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1) a^2 + b^2 + 2a - 2b - 2ab = (a^2 - 2ab + b^2) + (2a-2b) = (a-b)^2 + 2(a-b) = (a-b)(a-b+2)
2) 4a^2 - 4b^2 - 4a + 1 = ( 4a^2 - 4a +1) - 4b^2 = (2a-1)^2 - 4b^2 = (2a-1-2b)(2a-1+2b)
3) a^3+6a^2+12a+8= (a^3+8)+(6a^2+12a)= (a+2)(a^2-2a+4)+6a(a+2)=(a+2)(a^2-2a+4+6a)=(a+2)(a^2+4a+4)=(a+2)(a+2)^2=(a+2)^3
\(a,\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\) =\(a^2+b^2+c^2-2ab-2bc+2ac-b^2+2bc-c^2+2ab-2ac\) =\(a^2\) b)\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\) =\(\left(3x+1\right)^2-2\left(3x+3-2\right)\left(3x+3+2\right)+\left(3x+5\right)^2\) =\(\left(3x+1\right)^2-2\left(\left(3x+3\right)^2-4\right)+\left(3x+5\right)^2\) =\(9x^2+6x+1-18x^2-36x-9+8+9x^2+30x+25\) =25 c)\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)\) =\(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)....\left(2^{64}+1\right)\) =\(\left(2^2-1\right)\left(2^2+1\right)...\left(2^{64}+1\right)\) =... =\(\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1\) \)
d)Tương tự
\(a,\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)
=\(a^2+b^2+c^2-2ab-2bc+2ac-b^2+2bc-c^2+2ab-2ac\)
=\(a^2\)
b)\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
=\(\left(3x+1\right)^2-2\left(3x+3-2\right)\left(3x+3+2\right)+\left(3x+5\right)^2\)
=\(\left(3x+1\right)^2-2\left(\left(3x+3\right)^2-4\right)+\left(3x+5\right)^2\)
=\(9x^2+6x+1-18x^2-36x-9+8+9x^2+30x+25\)
=25
c)\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)\)
=\(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)....\left(2^{64}+1\right)\)
=\(\left(2^2-1\right)\left(2^2+1\right)...\left(2^{64}+1\right)\)
=...
=\(\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1\)
d)Tương tự
a: \(=\left[a-\left(b-c\right)\right]^2-\left(b-c\right)^2+2ab-2ac\)
\(=a^2-2a\left(b-c\right)+\left(b-c\right)^2-\left(b-c\right)^2+2ab-2ac\)
\(=a^2-2ab+2ac+2ab-2ac=a^2\)
b: \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left(3x+1-3x-5\right)^2\)
\(=\left(-4\right)^2=16\)
c: \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{64}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(=2^{128}-1\)
d: \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)
\(=\dfrac{3^{64}-1}{2}\)
a. Ta có: a > b
4a > 4b ( nhân cả 2 vế cho 4)
4a - 3 > 4b - 3 (cộng cả 2 vế cho -3)
b. Ta có: a > b
-2a < -2b ( nhân cả 2 vế cho -2)
1 - 2a < 1 - 2b (cộng cả 2 vế cho 1)
d. Ta có: a < b
-2a > -2b ( nhân cả 2 vế cho -2)
5 - 2a > 5 - 2b (cộng cả 2 vế cho 5)
b) \(\left(x+1\right)^3+\left(x-1\right)^3+x^3-3x\left(x+1\right)\left(x-1\right)\)
\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1+x^3-3x\left(x^2-1\right)\)
\(=3x^3+6x-3x^3+3x\)
\(=3x\)
d) \(100^2-99^2+98^2-97^2+...+2^2-1\)
\(=\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+..+\left(2+1\right)\left(2-1\right)\)
\(=100+99+98+97+..+2+1\)
\(=\frac{\left(100+1\right)\cdot100}{2}=5050\)
Làm trc cho 2 câu cuối
c) \(a^2-b^2-4a+4b\)
\(=\left(a+b\right)\left(a-b\right)-4\left(a-b\right)\)
\(=\left(a-b\right)\left[\left(a+b\right)-4\right]\)
d) \(a^2+2ab+b^2-2a-2b+1\)
\(=\left(a+b\right)^2-2\left(a+b\right)+1\)
\(=\left(a+b\right)\left[\left(a+b\right)-2\right]+1\)