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\(\Leftrightarrow\left(3x-1\right)^2-4^2=0\)
\(\Leftrightarrow\left(3x-1-4\right)\left(3x-1+4\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\Leftrightarrow\orbr{\begin{cases}3x-5=0\\3x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}\)
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}
Câu a :
\(\left(x-5\right)^2+\left(x-5\right)\left(x+5\right)-\left(5-x\right)\left(2x+1\right)\)
\(=x^2-10x+25+x^2-25-10x-5+2x^2+x\)
\(=4x^2-19x-5\)
Câu b :
\(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=12x^2-9x-8x+6-2x+2+3x^2-3x-6x^2-6x+4x+4\)
\(=9x^2-24x+2\)
(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
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 )
\(9\left(x+1\right)^2-\left(3x-2\right)^2\)
\(=9\left(x^2+2x+1\right)-\left(9x^2-12x+4\right)\)
\(=9x^2+18x+9-9x^2+12x-4\)
\(=30x+5\)
\(=5\left(6x+1\right)\)
\(9\left(x+1\right)^2-\left(3x-2\right)^2\)
\(=\left[3\left(x+1\right)+3x-2\right]\left[3\left(x+1\right)-3x+2\right]\)
\(=\left(3x+3+3x-2\right)\left(3x+3-3x+2\right)\)
\(=5\left(6x+1\right)\)
\(\left(x^2+6x\right)\left(x^2+14x+40\right)+128\)
\(=\left(x^2+6x\right)\left(x^2+14x+40\right)\)
\(=x^4+20x^3+124x^2+240x\)
\(=x^4+20x^3+124x^2+240x+128\)
\(\left(x^2+6x\right)\left(x^2+14x+40\right)+128\)
\(=x^4+14x^3+40x^2+6x^3+84x^2+240x+128\)
\(=x^4+20x^3+124x^2+240x+128\)
mk chỉ biết đến đây thôi
( x + 2 ) ( x + 3 ) ( x + 4 ) ( x + 5 ) - 24
= ( x2 + 7x + 10 ) ( x2 + 7x + 12 ) - 24
Đặt x2 + 7x + 10 = y
Ta có :
y2 + 2y - 24 = ( y - 4 ) ( y + 6 ) = ( x2 + 7x + 6 ) ( x2 + 7x + 16 )
= ( x + 1 ) ( x + 6 ) ( x2 + 7x + 16 )
Đặt x2+7x+10=t
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=t\left(t+2\right)-24=t^2+2t-24\)
\(=\left(t^2+2t+1\right)-25=\left(t+1\right)^2-5^2=\left(t-4\right)\left(t+6\right)\)=(x2+7x+6)(x2+7x+16)
=(x2+x+6x+6)(x2+7x+16)=[x(x+1)+6(x+1)](x2+7x+16)=(x+1)(x+6)(x2+7x+16)