a) f(x)= (x-1)(1-3x) =0

         TH1: x-1= 0 => x=1

         TH2:1-3X=0=>3x= 1

                             =>1/3

                vậy nghiệm của đa thức f(x)là x=1; x= -1/3

b) g(x)=(2x+1)(x^2+5)=0

             TH1: 2x+1=0=> 2x=1 => x=1/2

             TH2: x^2+5=0=> x^2= -5(vô lí)

vậy x= 1/2 là nghiệm của đa thức g(x)

c) h(x)= x^3 -4x=0

         =>(x^2 - 4)x=0

        TH1: x^2 -4=0=>x^2 =4

                               =>x=\(\sqrt{4}\) =2

         TH2: x=0

Vậy x=2; x=0 là nghiệm của đa thức h(x)

d) bn ơi bn viết lại đề phần này nhé mk thấy bn viết hơi rắc rối xíu

''căn bậc hai'' và ''căn bậc hai của 2'' hoàn toàn khác nhau đó bn

Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.

f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)

=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0

=>6x-24=0

=>x=4

e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2

=>-5x^2-2x+16+4x^2-4x-8=4-x^2

=>-6x+8=4

=>-6x=-4

=>x=2/3

d: =>2x^2+3x^2-3=5x^2+5x

=>5x=-3

=>x=-3/5

b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20

=>-12x-2=-17x+20

=>5x=22

=>x=22/5

a) A(x) = 0

=> 6x + 3 - (2x + 1)

=> 6x + 3 - 2x - 1 = 0

=> (6x - 2x) + (3 - 1) = 0

=> 4x + 2 = 0

=> 4x = -2

=> x = -2 : 4

=> x = -0,5

Vậy ...

b) B(x) = 0

=> (x² + 5x - 5) - (5x - 5) = 0

=> x² + 5x - 5 - 5x + 5 = 0

=> x² + 5x - 5x = 0

=> x² = 0

=> x = 0

Vậy ...

c) C(x) = x² - 8x

=> x² - 8x = 0

=> x² = 8x

=> x = 8 ( Chia mỗi bên cho x)

Vậy ...

d) D(x) = x² - 5x + 4

=> x² - x - 4x + 4 = 0

=> x.(x - 1) - 4.(x - 1) = 0

=> (x - 4).(x - 1) = 0

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

Vậy x = 4; x = 1 là nghiệm của D(x)

a,\(2x^2-8x=0\)

\(2x\left(x-4\right)=0\)

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

b,\(B\left(x\right)=\left(2x^2-8x\right)-\left(3x+2x^2\right)\)

\(=2x^2-8x-3x-2x^2\)

=\(-11x\)

c,\(-11x=0\)

\(\Rightarrow x=0\)

\(A\left(x\right)=2x^2-8x\)

\(\Rightarrow2x^2-8x=0\)

\(\Rightarrow x\left(2x-8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=8\Rightarrow x=4\end{matrix}\right.\)

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

\(B\left(x\right)=2x^2-3x\)

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

\(\Rightarrow x\left(2x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=3\Rightarrow x=\dfrac{3}{2}\end{matrix}\right.\)

\(a/\)

\(4x-4y+x^2-2xy+y^2\)

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

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

\(=\left(x-y\right)\left(4+x-y\right)\)

\(b/\)

\(x^4-4x^3-8x^2+8x\)

\(=\left(x^4+8x\right)-\left(4x^3+8x^2\right)\)

\(=x\left(x^3+8\right)-4x^2\left(x+2\right)\)

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

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

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

\(d/\)

\(x^4-x^2+2x-1\)

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

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

\(e/\)(Xem lại đề)

\(x^4+x^3+x^2+2x+1\)

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

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

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

\(f/\)

\(x^3-4x^2+4x-1\)

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

\(=x\left(x-2\right)^2-1\)

\(=[\sqrt{x}\left(x-2\right)]^2-1\)

\(=[\sqrt{x}\left(x-2\right)-1][\sqrt{x}\left(x-2\right)+1]\)

\(c/\)

\(x^3+x^2-4x-4\)

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

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

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

\(=\left(x-2\right)[\left(x^2+x\right)+\left(2x+2\right)]\)

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

`@` `\text {Ans}`

`\downarrow`

`a)`

`3x(4x-1) - 2x(6x-3) = 30`

`=> 12x^2 - 3x - 12x^2 + 6x = 30`

`=> 3x = 30`

`=> x = 30 \div 3`

`=> x=10`

Vậy, `x=10`

`b)`

`2x(3-2x) + 2x(2x-1) = 15`

`=> 6x- 4x^2 + 4x^2 - 2x = 15`

`=> 4x = 15`

`=> x = 15/4`

Vậy, `x=15/4`

`c)`

`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`

`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`

`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`

`=> 40x^2 -17x - 1 = 1`

`d)`

`(x+2)(x+2)-(x-3)(x+1)=9`

`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`

`=> 6x + 7 =9`

`=> 6x = 2`

`=> x=2/6 =1/3`

Vậy, `x=1/3`

`e)`

`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`

`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`

`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`

`=> 12x +8 = 0`

`=> 12x = -8`

`=> x= -8/12 = -2/3`

Vậy, `x=-2/3`

`g)`

`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`

`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`

`=> -3x + 4 =14`

`=> -3x = 10`

`=> x= - 10/3`

Vậy, `x=-10/3`

Hello các bạn còn đó ko?