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1/ = x4 + 2x3 + 4x2 + 3x - 10 = (x4 - x3) + (3x3 - 3x2) + (7x2 - 7x) + (10x - 10)
= (x - 1)(x3 + 3x2 + 7x + 10) = (x - 1)[(x3 + 2x2) + (x2 + 2x) + (5x + 10)]
= (x - 1)(x + 2)(x2 + x + 5)
2/ = (x5 - 2x4) + (x4 - 2x3) + (x3 - 2x2) + (x2 - 2x) + (x - 2) = (x - 2)(x4 + x3 + x2 + x + 1)
\((x+5)^2+4(x+5)(x-5)+4(x^2-10x+25)=0\\\Rightarrow(x+5)^2+4(x+5)(x-5)+4(x^2-2\cdot x\cdot5+5^2)=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+4(x-5)^2=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+[2(x-5)]^2=0\\\Rightarrow[(x+5)+2(x-5)]^2=0\\\Rightarrow(x+5+2x-10)^2=0\\\Rightarrow(3x-5)^2=0\\\Rightarrow3x-5=0\\\Rightarrow3x=5\\\Rightarrow x=\frac53\\\text{#}Toru\)
phân tích đa thức thành nhân tử \(x^2\cdot\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
\(x^2\cdot\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\)
\(S=x^6-8\)
\(S=\left(x^2\right)^3-2^3\)
\(S=\left(x^2-2\right)\left(x^4+2x^2+4\right)\)
⇒ Chọn C
\(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x+4\right)^2\left(x^2-1\right)-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\)
\(=\left(x-1\right)\left(x+1\right)\left(x+4+1\right)\left(x+4-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+5\right)\left(x-3\right)\)
=.= hok tốt!!
\(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\)\(\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\)\(\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\)\(\left(3x-2\right)\left(3x-6\right)\)
\(=\)\(3\left(x-2\right)\left(3x-2\right)\)
Chúc bạn học tốt ~
ta có: \(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x+4\right)^2.\left(x^2-1\right)-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(\left(x+4\right)^2-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)
Cho mình nhé hihi!!!
x2(x+4)2-(x+4)2-(x2-1)
=(x+4)2 (x2-1)-(x2-1)
=(x2-1)(x2+8x+16-1)
=(x-1)(x+1)(x2+8x+15)