3a)

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

\(=\left(x-3\right)^3=\left(1-3\right)^3=-8\)

1a) \(36a^4-y^2\)

\(=\left(6a^2\right)^2-y^2\)

\(=\left(6a^2+y\right)\left(6a^2-y\right)\)

4x² - 6x

= 2x . (2x - 3)

Trả lời:

a, \(\left(xy+4\right)^2-4\left(x+y\right)^2\)

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

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

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

\(=\left[\left(xy-2x\right)-\left(2y-4\right)\right]\left[\left(xy+2x\right)+\left(2y+4\right)\right]\)

\(=\left[x\left(y-2\right)-2\left(y-2\right)\right]\left[x\left(y+2\right)+2\left(y+2\right)\right]\)

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

b, \(2x-\sqrt{x}=2.\sqrt{x}.\sqrt{x}-\sqrt{x}=\sqrt{x}.\left(2\sqrt{x}-1\right)\)

Ta chứng minh C chia hết cho 11 và 17 vì 11 và 17 nguyên tố cùng nhau

C chia hết cho 11 vì \(C=\left(16^n-5^n\right)+\left(12^n-1^n\right)⋮11+11⋮11\)

C chia hết cho 17 vì \(C=\left(16^n-1^n\right)+\left(12^n-5^n\right)⋮17+17⋮17\)

Ta có đpcm

cảm ơn bạn nha

a) 4x³-64x=4x.(x²-4²)=4x.(x-4).(x+4)

b)x³-27+(3-x).(9-4x)=x³-27+27-12x-9x+4x²=x³-21x+4x²=x.(x²-21+4x)

Bạn xem trong hình nhé. Nếu không thấy hình thì nhắn cho mình.

=>    

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=>   \(8x^2-6x+4x-3=0\)

=>  \(\left(8x^2+4x\right)-\left(6x+3\right)=0\)

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

=>  \(\orbr{\begin{cases}2x+1=0\\4x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{4}\end{cases}}}\)

Trả lời:

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

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

\(\Leftrightarrow\left[\left(3x-1\right)-\left(x-2\right)\right]\left[\left(3x-1\right)+\left(x-2\right)\right]=0\)

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

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

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

Vậy x = - 1/2; x = 3/4 là nghiệm của pt.

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

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

\(=\left(x^2-4x^2+3x^2\right)+\left(2x+4x\right)+\left(1-1-12\right)=6x-12=-6\)

Trả lời:

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

\(\Leftrightarrow3\left(3x^2+2x-1\right)=0\)

\(\Leftrightarrow3x^2+2x-1=0\)

\(\Leftrightarrow3x^2+3x-x-1=0\)

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

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

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

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

Vậy x = - 1; x = 1/3 là nghiệm của pt.