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\(9\left(x+1\right)^2-\left(3x-2\right)^2\)
\(=9\left(x^2+2x+1\right)-\left(9x^2-12x+4\right)\)
\(=9x^2+18x+9-9x^2+12x-4\)
\(=30x+5\)
\(=5\left(6x+1\right)\)
\(9\left(x+1\right)^2-\left(3x-2\right)^2\)
\(=\left[3\left(x+1\right)+3x-2\right]\left[3\left(x+1\right)-3x+2\right]\)
\(=\left(3x+3+3x-2\right)\left(3x+3-3x+2\right)\)
\(=5\left(6x+1\right)\)
\(=x^4-16x^2+100=x^4+20x^2+100-36x^2\)
\(=\left(x^2+10\right)^2-\left(6x\right)^2=\left(x^2-6x+10\right)\left(x^2+6x+10\right)\)
\(\left(x^2-8\right)^2+36\)
\(=x^4-16x^2+64+36\)
\(=x^4-16x^2+100\)
\(=x^4+20x^2+100-36x^2\)
\(=\left(x^2+10\right)^2-\left(6x\right)^2\)
\(=\left(x^2-6x+10\right)\left(x^2+6x+10\right)\)
Ta có (x^2 + y^2 )^3 + (z^2 – x^2 )^3 – (y^2 + z^2 )^3
= (x^2 + y^2 )^3 + (z^2 – x^2 )^3 + (-y^2 - z^2 )^3
Ta thấy x^2 + y^2 + z^2 – x^2 – y^2 – z^2 = 0
=> áp dụng nhận xét ta có: (x^2+y^2 )^3+ (z^2 -x^2 )^3 -y^2 -z^2 )^3
= 3(x^2 + y^2 ) (z^2 –x^2 ) (-y^2 – z^2 )
= 3(x^2+y^2 ) (x+z)(x-z)(y^2+z^2 )
PTĐTTNT ??? :)) bn phân tích rồi đấy, đề là tìm x thôi
Giải ( suỵt :), đừng ai nhìn thấy ... :v
\(\left(2x-10\right)\left(x+10\right)\left(x+\sqrt{3}\right)=0\)
TH1 : \(2x-10=0\Leftrightarrow x=5\)
TH2 : \(x+10=0\Leftrightarrow x=-10\)
TH3 : \(x+\sqrt{3}=0\Leftrightarrow x=-\sqrt{3}\)( vô lí )
Vậy x = {5;-10}
(x -y)3 - 1 - 3(x -y)(x - y - 1)
= (x -y)3 - 3(x -y)(x - y - 1) - 1
Đặt x - y = t, khi đó ta có:
t3 - 3t. (t - 1) - 1
= t3 - 3t2 + 3t - 1
= (t - 1)3
Thay t = x - y vào (t - 1)3 , ta có: ( x - y - 1)3
Vậy (x -y)3 - 1 - 3(x -y)(x - y - 1) = ( x - y - 1)3
\(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
\(6x^4-11x^2+3=6x^4-9x^2-2x^2+3\)
\(=3x^2\left(2x^2-3\right)-\left(2x^2-3\right)=\left(2x^2-3\right)\left(3x^2-1\right)\)
\(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
Chúc bạn học tốt!!!
\(\left(x^2+6x\right)\left(x^2+14x+40\right)+128\)
\(=\left(x^2+6x\right)\left(x^2+14x+40\right)\)
\(=x^4+20x^3+124x^2+240x\)
\(=x^4+20x^3+124x^2+240x+128\)
\(\left(x^2+6x\right)\left(x^2+14x+40\right)+128\)
\(=x^4+14x^3+40x^2+6x^3+84x^2+240x+128\)
\(=x^4+20x^3+124x^2+240x+128\)
mk chỉ biết đến đây thôi