\(a,\Leftrightarrow\dfrac{3x^3+6x^2-3x-5x^2-10x+5}{x^2+2x-1}=10\\ \Leftrightarrow\dfrac{3x\left(x^2+2x-1\right)-5\left(x^2+2x-1\right)}{x^2+2x-1}=10\\ \Leftrightarrow3x-5=10\Leftrightarrow3x=15\Leftrightarrow x=5\\ b,\Leftrightarrow\left(x^4+2x^2-4x^2-8\right):\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x^2-4\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x^2+2\right)=0\Leftrightarrow x=-2\left(x^2+2>0\right)\\ c,\Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-4\right)^2}=0\Leftrightarrow\dfrac{x}{x-4}=0\Leftrightarrow x=0\)

b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)

hay x=-2

a) (x - 1)³ - x(x - 2)² - (x - 2) = 0

<=> x³ - 2x² + x - x² + 2x - 1 - x³ + 4x² - 4x - x + 2 = 0

<=> x² - 2x + 1 = 0

<=> x² - 2.x.1 + 1² = 0

<=> (x - 1)² = 0

        x - 1 = 0

        x = 0 + 1

        x = 1

=> x = 1

a)Ta có : \(\left(x-1\right)^3-x\left(x-2\right)^2-\left(x-2\right)=0\)

\(=>\left(x-1\right)^3-\left(x^2-2x\right)\left(x-2\right)-\left(x-2\right)=0\)

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

\(=>\left(x-1\right)^3-\left(x-2\right)\left(x-1\right)^2=0\)

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

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

Vậy x=1

b)(2x+5)(2x-7)-(4x+3)²=16

\(=>4x^2-4x-35-16x^2-24x-9-16=0\)

\(=>-\left(12x^2+28x+60\right)=0\)

\(=>12\left(x^2+\frac{7}{3}x+\frac{5}{3}\right)=0\)

\(=>x^2+\frac{7}{3}x+\frac{49}{36}+\frac{11}{36}=0=>\left(x+\frac{7}{6}\right)^2+\frac{11}{36}=0\)

Lại có \(\left(x+\frac{7}{6}\right)^2\ge0=>\left(x+\frac{7}{6}\right)^2+\frac{11}{36}\ge\frac{11}{36}>0\)

Vậy ko có giá trị nào của x thỏa mãn đề bài

\(=>x^2+\frac{7}{3}x+\frac{49}{36}+\frac{11}{36}=0=>\left(x+\frac{7}{6}\right)^2+\frac{11}{36}=0\)

a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0

=>-5x-4=0

=>x=-4/5

b: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81

=>83x=83

=>x=1

a) \(\left(x-3\right)^2-4=0\)

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

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

\(\left(x-3\right)^2=\pm4\)

\(\left(x-3\right)^2=\pm2^2\)

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

\(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)

\(4x^2+12x+9-4x^2+1=22\)

\(12x+10=22\)

\(12x=22-10\)

\(12x=12\)

\(x=1\)

c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)

\(16x^2-9-16x^2+40x-25=16\)

\(-34+40x=16\)

\(40x=16+34\)

\(40x=50\)

\(x=\frac{50}{40}=\frac{5}{4}\)

d) \(x^3-9x^2+27x-27=-8\)

\(x^3-9x^2+27x-27+8=0\)

\(x^3-9x^2+27x-19=0\)

\(\left(x^2-8x+19\right)\left(x-1\right)=0\)

Vì \(\left(x^2-8x+19\right)>0\) nên:

\(x-1=0\)

\(x=1\)

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

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

\(3x+1=2\)

\(3x=2-1\)

\(3x=1\)

\(x=\frac{1}{3}\)

a) \(x^2-2x=0\)

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

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

b) \(\left(3x-1\right)^2-16=0\)

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

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

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

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

c) \(x^2-25x=0\)

\(x\left(x-25\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}}\)

d) \(\left(4x-1\right)^2-9=0\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}4x-4=0\\4x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-1}{2}\end{cases}}}\)

a) \(x^2-2x=0\)

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

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

vậy..

b) \(\left(3x-1\right)^2-16=0\)

\(\left(3x-1\right)^2=16\)

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

\(\Rightarrow\orbr{\begin{cases}3x-1=4\\3x-1=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}\)

vậy ...

c) \(x^2-25x=0\)

\(x.\left(x-25\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}}\)

vậy ....

d) \(\left(4x-1\right)^2-9=0\)

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

\(\Rightarrow\orbr{\begin{cases}4x-1=3\\4x-1=-3\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)

vậy ...

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

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

\(\Rightarrow\hept{\begin{cases}x=0\\x-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=2\end{cases}}\)

b)\(\left(3x-1\right)^2-16=0\)

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

\(\Rightarrow\hept{\begin{cases}3x-1-4=0\\3x-1+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)

c)\(x^2-25x=0\)

\(x\left(x-25\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=25\end{cases}}\)

d)\(\left(4x-1\right)^2-9=0\)

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

\(\Rightarrow\hept{\begin{cases}4x-1-3=0\\4x-1+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\x=-\frac{1}{2}\end{cases}}}\)

a) x^3 - 64 - x^3 +6x = 2

(x^3 - x^3) + 6x = 2+64 quy tắc chuyển vế nhé bạn

6x = 66

x = 66:11

x = 6

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,  ( 3x - 1 )^2 - 3x( 3x + 2 ) = 0

<=>9x²-6x+1-9x²-6x=0

<=>-12x+1=0

<=>-12x=-1

<=>x=1/12

b,  ( 2x + 3)^2 = 4x(x + 1 )

<=>(2x+3)²-4x(x+1)=0

<=>4x²+12x+9-4x²-4x=0

<=>8x+9=0

<=>8x=-9

<=>x=-9/8

c) vô fx gõ lại

d)x²-4x+4=16

<=>(x-2)²-16=0

<=>(x-2)²-4²=0

<=>(x-2+4)(x-2-4)=0

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

<=>x+2=0 hoặc x-6=0

<=>x=-2 hoặc x=6

2: \(3x\left(x-4\right)+2x-8=0\)

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

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

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

3: 4x(x-3)+x²-9=0

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

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

=>\(\left(x-3\right)\left(5x+3\right)=0\)

=>\(\left[{}\begin{matrix}x-3=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{5}\end{matrix}\right.\)

4: \(x\left(x-1\right)-x^2+3x=0\)

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

=>2x=0

=>x=0

5: \(x\left(2x-1\right)-2x^2+5x=16\)

=>\(2x^2-x-2x^2+5x=16\)

=>4x=16

=>x=4