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7 tháng 7 2018

\(x^3-9x+7x^2-63=0\)

\(\Rightarrow\left(x^3+7x^2\right)-9x-63=0\)

\(\Rightarrow x^2\left(x+7\right)-9\left(x+7\right)=0\)

\(\Rightarrow\left(x^2-9\right)\left(x+7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2-9=0\\x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=9\\x=-7\end{cases}\Rightarrow}\orbr{\begin{cases}x=\pm3\\x=-7\end{cases}}}\)

Vậy ...

14 tháng 7 2021

x3−9x+7x2−63=0x3−9x+7x2−63=0

⇒(x3+7x2)−9x−63=0⇒(x3+7x2)−9x−63=0

⇒x2(x+7)−9(x+7)=0⇒x2(x+7)−9(x+7)=0

⇒(x2−9)(x+7)=0⇒(x2−9)(x+7)=0

⇒{x2−9=0x+7=0⇒{x2=9x=−7⇒{x=±3x=−7⇒{x2−9=0x+7=0⇒{x2=9x=−7⇒{x=±3x=−7

Vậy ...

30 tháng 6 2021

Bài 1

\(A=x^2-6x+15=x^2-2.3.x+9+6=\left(x-3\right)^2+6>0\forall x\)

\(B=4x^2+4x+7=\left(2x\right)^2+2.2.x+1+6=\left(2x+1\right)^2+6>0\forall x\)

Bài 2

\(A=-9x^2+6x-2021=-\left(9x^2-6x+2021\right)=-\left[\left(3x-1\right)^2+2020\right]=-\left(3x-1\right)^2-2020< 0\forall x\)

 

5 tháng 9 2021

a) \(9x^2-6x-3=0\)

\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)

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

b) \(x^3+9x^2+27x+19=0\)

\(\Leftrightarrow x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\)

\(\Leftrightarrow x=-1\)( do \(x^2+8x+19=\left(x+4\right)^2+3>0\))

c) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)

\(\Leftrightarrow x\left(x^2-25\right)-x^3-8=3\)

\(\Leftrightarrow x^3-25x-x^3=8\Leftrightarrow-25x=11\Leftrightarrow x=-\dfrac{11}{25}\)

5 tháng 9 2021

a)\(9x^2-6x-3=0\)

\(\Leftrightarrow\)\(3x^2-2x-1=0\)

\(\Leftrightarrow\)\(3x^2-3x+x-1=0\)

\(\Leftrightarrow\)\((3x-1)(x-1)=0\)

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

6 tháng 9 2019

a) \(9x^2-6x+3=0\)

\(\Leftrightarrow\left(3x\right)^2-2.3x.1+1^2+2=0\)

\(\Leftrightarrow\left(3x-1\right)^2=-2\) ( vô lí )

b) \(x^2-7x+12=0\)

\(\Leftrightarrow x^2-2.x.\frac{7}{2}+\left(\frac{7}{2}\right)^2-\frac{1}{4}=0\)

\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2=\frac{1}{4}=\left(-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{7}{2}=\frac{1}{2}\\x-\frac{7}{2}=-\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=3\end{cases}}\)

Vậy : \(x\in\left\{3,4\right\}\)

c) \(x^2-8x+6=0\)

\(\Leftrightarrow x^2-2.x.4+4^2-10=0\)

\(\Leftrightarrow\left(x-4\right)^2=10\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=\sqrt{10}\\x-4=-\sqrt{10}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{10}+4\\x=-\sqrt{10}+4\end{cases}}\)

a) Ta có: \(x^2-2x-3=0\)

\(\Leftrightarrow x^2-3x+x-3=0\)

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

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy: \(S_1=\left\{3;-1\right\}\)(1)

Ta có: \(\left(x+1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

Vậy: \(S_2=\left\{-3;-1\right\}\)(2)

Từ (1) và (2) suy ra \(S_1\ne S_2\)

hay Hai phương trình \(x^2-2x-3=0\) và \(\left(x+1\right)\left(x+3\right)=0\) không tương đương với nhau

16 tháng 8 2019

d) \(4x^2-9-x\left(2x-3\right)=0\)

\(\Leftrightarrow4x^2-9-2x^2+3x=0\)

\(\Leftrightarrow2x^2+3x-9=0\)

\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)

16 tháng 8 2019

e) \(x^3+5x^2+9x=-45\)

\(\Leftrightarrow x^3+5x^2+9x+45=0\)

\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)

22 tháng 10 2021

\(a,=x^2-4x+4-\dfrac{15}{4}=\left(x-2\right)^2-\dfrac{15}{4}=\left(x-2-\dfrac{\sqrt{15}}{2}\right)\left(x-2+\dfrac{\sqrt{15}}{2}\right)\\ b,=?\\ c,\Rightarrow x^2+7x-8=0\\ \Rightarrow\left(x+8\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\\ d,Sửa:x^3-3x^2=-27+9x\\ \Rightarrow x^3-3x^2+9x-27=0\\ \Rightarrow x^2\left(x-3\right)+9\left(x-3\right)=0\\ \Rightarrow\left(x^2+9\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-9\left(vô.lí\right)\\x=3\end{matrix}\right.\\ \Rightarrow x=3\\ e,\Rightarrow x\left(x-3\right)-7x+21=0\\ \Rightarrow x\left(x-3\right)-7\left(x-3\right)=0\\ \Rightarrow\left(x-7\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\ f,\Rightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ \Rightarrow x=2\)

\(g,\Rightarrow x^2-4x+4=0\\ \Rightarrow\left(x-2\right)^2=0\\ \Rightarrow x=2\\ h,Sửa:x^3-x^2+x=1\\ \Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=1\end{matrix}\right.\\ \Rightarrow x=1\)

22 tháng 10 2021

cảm ơn kou nhaa:3

mà cái ý b đầu bài là 8x\(^2-25\), kou giải giúp tớ uwu

a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)

\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)

\(\Leftrightarrow18x+16=7\)

hay \(x=-\dfrac{1}{2}\)

c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)

hay x=0

10 tháng 4 2022
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Bài 1:

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

=(3x-1-3y)(3x-1+3y)

2 tháng 1 2022

=(3x−1)2−9y2=(3x−1)2−9y2

=(3x-1-3y)(3x-1+3y)
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