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=(x^2+8x)^2+23(x^2+8x)+135
Cái này ko phân tích được nha bạn
\(x^2-8x+15\)
\(=x^2-3x-5x+15\)
\(=x\left(x-3\right)-5 \left(x-3\right)\)
\(=\left(x-5\right)\left(x-3\right)\)
\(x^2-8x+15\)
\(=\left(x^2-3x\right)-\left(5x-15\right)\)
\(=x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(x-5\right)\)
Tham khảo nhé~
x^2-8x+7= x^2-x-7x-7=x(x-1)-7(x-1)=(x-7)(x-1)
1) \(\left(x^2+8x+7\right).\left(x+3\right).\left(x+5\right)+15\)
\(=\left(x^2+8x+7\right).\left(x^2+5x+3x+15\right)+15\)
\(=\left(x^2+8x+7\right).\left(x^2+8x+15\right)+15\)
Ta đặt: \(x^2+8x+7=n\)
\(=n.\left(n+8\right)+15\)
\(=n^2+8n+15\)
\(=n^2+3n+5n+15\)
\(=\left(n^2+3n\right)+\left(5n+15\right)\)
\(=n.\left(n+3\right)+5.\left(n+3\right)\)
\(=\left(n+3\right).\left(n+5\right)\)
\(=\left(x^2+8x+7+3\right).\left(x^2+8x+7+5\right)\)
\(=\left(x^2+8x+10\right).\left(x^2+8x+12\right)\)
\(=\left(x^2+8x+10\right).\left(x^2+2x+6x+12\right)\)
\(=\left(x^2+8x+10\right).[x.\left(x+2\right)+6.\left(x+2\right)]\)
\(=\left(x^2+8x+10\right).\left(x+2\right).\left(x+6\right)\)
2) \(x^2-2xy+3x-3y-10+y^2\)
\(=\left(x-y\right)^2+3.\left(x-y\right)-10\)
Ta đặt: \(x-y=n\)
\(=n^2+3n-10\)
\(=n^2-2n+5n-10\)
\(=\left(n^2-2n\right)+\left(5n-10\right)\)
\(=n.\left(n-2\right)+5.\left(n-2\right)\)
\(=\left(n-2\right).\left(n+5\right)\)
\(=\left(x-y-2\right).\left(x-y+5\right)\)
x4 - 4x3 - 8x2 + 8x
= x(x3 - 4x2 - 8x + 8)
= x[x3 + 8 - 4x(x + 2)]
= x[(x + 2)(x2 - 2x + 4) - 4x(x + 2)]
= x(x + 2)(x2 - 6x + 4)
= x(x + 2)(x2 - 6x + 9 - 5)
= \(x\left(x+2\right)\left[\left(x-3\right)^2-5\right]=x\left(x+2\right)\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)\)
\(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left(x^3-6x^2+2x^2+4x-12x+8\right)\)
\(=x\left[\left(x^3-6x^2+4x\right)+\left(2x^2-12x+8\right)\right]\)
\(=x\left[x\left(x^2-6x+4\right)+2\left(x^2-6x+4\right)\right]\)
\(=x\left(x^2-6x+4\right)\left(x+2\right)\)
\(=x\left[\left(x-3\right)^2-\left(\sqrt{5}\right)^2\right]\left(x+2\right)\)
\(=x\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\left(x+2\right)\)
\(=\left(x+4\right)^2-1=\left(x+4-1\right)\left(x+4+1\right)=\left(x+3\right)\left(x+5\right)\)
\(x^2+8x+15=x\left(x+3\right)+5\left(x+3\right)=\left(x+3\right)\left(x+5\right)\)
\(x^2-8x+12=\left(x^2-6x\right)-\left(2x-12\right)=x\left(x-6\right)-2\left(x-6\right)=\left(x-2\right)\left(x-6\right)\)
\(=x^2\left(x-1\right)-4\left(x-1\right)^2=\left(x-1\right)\left[x^2-4\left(x-1\right)\right]\\ =\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)
x³ - 3x²y + 3xy² - y³ - z³
= (x³ - 3x²y + 3xy² - y³) - z³
= (x - y)³ - z³
= (x - y - z)[(x - y)² + (x - y)z + z²]
= (x - y - z)(x² - 2xy + y² + xz - yz + z³)
--------------------
x² - y² + 8x + 6y + 7
= (x² + 8x + 16) - (y² - 6y + 9)
= (x + 4)² - (y - 3)²
= (x + 4 - y + 3)(x + 4 + y - 3)
= (x - y + 7)(x + y + 1)
a: \(=\left(x^3-3x^2y+3xy^2-y^3\right)-z^3\)
\(=\left(x-y\right)^3-z^3\)
\(=\left(x-y-z\right)\left[\left(x-y\right)^2+z\left(x-y\right)+z^2\right]\)
\(=\left(x-y-z\right)\left(x^2-2xy+y^2+xz-yz+z^2\right)\)
b: \(=x^2+8x+16-y^2+6y-9\)
=(x+4)^2-(y-3)^2
=(x+4+y-3)(x+4-y+3)
=(x+y+1)(x-y+7)
( x2 + 8x + 7 ) ( x2 + 8x + 15 ) + 15
Đặt x2 + 8x + 7 = y ta có:
y ( y + 8 ) + 15
= y2 + 8y + 15
= ( y + 3 ) ( y + 5 )
= ( x2 + 8x + 10 ) ( x2 + 8x + 12 )
= ( x2 + 8x + 10 ) ( x + 2 ) ( x + 6 )
Đặt x2 + 8x + 7 = y ta có:
y ( y + 8 ) + 15
= y2 + 8y + 15
= ( y + 3 ) ( y + 5 )
= ( x2 + 8x + 10 ) ( x2 + 8x + 12 )
= ( x2 + 8x + 10 ) ( x + 2 ) ( x + 6 )