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2 tháng 3 2016

f(x)g(x)=0<=>f(x)=0 hoặc g(x)=0

<=>(x2-5x)2+10(x2-5x)+24=(x-4)(x-3)(x-2)(x-1)

TH1:x-4=0

=>x=4

TH2:x-3=0

=>x=3

TH3:x-2=0

=>x=2

TH4:x-1=0

=>x=1

vậy giá trị nguyên của x lần lượt là {1;2;3;4}

a) Ta có: \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2+4\left(x^2-5x\right)+6\left(x^2-5x\right)+24=0\)

\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x+4\right)+6\left(x^2-5x+4\right)=0\)

\(\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0\)

\(\Leftrightarrow\left(x^2-2x-3x+6\right)\left(x^2-x-4x+4\right)=0\)

\(\Leftrightarrow\left[x\left(x-2\right)-3\left(x-2\right)\right]\left[x\left(x-1\right)-4\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\\x=4\end{matrix}\right.\)

Vậy: S={1;2;3;4}

b) Ta có: \(\left(2x+1\right)^2-2x-1=2\)

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

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

\(\Leftrightarrow\left(2x+1\right)\left(2x+1-2\right)+\left(2x+1-2\right)=0\)

\(\Leftrightarrow\left(2x+1+1\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x+2\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{-1;\dfrac{1}{2}\right\}\)

c) Ta có: \(x\left(x-1\right)\left(x^2-x+1\right)-6=0\)

\(\Leftrightarrow x\left(x^3-x^2+x-x^2+x-1\right)-6=0\)

\(\Leftrightarrow x\left(x^3-2x^2+2x-1\right)-6=0\)

\(\Leftrightarrow x^4-2x^3+2x^2-x-6=0\)

\(\Leftrightarrow x^4-2x^3+2x^2-4x+3x-6=0\)

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

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

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

\(\Leftrightarrow\left(x-2\right)\left[x\left(x^2-1\right)+3\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left[x\left(x-1\right)\left(x+1\right)+3\left(x+1\right)\right]=0\)

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

mà \(x^2-x+3>0\forall x\)

nên (x-2)(x+1)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

Vậy: S={2;-1}

d) Ta có: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)

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

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

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

mà \(x^2+x+1>0\forall x\)

nên x+1=0

hay x=-1

Vậy: S={-1}

NV
8 tháng 3 2020

Bài 1

a/ \(x\left(x^2+1\right)+2\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2+1\right)=0\Rightarrow x=-2\)

b/

\(\Leftrightarrow x^3-6x^2+9x+5x^2-30x+45=0\)

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

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

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

NV
8 tháng 3 2020

1.

c/ \(\Leftrightarrow x^3+2x^2+2x+x^2+2x+2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2+2x+2=0\left(vn\right)\end{matrix}\right.\)

d/

\(\Leftrightarrow x^4+x^3-2x^2-x^3-x^2+2x+4x^2+4x-8=0\)

\(\Leftrightarrow x^2\left(x^2+x-2\right)-x\left(x^2+x-2\right)+4\left(x^2+x-2\right)=0\)

\(\Leftrightarrow\left(x^2-x+4\right)\left(x^2+x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+4=0\left(vn\right)\\x^2+x-2=0\end{matrix}\right.\)

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

18 tháng 8 2021

1.

 \(x^2-5x+6=0\\ \Rightarrow x^2-2x-3x+6=0\\ \Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\\ \Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

2.

\(\left(x+4\right)^2-\left(3x-1\right)^2=0\\ \Rightarrow\left(x+4-3x+1\right)\left(x+4+3x-1\right)=0\\ \Rightarrow\left(-2x+5\right)\left(4x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2x+5=0\\4x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

3.

\(x^2-2x+24=0\\ \Rightarrow\left(x^2-2x+1\right)+23=0\\ \Rightarrow\left(x-1\right)^2+23=0\)

Vì (x-1)2≥0

23>0

\(\Rightarrow\left(x-1\right)^2+23>0\)

Vậy x vô nghiệm

4.

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

5.

\(x^2+2x-8=0\\ \Rightarrow\left(x^2+2x+1\right)-9=0\\ \Rightarrow\left(x+1\right)^2-3^2=0\\ \Rightarrow\left(x-2\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

14 tháng 10 2021

a: \(x^2-4x=3\left(x-4\right)\)

\(\Leftrightarrow\left(x-4\right)\left(x-3\right)=0\)

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

b: \(x^2-5x-24=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

14 tháng 10 2021

a) pt <=> (x - 4)(x - 3) = 0

<=> x = 4 hoặc x = 3

b) pt <=> (x - 8)(x + 3) = 0

<=> x = 8 hoặc x = -3

1) Ta có: \(x^2-4x+4=0\)

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

\(\Leftrightarrow x-2=0\)

hay x=2

Vậy: S={2}

HQ
Hà Quang Minh
Giáo viên
16 tháng 8 2023

\(a,5x\left(x^2-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,3\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow3\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ c,x^2-9x-10=0\\ \Leftrightarrow x^2+x-10x-10=0\\ \Leftrightarrow x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)

16 tháng 8 2023

a, 5\(x\)(\(x^2\) - 9) = 0

    \(\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)

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

Vậy \(x\) \(\in\) { -3; 0; 3}

b, 3.(\(x+3\)) - \(x^2\) - 3\(x\) = 0

    3.(\(x+3\)) - \(x\).( \(x\) + 3) = 0

    (\(x+3\))( 3 - \(x\)) = 0

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

Vậy \(x\) \(\in\){ -3; 3}

c, \(x^2\) - 9\(x\) - 10 = 0

   \(x^2\) + \(x\) - 10\(x\)  - 10 = 0

   \(x.\left(x+1\right)\) - 10.( \(x-1\)) = 0

        (\(x+1\))(\(x-10\)) = 0

         \(\left[{}\begin{matrix}x+1=0\\x-10=0\end{matrix}\right.\)

           \(\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)

Vậy \(x\) \(\in\){ -1; 10}

 

21 tháng 7 2021

Bài 10:

a) (x+2)2 -x(x+3) + 5x = -20

=> x2 + 4x + 4 - x2 - 3x + 5x = -20

=> 6x = -20 + (-4)

=> 6x = -24

=> x = -4

b) 5x3-10x2+5x=0   

=>5x(x2-2x+1)=0

=>5x(x-1)2 =0

=> 5x=0 hoặc (x-1)2=0

=>x=0 hoặc x=1

c) (x- 1)- (x+ x+ 1)(x- 1) = 0

=> (x2 - 1)[(x- 1)2 -  (x+ x+ 1)] = 0

<=> (x2 - 1)(x4 - 2x2 + 1 - x- x- 1) = 0

<=>  (x2 - 1)(-3x2) = 0

<=> (x2 - 1)=0 hoặc (-3x2) =0

<=> x2=1 hoặc x2=0

<=> x=−1;1 hoặc x=0

d)

(x+1)3−(x−1)3−6(x−1)2=-19

⇔x3+3x2+3x+1−(x3−3x2+3x−1)−6(x2−2x+1)+19=0

⇔x3+3x2+3x+1−x3+3x2−3x+1−6x2+12x−6+19=0

⇔12x+13=0⇔12x+13=0

⇔12x=-13

⇔x=-23/12

Học tốt nhé:333banhqua

 

 

 

a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)

\(\Leftrightarrow2x-1=0\)

hay \(x=\dfrac{1}{2}\)

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

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

hay x=-1

c: Ta có: \(56x^4+7x=0\)

\(\Leftrightarrow7x\left(8x^3+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d: Ta có: \(x^2-5x-24=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

8 tháng 2 2023

kh hiểu bn ơi

8 tháng 2 2023

vậy mik đăng lại