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1) \(x^6+1\)
\(=x^6+x^4-x^4+x^2-x^2+1\)
\(=\left(x^6-x^4+x^2\right)+\left(x^4-x^2+1\right)\)
\(=x^2\left(x^4-x^2+1\right)+\left(x^4-x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
2) \(x^6-y^6\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
a) =(a-b-c +a-b+c)( a-b-c -a+b-c)
= 2(a-b)(-2c)= -4c(a-b)
làm tặng câu a) thui
\(\left(a-b-c\right)^2-\left(a-b+c\right)^2\)
\(=\left(a-b-c-a+b-c\right)\left(a-b-c+a-b+c\right)\)
\(=\left(-2c\right)\left(-2b+2a\right)\)
\(=2\left(a-b\right)\left(-2c\right)\)
\(=-4c\left(a-b\right)\)
a) Mình không hiểu đề cho lắm 
b) \(3x\left(x-1\right)^2-2x\left(x+3\right)\left(x-3\right)+4x\left(x-4\right)\)
\(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x\left(x-4\right)\)
\(=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)
\(=x^3-2x^2+5x\)
c) \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)
\(=2\left(2x+5\right)^2+3\left(4x+1\right)\left(4x-1\right)\)
\(=2\left(4x^2+20x+25\right)+3\left(16x^2-1\right)\)
\(=8x^2+40x+50+48x^2-3\)
\(=56x^2+40x+47\)
d) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)-\left(x^4-1\right)\)
\(=x^3-16x-x^4+1\)
e) \(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
\(=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\)
\(=y^4-81-y^4+4\)
\(=-77\)
a)
(x3 – 3x2+3x-1) – (x3+9) +(3x2-12) = 2
3x-22 = 2
3x =24
=> x= 8
b)
x3+6x2+12x+8-6x2-x3-x2+x2 = 5
12x+8 = 5
12x = -3
=>x = -1/4
a) ( x - 1 )3 - ( x + 3 )(x2 - 3x + 9 ) + 3( x2 - 4 ) = 2
(x3-3x2\(\times\)1+3x\(\times\)12-13)-(x3+33)+(3x2-3\(\times\)4)=2
x3-3x2+3x-1-x3-9+3x2-12=2
3x-40=2
3x=42
x=14
b ) ( x + 2 )3 - 6x2 - x2 ( x + 1 ) + x2 = 5
(x3+3\(\times\)2x2+3x\(\times\)22+23)-6x2-(x3+x2)+x2=5
x3+6x2+12x+8-6x2-x3-x2+x2=5
12x+8=5
12x=\(-\)3
x=\(-\frac{1}{4}\)
bài dễ mà 
Bài 1:
a) \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)
\(\Rightarrow x^3-3x^2+3x-1+2^3-x^3+3x^2+6x=17\)
\(\Rightarrow9x+7=17\)
\(\Rightarrow9x=17-7=10\)
\(\Rightarrow x=\dfrac{10}{9}\)
b) \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
\(\Rightarrow x^3+2^3-x^3+2x=15\)
\(\Rightarrow8+2x=15\)
\(\Rightarrow2x=15-8=7\)
\(\Rightarrow x=\dfrac{7}{2}\)
c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
\(\Rightarrow x^3-3x^2.3+3x.3^2-3^3-x^3+3^3+9\left(x^2+2x+1\right)=15\)
\(\Rightarrow-9x^2+27x+9x^2+18x+9=15\)
\(\Rightarrow45x+9=15\)
\(\Rightarrow45x=6\)
\(\Rightarrow x=\dfrac{6}{45}=\dfrac{2}{15}\)
d) \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
\(\Rightarrow x\left(x^2-5^2\right)-x^3-2^3=3\)
\(\Rightarrow x^3-25x-x^3-8=3\)
\(\Rightarrow-25x-8=3\)
\(\Rightarrow-25x=3+8=11\)
\(\Rightarrow x=-\dfrac{11}{25}\)
Bài 2:
a) Ta có:
\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\)
\(B=2^{16}-1\)
Vì 216 - 1 < 216
=> B < A
b) Ta có:
\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{128}-1\right)\)
Vì 1/2( 3128 - 1) < 3128 - 1
=> A < B
a3 + b3=(a+b)(a2-ab+b2)
(a + b)3 =a3+b3+3ab(a+b)
a2 + b2=a2+2ab+b2
SGK TOÁN 8 TẬP 1