Viết về dạng tích
4+4+4+4-4+4+4
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\(a,=5^{40}\cdot5^{12}=5^{52}\\ b,=x^{7+4+3}=x^{14}\\ c,=\left(3\cdot4\right)^6=12^6\)
\(\begin{array}{l}{3^3}{.3^4} = {3^{3 + 4}} = {3^7};\\{10^4}{.10^3} = 10^{4+3}= {10^7};\\{x^2}.{x^5} = x^{2+5} = {x^7}.\end{array}\
\(a,\left(-7\right)^6\)
\(b,\left(-4\right)^3.\left(-5\right)^3=\left[\left(-4\right).\left(-5\right)\right]^3=20^3\)
\(\frac{4}{9}x^4-16x^2=\left(\frac{2}{3}x^2\right)^2-\left(4x\right)^2=\left(\frac{2}{3}x^2+4x\right).\left(\frac{2}{3}x^2-4x\right)\)
a, \(x^2-6x+9=\left(x-3\right)^2\)
b, \(x^2-12x+36=\left(x-4\right)^2\)
c, \(9x^2-25=\left(3x-5\right)\left(3x+5\right)\)
d, \(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)
e, \(x^4-8x^2+16=\left(x^2-4\right)^2=\left[\left(x-2\right)\left(x+2\right)\right]^2\)
f, \(x^4-81=\left(x^2-9\right)\left(x^2+9\right)=\left(x-3\right)\left(x+3\right)\left(x^2+9\right)\)
g, \(\left(4x+5\right)^2-\left(5x+4\right)^2=\left(4x+5-5x-4\right)\left(4x+5+5x+4\right)=9\left(1-x\right)\left(x+1\right)\)
h, \(\left(2x-3\right)^2-2\left(2x-3\right)\left(x+2\right)+\left(-x-2\right)^2\)
\(=\left(2x-3\right)^2-2\left(2x-3\right)\left(x+2\right)+\left(x+2\right)^2\)
\(=\left(2x-3-x-2\right)^2=\left(x-5\right)^2\)
4 + 4 + 4 + 4 - 4 + 4 + 4
= 4 + 4 + 4 + 4 + 4
= 5 × 4
Bằng 31