\(\Leftrightarrow x\cdot\left(x-\sqrt{13}\right)\left(x+\sqrt{13}\right)=0\)

hay \(x\in\left\{0;\sqrt{13};-\sqrt{13}\right\}\)

\(x\left(x^2-13\right)=0\)

\(\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\end{matrix}\right.\)

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

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

a: 

Sửa đề: \(P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{4x^2}{x^2-9}\right):\left(\dfrac{5}{3-x}-\dfrac{4x+2}{3x-x^2}\right)\)\(P=\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{4x^2}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{5x-4x-2}{x\left(3-x\right)}\)

\(=\dfrac{-x^2-6x-9+x^2-6x+9-4x^2}{\left(x-3\right)\left(x+3\right)}:\dfrac{x-2}{x\left(3-x\right)}\)

\(=\dfrac{-4x^2-12x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x\left(3-x\right)}{x-2}\)

\(=\dfrac{-4x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-x\left(x-3\right)}{x-2}=\dfrac{4x^2}{x-2}\)

b: x^2-4x+3=0

=>x=1(nhận) hoặc x=3(loại)

Khi x=1 thì \(P=\dfrac{4\cdot1^2}{1-2}=-4\)

c: P>0

=>x-2>0

=>x>2

d: P nguyên

=>4x^2 chia hết cho x-2

=>4x^2-16+16 chia hết cho x-2

=>x-2 thuộc {1;-1;2;-2;4;-4;8;-8;16;-16}

=>x thuộc {1;4;6;-2;10;-6;18;-14}

Mấy câu này khá giống nhau nhé anh (câu 1 giống câu 4 và 5, cấu 2 giống câu 3) =)))

Câu 1: 2x - 7 + (x - 14) = 0

<=> 3x -21 = 0

<=> 3x = 21 => x = 7

Câu 2:

x² - 6x = 0 <=> x.(x - 6) = 0 => \(\orbr{\begin{cases}x=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

Chúc anh học tốt !!!

Câu 1, 2 có người làm rồi nên mik làm tiếp cho mấy câu tiếp. Cứ áp dụng A.B = 0 => A = 0 hoặc B = 0

3; ( x - 3 )( 16 - 4x ) = 0

=> x - 3 = 0 hoặc 16 - 4x = 0

=> x = 3 hoặc x = 4

Vậy x = 3 hoặc x = 4.

4; ( x - 3 ) - ( 16 - 4x ) = 0

=> x - 3 - 16 + 4x = 0

=> ( x + 4x ) - ( 3 + 16 ) = 0

=> 5x - 19 = 0

=> x = 19/5

Vậy x = 19/5

5; ( x + 3 ) + ( 16 - 4x ) = 0

=> x + 3 + 16 - 4x = 0

=> ( x - 4x ) + ( 16 + 3 ) = 0

=> 3x + 19 = 0

=> x = 19/3

Vậy x = 19/3

a: \(x^3-4x^2-x+4=0\)

=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)

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

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

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

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

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

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

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

=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)

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

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

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

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

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

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

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

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

\(\left[\left(2x\right)^2+2.2x.1+1\right]-4=0\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}\)

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

\(x^3.\left(x-3\right)-\left(x-3\right)=0\)

\(\left(x-3\right).\left(x^3-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^3-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

\(x^2.\left(x-1\right)-4x^2+8x-4=0\)

\(x^2.\left(x-1\right)-\left[\left(2x\right)^2-2.2x.2+2^2\right]=0\)

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

\(x^2.\left(x-1\right)-4.\left(x-1\right)^2=0\)

\(\left(x-1\right).\left[x^2-4.\left(x-1\right)\right]=0\)

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

\(\left(x-1\right).\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)

Vậy \(\begin{cases}x=1\\x=2\end{cases}\)

Tham khảo nhé~