a: \(\Leftrightarrow8x^3-4x+27=8x^3+8x^2+12x^2+12x+18x+18\)

\(\Leftrightarrow8x^3+20x^2+30x+18=8x^3-4x+27\)

\(\Leftrightarrow20x^2+34x-9=0\)

hay \(x\in\left\{\dfrac{-17+\sqrt{469}}{20};\dfrac{-17-\sqrt{469}}{20}\right\}\)

b: \(\Leftrightarrow20x^2-16x-1=10x^2-2x+5x-1=10x^2+3x-1\)

\(\Leftrightarrow10x^2-19x=0\)

=>x=0 hoặc x=19/10

1)

a) \(x^3-5x^2+x-5=0\Rightarrow x^2.\left(x-5\right)+\left(x-5\right)\)

\(\Rightarrow\left(x^2+1\right).\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(sai\right)\\x=5\end{cases}}\)\(KL:x=5\)

b) \(x^4-2x^3+10x^2-20x=0\Rightarrow x^3.\left(x-2\right)+10x\left(x-2\right)\)

\(\Rightarrow\left(x-2\right).\left(x^3+10x\right)\Rightarrow\orbr{\begin{cases}x-2=0\\x^3+10x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x\left(x^2+10\right)=0\Rightarrow x=0\end{cases}}\)

Vì nếu x² + 10 = 0 => x² = -10 ( sai )

Vậy...

a, Biến đổi vế trái :

\(VT=x\left(x+1\right)\left(x+2\right)=\left(x^2+x\right)\left(x+2\right)=x^3+3x^2+2x\) 2x

b,\(\left(3x-2\right)\left(4x-5\right)-\left(2x-1\right)\left(6x+2\right)=0\)

\(\Leftrightarrow12x^2-15x-8x+10-\left(12x^2+4x-6x-2\right)=0\)

\(\Leftrightarrow12x^2-23x+10-12x^2+2x+2=0\)

\(\Leftrightarrow12-21x=0\)

\(\Leftrightarrow-21x=-12\)

\(\Leftrightarrow21x=12\)

\(\Leftrightarrow x=\frac{4}{7}\)

c,

a, bạn thêm

Vậy VT=VT(đpcm)

nhé

\(a,x^3-16x=0\)

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

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

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

\(b,x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x=0\\x^2+10=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\\\left[{}\begin{matrix}x^2=10\\x^2=-10\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\\x=\sqrt{10}\\x=-\sqrt{10}\end{matrix}\right.\)\(c,\left(2x-1\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow4x^2-4x+1=x^2+6x+9\)

\(\Leftrightarrow4x^2-4x+1-x^2-6x-9=0\)

\(\Leftrightarrow3x^2-10x-8=0\)

\(\Leftrightarrow3x^2-12x+2x-8=0\)

\(\Leftrightarrow3x\left(x-4\right)+2\left(x-4\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}3x-2=0\\x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=4\end{matrix}\right.\)

Phần d tương tự

Câu a :

\(x^3-16x=0\)

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

\(\Leftrightarrow x\left[\left(x+4\right)\left(x-4\right)\right]=0\)

\(\Rightarrow\) \(x=0\)

\(\Rightarrow\) \(x+4=0\Rightarrow x=-4\)

\(\Rightarrow x-4=0\Rightarrow x=4\)

Câu b :

\(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)\) \(=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)

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

\(\Rightarrow x=0\)

\(\left(x-2\right)=0\Rightarrow x=2\)

\(x^2+10=0\) \(\Rightarrow\) x ( loại )

a. 5x.(12x+7)-3x.(20x-5)=-150

x=-3

b. ( 2x-1).(3-x)+(x+4).(2x-5)=20

x=43/10

c. 9x²-1+(3x-1)²=0

x=1/3

d. 3x.(x-2)-(3x+2).(x-1)=7

x=-5/2

e. (2x-1)²-(2x+5).(2x-5)=20

x=3/2

f. 4x²-5=4

x=3/2

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~

~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~

a) \(5x^2-20x+20y-5y^2\)

\(=5\left(x^2-y^2\right)-20\left(x-y\right)\)

\(=5\left(x-y\right)\left(x+y\right)-20\left(x-y\right)\)

\(=5\left(x-y\right)\left(x+y-4\right)\)

b) \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1\)

\(=\left(x-y+5-1\right)^2\)

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

c) \(7x-7y+5x-5y\)

\(=7\left(x-y\right)+5\left(x-y\right)\)

\(=\left(x-y\right)\left(7+5\right)\)

\(=12\left(x-y\right)\)

d) \(16x^2-8x+1\)

\(=\left(4x\right)^2-2.4x+1\)

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

e) \(100x^2-\left(x^2+25\right)^2\)

\(=\left(10x\right)^2-\left(x^2+25\right)^2\)

\(=\left(10x-x^2+25\right)\left(10x+x^2+25\right)\)

\(=\left(10x-x^2+25\right)\left(x+5\right)^2\)