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\((x+5)^2+4(x+5)(x-5)+4(x^2-10x+25)=0\\\Rightarrow(x+5)^2+4(x+5)(x-5)+4(x^2-2\cdot x\cdot5+5^2)=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+4(x-5)^2=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+[2(x-5)]^2=0\\\Rightarrow[(x+5)+2(x-5)]^2=0\\\Rightarrow(x+5+2x-10)^2=0\\\Rightarrow(3x-5)^2=0\\\Rightarrow3x-5=0\\\Rightarrow3x=5\\\Rightarrow x=\frac53\\\text{#}Toru\)
=(x^2+8x)^2+23(x^2+8x)+135
Cái này ko phân tích được nha bạn
x3+27+(x+3)(x+9)
= (x+3)(x2-3x+9)+(x+3)(x+9)
= (x+3)(x2-3x+9+x+9)
=(x+3)(x2-2x+18)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\\ =\left(x+3\right)\left(x^2-3x+9+x-9\right)\\ =\left(x+3\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x+3\right)\)
`(x+3)^4+(x+5)^4-2`
`={[(x+3)^2]^2-1^2}+{[(x+5)^2]^2 -1^2}`
`=[(x+3)^2-1^2][(x+3)^2+1]+[(x+5)^2-1^2][(x+5)^2+1]`
`=(x+3-1)(x+3+1)[(x+3)^2+1]+(x+5-1)(x+5+1)[(x+5)^2+1]`
`=(x+2)(x+4)[(x+3)^2+1]+(x+4)(x+6)[(x+5)^2+1]`
`=(x+4){(x+2)[(x+3)^2+1]+(x+6)[(x+5)^2+1]}`
`=(x+4)(2x^3+24x^2+108x+176)`
Bạn gì ơi hình như phải ra \(2\left(t+4\right)^2\left(x^2+8x+22\right)\)chứ nhỉ???
phân tích đa thức thành nhân tử \(x^2\cdot\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
\(x^2\cdot\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\)
\(\left(x+2\right)^2-\left(x-2\right)^2\)
\(=\left(x+2-x+2\right)\left(x+2+x-2\right)\)
\(=4.2x\)
\(=8x\)
\(=\left[\left(x+2\right)-\left(x-2\right)\right]\cdot\left[\left(x+2\right)+\left(x-2\right)\right]\)
\(2\left(x^2+x+1\right)^2-\left(2x+1\right)^2-\left(x^2+2x\right)^2\)
\(=2.\left[x^4+x^2+1+2x^3+2x+2x^2\right]-\left(4x^2+4x+1\right)-\left(x^4+4x^3+4x^2\right)\)
\(=x^4-2x^2+1=\left(x^2-1\right)^2=\left(x-1\right)^2\left(x+1\right)^2\)
Chúc bạn học tốt.
\(=\left[\left(x+2\right)+\left(x-2\right)\right]\left[\left(x+2\right)-\left(x-2\right)\right]\)
\(=2x4=8x\)
\(\left(x+2\right)^2-\left(x-2\right)^2\)
\(=x^2+4x+4-x^2+4x-4\)
\(=8x\)