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\(4x^4-16-4x^2-16x\)
\(=4x^2\left(x^2-1\right)-16\left(1+x\right)\)
\(=4x^2\left(x+1\right)\left(x-1\right)-16\left(x+1\right)\)
\(=\left(x+1\right)\left[4x^2\left(x-1\right)-16\right]\)
\(=\left(x+1\right)4\left[x^2\left(x-1\right)-4\right]\)
Nguyễn Văn Tuấn AnhNs r, không biết thì not làm
\(4x^4-16-4x^2-16x\)
\(=4x^2\left(x^2-1\right)-16\left(x+1\right)\)
\(=4x^2\left(x-1\right)\left(x+1\right)-16\left(x+1\right)\)
\(=\left(x+1\right)\left[4x^2\left(x-1\right)-16\right]\)
\(=4\left(x+1\right)\left[x^2\left(x-1\right)-4\right]\)
\(=4\left(x+1\right)\left[x^3-x^2-4\right]\)
\(=4\left(x+1\right)\left[x^3+x^2+2x-2x^2-2x-4\right]\)
\(=4\left(x+1\right)\left[x\left(x^2+x+2\right)-2\left(x^2+x+2\right)\right]\)
\(=4\left(x+1\right)\left(x-2\right)\left(x^2+x+2\right)\)
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Lời giải:
$(4x-2)^2-16x+16x=(4x-2)^2$
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$(3x-4)(3x+4)-(20x^2y-15xy^2):(5xy)$
$=(3x-4)(3x+4)-(4x-3y)$ không phân tích được thành nhân tử.
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$(x-2)(3x^2+6x+12)-(120^2x^2y^2):(60xy^2)$
$=3(x-2)(x^2+2x+4)-240x$
$=3(x^3-2^3)-240x=3x^3-240x-24$
$=3(x^3-80x-8)$
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https://h7.net/hoi-dap/toan-8/phan-h-da-thuc-x-4-16-thanh-nhan-tu-faq324398.html
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Sửa đề: \(8x^2-16x+8-32y^2\)
\(=8\left(x^2-2x+1-4y^2\right)\)
\(=8\left[\left(x^2-2x+1\right)-4y^2\right]\)
\(=8\left[\left(x-1\right)^2-\left(2y\right)^2\right]\)
\(=8\left(x-1-2y\right)\left(x-1+2y\right)\)
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16 - 5x2 - 3
= -5x2 + 15x + x - 3
= -5x(x - 3) + (x - 3)
= (1 - 5x)(x - 3)
16x-5x^2-3=-5x^2+15x+x-3=5x(3-x)-(3-x) =(3-x)(5x-1) (xy+1)^2-x^2-2xy-y^2=(xy+1)^2-(x^2+2xy+y^2) =(xy+1)^2-(x+y)^2 =(xy+1-x-y)(xy+1+x+y) =[(xy-x)-(y-1)][(xy+x)+(y+1)] =[x(y-1)-(y-1)][x(y+1)+(y+1)] =(y-1)(x-1)(y+1)(x+1)
\(=\left(4x^2-9\right)^2=\left(2x-3\right)^2\left(2x+3\right)^2\)
\(\left(4x^2-9\right)^2=\left(2x-3\right)^2\left(2x+3\right)^2\)