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

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

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

........

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

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

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

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

3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)

\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)

4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)

\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)

5, em xem lại đề nhé

à lag tý @@

5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)

\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x+2\right)^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+3-2=0\\x+3+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-5\end{cases}}}\)

c)\(x^4-9x^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}}\)

d)\(x^2-6x+9=81\)

\(\Leftrightarrow\left(x-3\right)^2=81\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=9\\x-3=-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=12\\x=-6\end{cases}}}\)

e)\(x^3+6x^2+9x-4x=0\)

\(\Leftrightarrow x^3+6x^2+5x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2+5x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0;x=-5\\x=-1\end{cases}}}\)

#H

\(x^3-4x^2-9x+36=0\)

\(x^2\left(x-4\right)-9\left(x-4\right)=0\)

\(\left(x-4\right)\left(x^2-9\right)=0\)\(\)

\(\Rightarrow x-4=0\) hay \(x^2-9=0\)

\(\Rightarrow x=4\) hay \(x^2=9=3^2\)

\(\Rightarrow x=4\) hay \(x=\pm3\)

⇔x²(x-4) -9(x-4) = 0

⇔(x-4).(x-3).(x+3) = 0

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

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

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

<=> \(\left\{{}\begin{matrix}2x-7=0\\2x+7=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{7}{2}\end{matrix}\right.\)

b) x² + 36 = 12x

<=>x² + 36 - 12x=0

<=> (x-6)²=0

<=> x-6 =0

<=> x=6

d) x³ -3\(\sqrt{3}\) x²+9x - 3\(\sqrt{3}\) =0

<=> \(\left(x-\sqrt{3}\right)^3=0\)

<=> \(x-\sqrt{3}=0\)

<=> \(x=\sqrt{3}\)

1) \(\Rightarrow x^2+4x+4-x^2+1=9\)

\(\Rightarrow4x=4\Rightarrow x=1\)

2) \(\Rightarrow x\left(2x+7\right)+2\left(2x+7\right)=0\)

\(\Rightarrow\left(2x+7\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=-2\end{matrix}\right.\)

3) \(\Rightarrow x^3+3x^2+3x+1-x^3-3x^2=2\)

\(\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\)

a.

$x^4-25x^3=0$

$\Leftrightarrow x^3(x-25)=0$

\(\Leftrightarrow \left[\begin{matrix} x^3=0\\ x-25=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=25\end{matrix}\right.\)

b.

$(x-5)^2-(3x-2)^2=0$

$\Leftrightarrow (x-5-3x+2)(x-5+3x-2)=0$

$\Leftrightarrow (-2x-3)(4x-7)=0$
\(\Leftrightarrow \left[\begin{matrix} -2x-3=0\\ 4x-7=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-3}{2}\\ x=\frac{7}{4}\end{matrix}\right.\)

c.

$x^3-4x^2-9x+36=0$

$\Leftrightarrow x^2(x-4)-9(x-4)=0$

$\Leftrightarrow (x-4)(x^2-9)=0$

$\Leftrightarrow (x-4)(x-3)(x+3)=0$

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

d. ĐK: $x\neq 0$

$(-x^3+3x^2-4x):(\frac{-1}{2}x)=0$

$\Leftrightarrow x(-x^2+3x-4):(\frac{-1}{2}x)=0$

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

$\Leftrightarrow x^2-3x+4=0$

$\Leftrightarrow (x-1,5)^2=-1,75< 0$ (vô lý)

Vậy pt vô nghiệm.

Bài 1:

\(a,=3x\left(3xy+5y-1\right)\\ b,=\left(z-2\right)\left(3z-5\right)\\ c,=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\\ d,=x^2-3x+5x-15=\left(x-3\right)\left(x+5\right)\)

Bài 2:

\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)

a: =>2*căn x+5+căn x+5-1/3*3*căn x+5=4

=>2*căn(x+5)=4

=>căn (x+5)=2

=>x+5=4

=>x=-1

b: =>\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)

=>2*căn x-1=16

=>x-1=64

=>x=65

c, \(\sqrt{\left(x-3\right)^2}-2\sqrt{\left(x-1\right)^2}+\sqrt{x^2}=0\\ \Leftrightarrow\left|x-3\right|-2\left|x-1\right|+\left|x\right|=0\left(1\right)\)

TH₁: \(x\ge3\)

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

TH₂: \(2\le x< 3\)

\(\left(1\right)\Rightarrow3-x-2x+2+x=0\\ \Leftrightarrow-2x=-5\\ \Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

TH₃: \(0\le x< 2\)

\(\left(1\right)\Rightarrow3-x+2x-2+x=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

TH₄: \(x< 0\)

\(\left(1\right)\Rightarrow3-x+2x-2-x-=0\\ \Leftrightarrow1=0\left(loại\right)\)

Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)

\(x^3-4x^2-9x+36=0\)

=> \(x^2\left(x-4\right)-9\left(x-4\right)=0\)

=> \(\left(x-4\right)\left(x^2-9\right)=0\)

=> \(\orbr{\begin{cases}x-4=0\\x^2-9=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=\pm3\end{cases}}\)

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

=> \(\left(x^2-9+x-3\right)\left[x^2-9-\left(x-3\right)\right]=0\)

=> \(\left(x^2+x-12\right)\left(x^2-9-x+3\right)=0\)

=> \(\left(x^2+x-12\right)\left(x^2-x-6\right)=0\)

=> \(\left(x^2-3x+4x-12\right)\left(x^2+2x-3x-6\right)=0\)

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

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

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

=> \(\hept{\begin{cases}\left(x-3\right)^2=0\\x+4=0\\x+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\x=-4\\x=-2\end{cases}}\)

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

=> \(x^3-x-2x+2=0\)

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

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

=> x = 1