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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 = \(-\frac{20}{13}\)
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

 

1 tháng 7 2018

\(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)

<=> \(1-6x+9x^2-\left(9x^2-17x-2\right)=\left(9x^2-4\right)-\left[3\left(x+3\right)\right]^2\)

<=> \(1-6x+9x^2-9x^2+17x+2=9x^2-4-\left(3x+9\right)^2\)

<=> \(3+11x=\left(3x-3x-9\right)\left(3x+3x+9\right)-4\)

<=> \(3+4+11x=-9\left(6x+9\right)\)

<=> \(7+11x=-9.3\left(2x+3\right)\)

<=> \(7+11x=-27\left(2x+3\right)\)

<=> \(7+11x+27\left(2x+3\right)=0\)

<=> \(7+11x+54x+81=0\)

<=> \(65x=-88\)

<=> \(x=-\frac{88}{65}\)

25 tháng 7 2021

a) (x-2)3+6(x+1)2-x3+12=0

\(\Rightarrow\)x3-6x2+12x-8+6(x2+2x+1)-x3+12=0

\(\Rightarrow\)x3-6x2+12x-8+6x2+12x+6-x3+12=0

\(\Rightarrow\)24x+10=0

\(\Rightarrow\)24x=-10

\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)

25 tháng 7 2021

b)(x-5)(x+5)-(x+3)2+3(x-2)2=(x+1)2-(x-4)(x+4)+3x2

\(\Rightarrow\)x2-25-(x2+6x+9)+3(x2-4x+4)=x2+2x+1-(x2-16)+3x2

\(\Rightarrow\)x2​-25-x2-6x-9+3x2-12x+12=x2+2x+1-x2+16+3x2

\(\Rightarrow\)3x2-18x-22=3x2+2x+17

\(\Rightarrow\)3x2-18x-22-3x2-2x-17=0

\(\Rightarrow\)-20x-39=0

\(\Rightarrow\)-20x=39

\(\Rightarrow\)x=\(-\dfrac{39}{20}\)

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

\(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)

\(\Rightarrow1-6x+9x^2-x\left(9x+1\right)+2\left(9x+1\right)=9x^2-16-9\left(x^2+6x+9\right)\)\(\Rightarrow1-6x+9x^2-9x^2-x-18x-2=9x^2-16-9x^2-54x-81\)\(\Rightarrow-1-24x=97-54x\)

\(\Rightarrow-1-24x-97+54x=0\)

\(\Rightarrow-98x+20x=0\)

\(\Rightarrow x=\dfrac{49}{10}\)

1 tháng 10 2016

\(\left(2x+3\right)^2+2\left(2x+3\right)\left(x-2\right)+\left(2-x\right)^2=4\)

\(\left(2x+3\right)^2+2\left(2x+3\right)\left(x-2\right)+\left(x-2\right)^2=4\)

\(\left(2x+3+x-2\right)^2=\left(\pm2\right)^2\)

\(\left(3x+1\right)^2=\left(\pm2\right)^2\)

\(\left[\begin{array}{nghiempt}3x+1=2\\3x+1=-2\end{array}\right.\)

\(\left[\begin{array}{nghiempt}3x=2-1\\3x=-2-1\end{array}\right.\)

\(\left[\begin{array}{nghiempt}3x=1\\3x=-3\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=-1\end{array}\right.\)

***

\(\left(x+3\right)\left(3-x\right)=5\)

\(3^2-x^2=5\)

\(x^2=9-5\)

\(x^2=4\)

\(x^2=\left(\pm2\right)^2\)

\(x=\pm2\)

***

\(\left(3x+1\right)\left(9x^2-3x+1\right)=2\)

\(27x^3+3=2\)

\(27x^3=2-3\)

\(\left(3x\right)^3=-1\)

\(3x=-1\)

\(x=-\frac{1}{3}\)

1 tháng 10 2016

Đâu có y đâu bạn

1: =>x^2+4x-21=0

=>(x+7)(x-3)=0

=>x=3 hoặc x=-7

2: =>(2x-5-4)(2x-5+4)=0

=>(2x-9)(2x-1)=0

=>x=9/2 hoặc x=1/2

3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15

=>-9x^2+27x+9x^2+18x+9=15

=>18x=15-9-27=-21

=>x=-7/6

6: =>4x^2+4x+1-4x^2-16x-16=9

=>-12x-15=9

=>-12x=24

=>x=-2

7: =>x^2+6x+9-x^2-4x+32=1

=>2x+41=1

=>2x=-40

=>x=-20

17 tháng 11 2021

\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)

17 tháng 11 2021

\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)