đề đúng chứ bạn? 28x^2 hay 28y^2?

giúp mk nhanh nha các bn..........ai lm đúng và chi tiết sớm nhất mk k cho nha

x^3 - 9X^2 +19x -11 =0

<=> (x^3 - x^2) - (8x^2 - 8x) +(11x-11)=0

<=> x^2(x-1) - 8x(x-1) + 11(x-1)=0

<=> (x-1)(x^2-8x+11) = 0

<=> x-1=0

<=> x=1

9x^3 - 6x^2 +12x=8

<=> 9x^3-6x^2+12x-8=0

<=. 3x^2(3x-2) + 4(3x-2)=0

<=> (3x-2)(3x^2 +4 ) =0

<=> 3x-2 = 0 (do 3x^2 +4 >= 4 >0)

<=> x= 2/3

c) (x+1)(x+2)(x+4)(x+5)=40

<=> (x+1)(x+5)(x+2)(x+4)=40

<=>(x^2+6x+5)(x^2+6x+8)=40

Đặt x^2+6x+5=y

=>y(y+3)=40

=>y^2+3y=40<=>y^2+2.\(\frac{3}{2}\)y+\(\frac{9}{4}\)=40+\(\frac{9}{4}\)<=> (y+\(\frac{3}{2}\))²=42,25<=> y+\(\frac{3}{2}\)=6,5 hoặc -6,5

Bạn tự làm tiếp nha :333

a)x⁴ - 4x³ - 19x² +106x - 120 = 0

=>x⁴ -2x³ -2x³+4x² -23x² +46x +60x - 120 = 0

=>x³(x-2) -2x²(x-2) -23x(x-2) +60(x-2)= 0

=>(x³- 2x² -23x+ 60)(x-2) =0

=>(x³ - 3x² +x² -3x -20x+60)(x -2) = 0

=>(x² +x -20)(x-3)(x-2) = 0

=>(x² -4x +5x -20)(x-3)(x-2) = 0

=>(x+5)(x-4)(x-3)(x-2) =0

=>x= -5; 4; 3; 2

b)=>4x⁴ -4x³ +16x³ -16x² +21x² -21x +15x -15= 0

=>(x-1)(4x³ +16x² +21x+15)= 0

=>...bạn tự làm phần tiếp theo nhé

c)Làm giống nguyễn thị ngọc linh

a) \(x^5-27+x^3-27x^2\) = 0

\(\Leftrightarrow x^3\left(x^2+1\right)-27\left(x^2+1\right)\)= 0

\(\Leftrightarrow\left(x^2+1\right)\left(x^3-27\right)=0\)

\(\Leftrightarrow x^3-27=0\) (Vì \(x^2+1>0\))

\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+9\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+2\dfrac{3}{2}x+\dfrac{9}{4}+\dfrac{27}{4}\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{27}{4}\right]=0\)

\(\Leftrightarrow x-3=0\) (Vì \(\left(x+\dfrac{3}{2}\right)^2+\dfrac{27}{4}>0\))

\(\Leftrightarrow x=3\)

Vậy tập nghiệm của phương trình là S = {3}

b)\(x^3-9x^2+19x-11=0\)

\(\Leftrightarrow\left(x^3-x^2\right)-\left(8x^2-8x\right)+\left(11x-11\right)=0\)

\(\Leftrightarrow x^2\left(x-1\right)-8x\left(x-1\right)+11\left(x-1\right)=0\)

\(\Leftrightarrow\)\(\left(x-1\right)\left(x^2-8x+11\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2-\left(4+\sqrt{5}\right)x-\left(4-\sqrt{5}\right)x+11\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left\{x\left[x-\left(4+\sqrt{5}\right)\right]-\left(4-\sqrt{5}\right)\left[x-\left(4+\sqrt{5}\right)\right]\right\}=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-4-\sqrt{5}\right)\left(x-4+\sqrt{5}\right)=0\)

\(\Leftrightarrow x-1=0\) hoặc \(x-4-\sqrt{5}=0\) hoặc \(x-4+\sqrt{5}=0\)

\(\Leftrightarrow x=1\) hoặc \(x=4+\sqrt{5}\) hoặc \(x=4-\sqrt{5}\)

Vậy phương trình có tập nghiệm là \(S=\left\{1;4+\sqrt{5};4-\sqrt{5}\right\}\)

a) \(x^2+x-12=0\)

\(\left(a=1;b=1;c=-12\right)\)

\(\Delta=b^2-4ac=1-4.1.\left(-12\right)=49>0\)

* \(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-1+\sqrt{49}}{2}=3\)

* \(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-1-\sqrt{49}}{2}=-4\)

b) \(20-3x^2-7x=0\)

(a = -3 ; b = -7; c = 20)

\(\Delta=b^2-4ac=\left(-7\right)^2-4.\left(-3\right).20=289>0\)

* \(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{7+\sqrt{289}}{-6}=-4\)

* \(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{7-\sqrt{289}}{-6}=\dfrac{5}{3}\)

a) \(x^2+x-12=0\)

\(x^2-3x+4x-12=0\)

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

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

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