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câu c:x^4-2x^3-x^2+x^3-2x^2-x+5x^2-10x-5=x^2(x^2-2x-1)+x(x^2-2x-1)+5(x^2-2x-1)=(x^2-2x-1)(x^2+x+5)
Cô hướng dẫn nhé.
1. Nhẩm nghiệm để suy ra nhân tử .
\(27x^3-27x^2+18x-4=27x^3-9x^2-18x^2+6x+12x-4\)
\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)
Xem lại đề câu b, nếu ko ta dùng công thức Cardano.
2.
a. Đặt ẩn phụ.
b. \(B=\left(x+y\right)^2-\left(x+y\right)-12\). Sau đó lại đặt ẩn phụ.
c. Đặt \(x^2+x+1=t\)
d. Ghép: \(\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)+24=\left(x^2+7x+10\right)\left(x^2+7x+12\right)+24\)
Đặt \(x^2+7x+10=t\)
2a. Đặt \(x^2+x=t\Rightarrow A=t^2-2t-15=t^2-5t+3t-15=\left(t-5\right)\left(t+3\right)\)
Quay lại biến x , ta có \(\left(x^2+x-5\right)\left(x^2+x+3\right)\)
Theo đề bài ta có :
\(\frac{x\left(3-x\right)}{x+1}\cdot\left(x+\frac{\left(3-x\right)}{x+1}\right)=2\)
=> \(\frac{\left(3x-x^2\right)}{x+1}\cdot\frac{\left(3-x+x^2+x\right)}{x+1}=2\)
=> \(\left(3x-x^2\right)\left(x^2+3\right)=2\left(x+1\right)^2\)
=> \(3x^3+9x-x^4-3x^2=2x^2+4x+2\)
=> \(3x^3+\left(9x-4x\right)+\left(-3x^2-2x^2\right)-x^4-2=0\)
=> \(3x^3+5x-5x^2-x^4-2=0\)
=> \(5x\left(1-x\right)+x^3\left(1-x\right)+2\left(x^3-1\right)=0\)
=> \(5x\left(1-x\right)+x^3\left(1-x\right)+2\left(x-1\right)\left(x^2+x+1\right)=0\)
=> \(5x\left(1-x\right)+x^3\left(1-x\right)-2\left(1-x\right)\left(x^2+x+1\right)=0\)
=> \(\left(1-x\right)\left(5x+x^3-2x^2-2x-2\right)=0\)
=> \(\left(1-x\right)\left(3x+x^3-2x^2-2\right)=0\)
=> \(\left(1-x\right)\left(x^3-x^2-x^2+x+2x-2\right)=0\)
=> \(\left(1-x\right)\left(x^2\left(x-1\right)-x\left(x-1\right)+2\left(x-1\right)\right)=0\)
=> \(\left(1-x\right)\left(x-1\right)\left(x^2-x+2\right)=0\)
Ta Thấy :
\(\left(x^2-x+2\right)=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}>0\)
=> \(\hept{\begin{cases}1-x=0\\x-1=0\end{cases}}\)
=> x = 1
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^6-x^4-9x^3+9x^2\)
\(=x^4\left(x^2-1\right)-9x^2\left(x-1\right)\)
\(=x^4\left(x-1\right)\left(x+1\right)-9x^2\left(x-1\right)\)
\(=x^2\left(x-1\right)\left[x^2\left(x+1\right)-9\right]\)
\(=x^2\left(x-1\right)\left(x^3+x^2-9\right)\)
\(x^4-4x^3+8x^2-16x+16\)
\(=\left(x^2+4\right)^2-4x\left(x^2+4\right)\)
\(=\left(x^2+4\right)\left(x^2+4-4x\right)\)
\(=\left(x^2+4\right)\left(x-2\right)^2\)
\(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2-4c^2\right)\)
\(=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]\)
\(=3\left(a-b+2c\right)\left(a-b-2c\right)\)
`@` `\text {Ans}`
`\downarrow`
`4x^3 - 4x^2 - 9x + 9`
`= (4x^3 - 4x^2) - (9x - 9)`
`= 4x^2(x - 1) - 9(x - 1)`
`= (4x^2 - 9)(x - 1)`
____
`x^3 + 6x^2 + 11x + 6`
`= x^3 + x^2 + 5x^2 + 5x + 6x + 6`
`= (x^3 + x^2) + (5x^2 + 5x) + (6x + 6)`
`= x^2*(x + 1) + 5x(x + 1) + 6(x + 1)`
`= (x^2 + 5x + 6)(x+1)`
____
`x^2y - x^3 - 9y + 9x`
`= (x^2y - 9y) - (x^3 - 9x)`
`= y(x^2 - 9) - x(x^2 - 9)`
`= (y - x)(x^2 - 9)`
b: =x^3+x^2+5x^2+5x+6x+6
=(x+1)(x^2+5x+6)
=(x+1)(x+2)(x+3)
c: =x^2(y-x)-9(y-x)
=(y-x)(x^2-9)
=(y-x)(x-3)(x+3)
a: =(4x^3-4x^2)-(9x-9)
=4x^2(x-1)-9(x-1)
=(x-1)(4x^2-9)
=(x-1)(2x-3)(2x+3)