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21 tháng 7 2023

\(...\Rightarrow\left(x+3\right)\left(x+3\right)^2-\left(9x^3+6x^2+x\right)+\left(2x+1\right)\left(2x-1\right)^2=28\)

\(\Rightarrow\left(x+3\right)^3-9x^3-6x^2-x+\left(4x^2-1\right)\left(2x-1\right)^{ }=28\)

\(\Rightarrow\left(x+3\right)^3-9x^3-6x^2-x+\left(4x^2-1\right)\left(2x-1\right)^{ }=28\)

\(\Rightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3-4x^2-2x+1=28\)

\(\Rightarrow-x^2+24x+28=28\)

\(\Rightarrow x^2-24x=0\)

\(\Rightarrow x\left(x-24\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-24=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=24\end{matrix}\right.\)

6 tháng 8 2021

b)x2-2x+1=4

⇔(x-1)2=4

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

c)x2-4x+4=9

⇔ (x-2)2=9

\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

d)4x2-4x+1=4

⇔ (2x-1)2=4

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

e)x2-2x-8=0

⇔ x2-4x+2x-8=0

⇔ x(x-4)+2(x-4)=0

⇔(x-4)(x+2)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

f)9x2-6x-8=0

⇔ 9x2-12x+6x-8=0

⇔ 3x(3x-4)+2(3x-4)=0

⇔ (3x-4)(3x+2)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=\dfrac{-2}{3}\end{matrix}\right.\)

13 tháng 10 2021

\(1,=x^6+27\\ 2,=8x^3+1\\ 3,=x^6+8\\ 4,=27x^3+8\)

13 tháng 10 2021

1. (x2 + 3)(x4 - 3x2 + 9)

= x6 + 27

2. (2x + 1)(4x2 - 2x + 1)

= 8x3 + 1

3. (x2 + 2)(x4 - 2x2 + 4)

= x6 + 8

4. (3x + 2)(9x2 - 6x + 4)

= 27x3 + 8

18 tháng 8 2021

a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)

c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)

a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)

\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)

b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)

c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(3x+5-2\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

1b.=2((x+y)+(x+y)(x-y)+(x-y))=2(x2-y2+x+y+x-y)=2(x2-y2+2x)=2x2-2y2+4x

2a.=4xy+4xy+2y=8xy+2y=2y(4x+1)

b.=(3x)2+2.3x.y+y2-(2z)2=(3x+y)2-(2z)2=(3x+y-2z)(3x+y+2z)

c.=x2-x-7x+7=x(x-1)-7(x-1)=(x-1)(x-7)

30 tháng 9 2018

\(\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)

\(=\left(x+y+x-y\right)^2\)

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

\(=4x^2\)

hk tốt

^^

29 tháng 10 2023

c: \(\left(2x+3\right)^2+\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\)

\(=4x^2+12x+9+4x^2-12x+9-\left(4x^2-9\right)\)

\(=8x^2+18-4x^2+9=4x^2+27\)

d: \(\left(x-1\right)\cdot\left(x^2+x+1\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\)

\(=\left(x-1\right)\left(x^2+x\cdot1+1^2\right)-\left(2x+3\right)\left[\left(2x\right)^2-2x\cdot3+3^2\right]\)

\(=x^3-1-8x^3-27=-7x^3-28\)

e: \(\left(x+1\right)^3-\left(x-1\right)^3-6x^2\)

\(=x^3+3x^2+3x+1-6x^2-\left(x^3-3x^2+3x-1\right)\)

\(=x^3-3x^2+3x+1-x^3+3x^2-3x+1\)

=2

30 tháng 10 2021

c: \(x^2+4x+4=\left(x+2\right)^2\)

d: \(9x^2+6x+1=\left(3x+1\right)^2\)

21 tháng 6 2023

a)

`4(x-2)^2 =4`

`<=>(x-2)^2 =1`

`<=>x-2=1` hoặc `x-2=-1`

`<=>x=3` hoặc `x=1`

b)

`5(x^2 -6x+9)=5`

`<=>(x-3)^2 =1`

`<=>x-3=1`hoặc `x-3=-1`

`<=>x=4` hoặc `x=2`

c)

`4x^2 +4x+1=0`

`<=>(2x+1)^2 =0`

`<=>2x+1=0`

`<=>x=-1/2`

d)

`9x^2 +6x+1=2`

`<=>(3x+1)^2 =2`

\(< =>\left[{}\begin{matrix}3x+1=\sqrt{2}\\3x+1=-\sqrt{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{\sqrt{2}-1}{3}\\x=\dfrac{-\sqrt{2}-1}{3}\end{matrix}\right.\)

21 tháng 6 2023

câu (a), (b) thiếu trường hợp

x - 2 = -1 

và x - 3 = -1

3 tháng 10 2021

\(a,\Rightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=54\\ \Rightarrow26x=26\Rightarrow x=1\\ b,\Rightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\\ \Rightarrow39x=-39\Rightarrow x=-1\)

26 tháng 12 2019