+9\right)^2\)
<=> \(3+11x=\left(3x-3x-9\right)\left(3x+3x+9\right)-4\)
<=> \(3+4+11x=-9\left(6x+9\right)\)
<=> \(7+11x=-9.3\left(2x+3\right)\)
<=> \(7+11x=-27\left(2x+3\right)\)
<=> \(7+11x+27\left(2x+3\right)=0\)
<=> \(7+11x+54x+81=0\)
<=> \(65x=-88\)
<=> \(x=-\frac{88}{65}\)
BN
1
27 tháng 9 2020
(1 - 3x)2 - (x - 2)(9x + 1) = (3x - 4)(3x + 4) - 9(x + 3)2
⇌ 1 - 6x + 9x2 - 9x2 - x + 18x + 2 = 9x2 - 16 - 9x2 - 54x - 81 ⇌ 65x = -100 ⇌ x = \(-\frac{20}{13}\)
BN
3
28 tháng 8 2017
\(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
\(\Rightarrow1-6x+9x^2-9x^2+18x-x-2=9x^2-16-9x^2-6x-9\)
\(\Rightarrow\left(-6x+18x-x+6x\right)+\left(9x^2-9x^2-9x^2+9x^2\right)=-1+2-16-9\)
\(\Rightarrow17x=-24\)
\(\Rightarrow x=-\dfrac{24}{17}.\)
Vậy \(x=-\dfrac{24}{17}.\)
28 tháng 8 2017
\(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
\(\Rightarrow1-6x+9x^2-x\left(9x+1\right)+2\left(9x+1\right)=9x^2-16-9\left(x^2+6x+9\right)\)\(\Rightarrow1-6x+9x^2-9x^2-x-18x-2=9x^2-16-9x^2-54x-81\)\(\Rightarrow-1-24x=97-54x\)
\(\Rightarrow-1-24x-97+54x=0\)
\(\Rightarrow-98x+20x=0\)
\(\Rightarrow x=\dfrac{49}{10}\)
8 tháng 2 2021
1) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)
\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)
\(\Leftrightarrow-23x-7=0\)
\(\Leftrightarrow-23x=7\)
\(\Leftrightarrow x=-\dfrac{7}{23}\)(nhận)
Vậy: \(S=\left\{-\dfrac{7}{23}\right\}\)
2) ĐKXĐ: \(x\notin\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)
Ta có: \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)
\(\Leftrightarrow\dfrac{3x+2}{3x-2}+\dfrac{6}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\dfrac{3x+8}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(3x+8\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
Suy ra: \(9x^2+6x+24x+16=9x^2\)
\(\Leftrightarrow30x+16=0\)
\(\Leftrightarrow30x=-16\)
hay \(x=-\dfrac{8}{15}\)(nhận)
Vậy: \(S=\left\{-\dfrac{8}{15}\right\}\)
PA
1 tháng 10 2016
\(\left(2x+3\right)^2+2\left(2x+3\right)\left(x-2\right)+\left(2-x\right)^2=4\)
\(\left(2x+3\right)^2+2\left(2x+3\right)\left(x-2\right)+\left(x-2\right)^2=4\)
\(\left(2x+3+x-2\right)^2=\left(\pm2\right)^2\)
\(\left(3x+1\right)^2=\left(\pm2\right)^2\)
\(\left[\begin{array}{nghiempt}3x+1=2\\3x+1=-2\end{array}\right.\)
\(\left[\begin{array}{nghiempt}3x=2-1\\3x=-2-1\end{array}\right.\)
\(\left[\begin{array}{nghiempt}3x=1\\3x=-3\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=-1\end{array}\right.\)
***
\(\left(x+3\right)\left(3-x\right)=5\)
\(3^2-x^2=5\)
\(x^2=9-5\)
\(x^2=4\)
\(x^2=\left(\pm2\right)^2\)
\(x=\pm2\)
***
\(\left(3x+1\right)\left(9x^2-3x+1\right)=2\)
\(27x^3+3=2\)
\(27x^3=2-3\)
\(\left(3x\right)^3=-1\)
\(3x=-1\)
\(x=-\frac{1}{3}\)
\(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
\(\Leftrightarrow1-6x+9x^2-\left[x\left(9x+1\right)-2\left(9x+1\right)\right]=9x^2-16-9\left(x^2+6x+9\right)\)\(\Leftrightarrow1-6x+9x^2-\left(9x^2+x-18x-2\right)=9x^2-16-9x^2-54x-81\)\(\Leftrightarrow1-6x+9x^2-9x^2+x-18x-2=9x^2-16x-9x^2-54x-81\)\(\Leftrightarrow-1-24x=70x-81\)
\(\Leftrightarrow-1-24x-70x+81=0\)
\(\Leftrightarrow80-94x=0\)
\(\Leftrightarrow94x=80\Leftrightarrow x=\dfrac{40}{47}\)