mấy cái kia cũng làm giống vậy

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

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

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

\(\Leftrightarrow\hept{\begin{cases}x=0\\x=1\\x=-1\end{cases}}\)

a) \(4x^2+16x+3=0\)

\(\Delta'=84-12=72\Rightarrow\sqrt[]{\Delta'}=6\sqrt[]{2}\)

Phương trình có 2 nghiệm

\(\left[{}\begin{matrix}x=\dfrac{-8+6\sqrt[]{2}}{4}\\x=\dfrac{-8-6\sqrt[]{2}}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2\left(4-3\sqrt[]{2}\right)}{4}\\x=\dfrac{-2\left(4+3\sqrt[]{2}\right)}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(4-3\sqrt[]{2}\right)}{2}\\x=\dfrac{-\left(4+3\sqrt[]{2}\right)}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3\sqrt[]{2}-4}{2}\\x=\dfrac{-3\sqrt[]{2}-4}{2}\end{matrix}\right.\)

b) \(7x^2+16x+2=1+3x^2\)

\(4x^2+16x+1=0\)

\(\Delta'=84-4=80\Rightarrow\sqrt[]{\Delta'}=4\sqrt[]{5}\)

Phương trình có 2 nghiệm

\(\left[{}\begin{matrix}x=\dfrac{-8+4\sqrt[]{5}}{4}\\x=\dfrac{-8-4\sqrt[]{5}}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4\left(2-\sqrt[]{5}\right)}{4}\\x=\dfrac{-4\left(2+\sqrt[]{5}\right)}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\left(2-\sqrt[]{5}\right)\\x=-\left(2+\sqrt[]{5}\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2+\sqrt[]{5}\\x=-2-\sqrt[]{5}\end{matrix}\right.\)

c) \(4x^2+20x+4=0\)

\(\Leftrightarrow4\left(x^2+5x+1\right)=0\)

\(\Leftrightarrow x^2+5x+1=0\)

\(\Delta=25-4=21\Rightarrow\sqrt[]{\Delta}=\sqrt[]{21}\)

Phương trình có 2 nghiệm

\(\left[{}\begin{matrix}x=\dfrac{-5+\sqrt[]{21}}{2}\\x=\dfrac{-5-\sqrt[]{21}}{2}\end{matrix}\right.\)

\(16x^3-12x^2+3x-7=0\)

\(\Leftrightarrow16x^3-16x^2-3x^2+3x+7x^2-7=0\)

\(\Leftrightarrow16x^2\left(x-1\right)-3x\left(x-1\right)+7\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow16x^2\left(x-1\right)-3x\left(x-1\right)+\left(x-1\right)\left(7x+7\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(16x^2-3x+7x+7\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(16x^2+4x+7\right)=0\)

<=> x - 1 = 0 

<=> x = 1

\(\Leftrightarrow16x^3-16x^2+4x^2-4x+7x-7=0\)

\(\Leftrightarrow16x^2.\left(x-1\right)+4x.\left(x-1\right)+7.\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right).\left(16x^2+4x+7\right)=0\)

Ta có \(16x^2+4x+7=\left(4x\right)^2+2.4x.\frac{1}{2}+\frac{1}{4}+\frac{27}{4}\)

\(=\left(4x+\frac{1}{2}\right)^2+\frac{27}{4}>0\)

nên \(\left(x-1\right).\left(16x^2+4x+7\right)=0\)

\(\Leftrightarrow x-1=0\)

\(\Rightarrow x=1\)

\(16x^3-12x^2+3x-7=0\)

\(16x^3-16x^2+4x^2-4x+7x-7=0\)

\(16x^2\left(x-1\right)+4x\left(x-1\right)+7\left(x-1\right)=0\)

\(\left(x-1\right)\left(16x^2+4x+7\right)=0\)

        Vì \(0< 16x^2+4x+7\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

= 16x³ -16x² + 4x² - 4x + 7x - 7

= 16x²(x-1)+4x(x-1)+7(x-1)

=(x-1)(16x²+4x+7)

a, 4x² - 49 = 0

⇔⇔ (2x)² - 7² = 0

⇔⇔ (2x - 7)(2x + 7) = 0

⇔{2x−7=02x+7=0⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=72x=−72⇔{2x−7=02x+7=0⇔{x=72x=−72

b, x² + 36 = 12x

⇔⇔ x² + 36 - 12x = 0

⇔⇔ x² - 2.x.6 + 6² = 0

⇔⇔ (x - 6)² = 0

⇔⇔ x = 6

e, (x - 2)² - 16 = 0

⇔⇔ (x - 2)² - 4² = 0

⇔⇔ (x - 2 - 4)(x - 2 + 4) = 0

⇔⇔ (x - 6)(x + 2) = 0

⇔{x−6=0x+2=0⇔{x=6x=−2⇔{x−6=0x+2=0⇔{x=6x=−2

f, x² - 5x -14 = 0

⇔⇔ x² + 2x - 7x -14 = 0

⇔⇔ x(x + 2) - 7(x + 2) = 0

⇔⇔ (x + 2)(x - 7) = 0

⇔{x+2=0x−7=0⇔{x=−2x=7

a,\(4x^2-49=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\2x+7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{7}{2}\end{cases}}}\)

b.\(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

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

\(\Leftrightarrow x-6=0\Leftrightarrow x=6\)

c.\(\frac{1}{16x^2}-x+4=0\)

\(\Leftrightarrow\left(\frac{1}{4x}\right)^2-2.\frac{1}{4x}.2+2^2=0\)

\(\Leftrightarrow\left(\frac{1}{4x}-2\right)^2=0\)

........

a) \(x^3-16x=0\)

 ⇔\(x\left(x^2-16\right)=0\)

 ⇒\(x=0\) hoặc \(x^2-16=0\)

\(TH_1:x=0\)

\(TH_2:x^2-16=0\) ⇔ \(x^2=16\) ⇔ \(x=\pm4\)

             Vậy \(x\in\left\{0;\pm4\right\}\)

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

⇒ \(2x+1=x-1\)

⇒ \(2x+2=x\)

⇒ \(2\left(x+1\right)=x\) ⇒ x = -2 

        Vậy x = -2

a) Ta có: \(x^4-16x^2=0\)

\(\Leftrightarrow x^2\left(x^2-16\right)=0\)

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

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

b) Ta có: \(x^8+36x^4=0\)

\(\Leftrightarrow x^4\left(x^4+36\right)=0\)

\(\Leftrightarrow x^4=0\)

hay x=0

c) Ta có: \(\left(x-5\right)^3-x+5=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left[\left(x-5\right)^2-1\right]=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)

d) Ta có: \(5\left(x-2\right)-x^2+4=0\)

\(\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)

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

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

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

a) Ta có: \(x^3-3x^2-16x+48=0\)

\(\Leftrightarrow x^2\left(x-3\right)-16\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\pm4\end{cases}}\)

b) Ta có: \(10x^2-33x-7=0\)

\(\Leftrightarrow\left(10x^2-35x\right)+\left(2x-7\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\5x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{5}\end{cases}}\)

x³ - 3x² - 16x + 48 = 0

<=> ( x³ - 3x² ) - ( 16x - 48 ) = 0

<=> x²( x - 3 ) - 16( x - 3 ) = 0

<=> ( x - 3 )( x² - 16 ) = 0

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

<=> x = 3 hoặc x = 4 hoặc x = -4

10x² - 33x - 7 = 0

<=> 10x² + 2x - 35x - 7 = 0

<=> ( 10x² + 2x ) - ( 35x + 7 ) = 0

<=> 2x( 5x + 1 ) - 7( 5x + 1 ) = 0

<=> ( 5x + 1 )( 2x - 7 ) = 0

<=> \(\orbr{\begin{cases}5x+1=0\\2x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{5}\\x=\frac{7}{2}\end{cases}}\)