chọn ý B nha 

\(2x\left(x-3\right)+5\left(x-3\right)=0\)

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

Chọn ( B )

\(A=\left(x-4\right)^2-\left(x+4\right)^2-16\left(x-2\right)\)

\(=x^2-8x+16-x^2-8x-16-16x+32\)

\(=-32x+32\)

Biểu thức phụ thuộc vào giá trị của biến

b) \(\left(x-3\right)^3-\left(x+3\right)^3+12\left(x+1\right)\left(x-1\right)\)

\(=\left(x^3-9x^2+27x-27\right)-\left(x^3+9x^2+27x+27\right)+12x^2-12\)

\(=-6x^2-66\)

Biểu thức này phụ thuộc vào giá trị của biến

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

\(\Leftrightarrow\left[\left(3x+5\right)-\left(2x+3\right)\right]^2=\left(x+2\right)^3\)

\(\Leftrightarrow\left(3x+5-2x-3\right)^2=\left(x+2\right)^3\)

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

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

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

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

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

Vậy tập nghiệm của phương trình là: \(S=\left\{-2;-1\right\}\)

a) x³ - 7x + 6 = 0

x³ - x - 6x + 6 = 0

(x³ - x) - (6x - 6) = 0

x(x² - 1) - 6(x - 1) = 0

x(x - 1)(x + 1) - 6(x - 1) = 0

(x - 1)[x(x + 1) - 6] = 0

(x - 1)(x² + x - 6) = 0

(x - 1)(x² - 2x + 3x - 6) = 0

(x - 1)[(x² - 2x) + (3x - 6)] = 0

(x - 1)[x(x - 2) + 3(x - 2)] = 0

(x - 1)(x - 2)(x + 3) = 0

x - 1 = 0 hoặc x - 2 = 0 hoăkc x + 3 = 0

*) x - 1 = 0

x = 1

*) x - 2 = 0

x = 2

*) x + 3 = 0

x = -3

Vậy x = -3; x = 1; x = 2

a: \(x^3-7x+6=0\)

=>\(x^3-x-6x+6=0\)

=>\(x\left(x^2-1\right)-6\left(x-1\right)=0\)

=>x(x-1)(x+1)-6(x-1)=0

=>(x-1)(x^2+x-6)=0

=>(x-1)(x+3)(x-2)=0

=>\(\left[\begin{array}{l}x-1=0\\ x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-3\\ x=2\end{array}\right.\)

b: \(x^4+4x^2-5=0\)

=>\(x^4+5x^2-x^2-5=0\)

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

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

=>\(x^2=1\)

=>\(\left[\begin{array}{l}x=1\\ x=-1\end{array}\right.\)

c: \(x^4+x^3-x^2-x=0\)

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

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

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

=>\(\left[\begin{array}{l}x=0\\ x+1=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-1\\ x=1\end{array}\right.\)

d: \(x^2+6x-x-6=0\)

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

=>(x+6)(x-1)=0

=>\(\left[\begin{array}{l}x+6=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-6\\ x=1\end{array}\right.\)

e: \(x^2-4x+5x-20=0\)

=>x(x-4)+5(x-4)=0

=>(x-4)(x+5)=0

=>\(\left[\begin{array}{l}x-4=0\\ x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-5\end{array}\right.\)

f: \(x^2-10x+2x-20=0\)

=>x(x-10)+2(x-10)=0

=>(x-10)(x+2)=0

=>\(\left[\begin{array}{l}x-10=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=10\\ x=-2\end{array}\right.\)

g: \(x^4-x^3-x^2+1=0\)

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

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

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

TH1: x-1=0

=>x=1

TH2: \(x^3-x-1=0\)

=>x≃1,32

h: \(x^5+x^4+x^3+x^2+x+1=0\)

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

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

mà \(x^2+x+1=\left(x+\frac12\right)^2+\frac34\ge\frac34>0\forall x\)

nên \(x^3+1=0\)

=>\(x^3=-1\)

=>x=-1

i: \(x^2-9+\left(x+3\right)\left(3x-5\right)=0\)

=>(x-3)(x+3)+(x+3)(3x-5)=0

=>(x+3)(x-3+3x-5)=0

=>(x+3)(4x-8)=0

=>4(x+3)(x-2)=0

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

=>\(\left[\begin{array}{l}x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=2\end{array}\right.\)

j: \(64x^2-9+8x+3=0\)

=>(8x+3)(8x-3)+(8x+3)=0

=>(8x+3)(8x-3+1)=0

=>(8x+3)(8x-2)=0

=>\(\left[\begin{array}{l}8x+3=0\\ 8x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac38\\ x=\frac28=\frac14\end{array}\right.\)

ANH HAY CHỊ ƠI LÀM GIÚP EM BAI LỚP 7 ĐI O DUOI DAY A

a) \(\left(x-3\right)^2-4=0\)

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

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

\(\Rightarrow x-3=2\)hoặc \(\left(x-3\right)=-2\)

\(\Rightarrow\hept{\begin{cases}x-3=2\\x-3=-2\end{cases}\Rightarrow\hept{\begin{cases}x=5\\x=-1\end{cases}}}\)

Vậy \(x\in\left\{5;-1\right\}\)

b) \(x^2-2x=24\)

\(\Rightarrow x.\left(x+2\right)=24\)

\(\Rightarrow x.\left(x+2\right)=4.6\)

\(\Rightarrow x=4\)

Vậy \(x=4\)