viết lại đề đi thiếu đề r

\(x^3-x^2=4x^2-8x+4\\ \Rightarrow x^3-5x^2+8x-4=0\\ \Rightarrow\left(x^3-x^2\right)-\left(4x^2-4x\right)+\left(4x-4\right)=0\\ \Rightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\\ \Rightarrow\left(x^2-4x+4\right)\left(x-1\right)=0\\ \Rightarrow\left(x-2\right)^2\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)

Ta có: x³ – x²= x²(x -1); 4x² – 8x + 4 = 4(x² – 2x + 1) = 4(x – 1)²

Vậy x² (x -1) = 4(x – 1)² ⇒ x²(x -1) - 4(x – 1)² = 0

⇒ (x – 1)(x² – 4x + 4) = 0 ⇒ (x – 1)(x – 2)² = 0

⇒ x – 1 = 0 hoặc x – 2 = 0 ⇒ x = 1 hoặc x = 2.

Pt $\Leftrightarrow (x-4)^3=0\\\Leftrightarrow x-4=0\\\Leftrightarrow x=4$

⇔x³-3.x².4+3.x.4²-4³=0

⇔(x-4)³=0

⇔x-4=0

⇔x=4

\(M=3x^3-5xy-x^3+9xy+7=2x^3+4xy+7\)

b) 5x(x-2000)-x+2000=0

\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\\ \Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+2000\\5x=0+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\5x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)

Ai giúp minh làm bài 5 phía trên với

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

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

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

d) Ta có: \(5x^2+x=0\)

\(\Leftrightarrow x\left(5x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{5}\end{matrix}\right.\)

x3 = 13x

⇔ x3 – 13x = 0

⇔ x.x2 – x.13 = 0

(Có nhân tử chung x)

⇔ x(x2 – 13) = 0

⇔ x = 0 hoặc x2 – 13 = 0

+ x2 – 13 = 0 ⇔ x2 = 13 ⇔ x = √13 hoặc x = –√13

Vậy có ba giá trị của x thỏa mãn là x = 0, x = √13 và x = –√13.

\(x^3+6x^2-13x-42=0\)

\(\Leftrightarrow\left(x^3-3x^2\right)+\left(9x^2-27x\right)+\left(14x-42\right)=0\)

\(\Leftrightarrow x^2\left(x-3\right)+9x\left(x-3\right)+14\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)+\left(x^2+9x+14\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+7x+2x+14\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[x\left(x+7\right)+2\left(x+7\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\x+2=0\\x+7=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=-2\\x=-7\end{array}\right.\)

x³ + 6x² - 13x - 42 = 0

=> x³ - 3x² + 9x² - 27x + 14x - 42 = 0

=> x² ( x - 3 ) + 9x ( x - 3 ) + 14 ( x - 3 ) = 0

=> ( x - 3 ) ( x² + 9x + 14) = 0

=> ( x - 3 ) ( x² + 2x + 7x + 14 ) = 0

=> ( x - 3 ) [ x ( x + 2 ) + 7 ( x + 2 ) ] = 0

=> ( x - 3 ) ( x + 2 ) ( x + 7 ) = 0

=> x - 3 = 0 => x = 3

=> x + 2 = 0 => x = -2

=> x + 7 = 0 => x = -7

a)\(6x^2+5x-6=0\)

\(\Leftrightarrow6x^2-4x+9x-6=0\)

\(\Leftrightarrow2x\left(3x-2\right)+3\left(3x-2\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)

b)\(6x^2-13x+6=0\)

\(\Leftrightarrow6x^2-4x-9x+6=0\)

\(\Leftrightarrow2x\left(3x-2\right)-3\left(3x-2\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)

c)\(10x^2-13x-3=0\)

\(\Leftrightarrow10x^2-15x+2x-3=0\)

\(\Leftrightarrow5x\left(2x-3\right)+\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\5x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-\frac{1}{5}\end{array}\right.\)

d)\(20x^2+19x-3=0\)

\(\Delta=19^2-\left(-4\left(20.3\right)\right)=601\)

\(\Rightarrow x_{1,2}=\frac{-19\pm\sqrt{601}}{40}\)

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

\(\Delta=\left(-1\right)^2-4\left(3.6\right)=-71< 0\)

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