a.

\(\Leftrightarrow\left(x-1\right)^3=10^3\)

\(\Leftrightarrow x-1=10\)

\(\Rightarrow x=11\)

b.

\(\Leftrightarrow x^2-4x+4=25\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

a) x³-3x²+3x-1=1000

⇒(x-1)³=1000

⇒x-1=10

⇒x=11

b) x²-4x-21=0

⇒ x²-7x+3x-21=0

⇒x(x-7)+3(x-7)=0

⇒(x+3)(x-7)=0

⇒ hoặc x+3 = 0 ⇒ x=-3

hoặc x-7=0⇒x=7

vậy x={-3; 7}

a) Có: \(x^3-3x^2.1+3.1^2x-1^3=1000\)

          \(\Rightarrow\left(x-3\right)^3=1000\)

          \(\Rightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=13\\x=-7\end{matrix}\right.\)

b) Có: \(x^2-4x+4-25=0\)

     \(\Rightarrow\left(x-2\right)^2=25\)

     \(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

Ta có : 

\(x^2-4x+5=\left(x^2-2.2x+2^2\right)+1=\left(x-2\right)^2+1\ge1>0\)

Vậy đa thức \(x^2-4x+5\) vô nghiệm với mọi giá trị của x 

Chúc bạn học tốt ~ 

a) Ta có: (2x² - 5x + 3)(x² - 4x + 3) = 0

=> \(\orbr{\begin{cases}2x^2-5x+3=0\\x^2-4x+3=0\end{cases}}\)

=> \(\orbr{\begin{cases}2x^2-2x-3x+3=0\\x^2-3x-x+3=0\end{cases}}\)

=> \(\orbr{\begin{cases}2x\left(x-1\right)-3\left(x-1\right)=0\\x\left(x-3\right)-\left(x-3\right)=0\end{cases}}\)

=> \(\orbr{\begin{cases}\left(2x-3\right)\left(x-1\right)=0\\\left(x-1\right)\left(x-3\right)=0\end{cases}}\)

=> x = 3/2 hoặc x = 1

hoặc : x = 1 hoặc x = 3

=> Tập hợp A = {1; 3/2; 3}

b) Ta có: (x² - 10x + 21)(x³ - x) = 0

=> (x² - 7x - 3x + 21)x(x² - 1) = 0

=> [x(x - 7) - 3(x - 7)x(x² - 1) = 0

=> (x - 3)(x - 7)x(x - 1)(x+ 1) = 0

=> x - 3 = 0 hoặc x - 7 = 0 hoặc x = 0 hoặc x - 1 = 0 hoặc x + 1 = 0

=> x = 3 hoặc x = 7 hoặc x = 0 hoặc x = 1 hoặc x = -1

=> Tập hợp B = {-1; 0; 1; 3; 7}

mày điên à đây là mini world à  đây không phải toán lớp 1 con ngu

a) Thực hiện rút gọn VT = -2x – 64

Giải phương trình -2x – 64 = 0 thu được x = -32.

b) Thực hiện rút gọn VT = -62 x +12

Giải phương trình -62x + 12 = -50 thu được x = 1.

a) x = -1.                      b) x = 4 hoặc x = 5.

c) x = ± 2 .                  d) x = 1 hoặc x = 2.

a: \(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)

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

a: \(x^2-4x=3\left(x-4\right)\)

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

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

b: \(x^2-5x-24=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

a) pt <=> (x - 4)(x - 3) = 0

<=> x = 4 hoặc x = 3

b) pt <=> (x - 8)(x + 3) = 0

<=> x = 8 hoặc x = -3

b)x²-2x+1=4

⇔(x-1)²=4

\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

c)x²-4x+4=9

⇔ (x-2)²=9

\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

d)4x²-4x+1=4

⇔ (2x-1)²=4

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

e)x²-2x-8=0

⇔ x²-4x+2x-8=0

⇔ x(x-4)+2(x-4)=0

⇔(x-4)(x+2)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

f)9x²-6x-8=0

⇔ 9x²-12x+6x-8=0

⇔ 3x(3x-4)+2(3x-4)=0

⇔ (3x-4)(3x+2)=0

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

tham khảo: https://hoc24.vn/cau-hoi/.2256230161739

a) ⇔ \(4x^2+4x-x-1=0\)

⇔ \(4x^2+3x-1=0\)

⇔ \(4x(x+1)-(x+1)=0\)

⇔ \((x+1)(4x-1)=0\)

⇒ \(\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy...

b) \(x^3-4x^2+4x=0\)

⇔ \(x^2(x-2)-2x(x-2)=0\)

⇔ \((x-2)(x^2-2x)=0\)

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

Vậy...

c) \(x^2-3x+2=0\)

⇔ \(x(x-2)-(x-2)=0\)

⇔ \((x-1)(x-2)=0\)

⇒ \(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy...

\(a,\Leftrightarrow\left(4x-8\right)\left(x+1\right)=0\\ \Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=-1\\ c,\Leftrightarrow x^2-2x-4x+8=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ d,\Leftrightarrow x^3-3x^2+3x-9x+2x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+x+2x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-2\end{matrix}\right.\)

a) \(\Rightarrow4\left(x+1\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

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

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

\(\Rightarrow x=-1\left(do.x^2+1\ge1>0\right)\)

c) \(\Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\)

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

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

d) \(\Rightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-1\end{matrix}\right.\)