5x3 +9 = ...
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\(\left\{5.\left[42:\left(5.3+50:10\right).3:2+4\right]\right\}+\left\{35.\left[63:9.\left(5^2\right)\right]\right\}\\ =\left\{5.\left[42:\left(15+5\right).1,5+4\right]\right\}+\left\{35.\left(9.25\right)\right\}\\ =\left\{5.\left[42:20.1,5+4\right]\right\}+\left\{35.225\right\}\\ =\left\{5.\left[2,1.1,5+4\right]\right\}+7875\\ =\left\{5.\left[3,15+4\right]\right\}+7875\\ =\left\{5.7,15\right\}+7875\\ =35,75+7875\\ =7910,75\)
{5.[42:(5.3+50:10).3:2+4]}+{35.[63:9.(52)]}={5.[42:(15+5).1,5+4]}+{35.(9.25)}={5.[42:20.1,5+4]}+{35.225}={5.[2,1.1,5+4]}+7875={5.[3,15+4]}+7875={5.7,15}+7875=35,75+7875=7910,75
P = \(\frac{5.3^{11}+4.3^{12}}{3^9.5^2-3^9.2^2}\)
= \(\frac{\left(5+4.3\right).3^{11}}{3^9.\left(5^2-2^2\right)}\)
=\(\frac{17.3^{11}}{3^9.21}\)
= \(\frac{17.3^2}{7.3}\)
= \(\frac{17.3}{7}=\frac{51}{7}\)
1) x³ + 2x² + x
= x(x² + 2x + 1)
= x(x + 1)²
2) 5x³ - 10x² + 5x
= 5x(x² - 2x + 1)
= 5x(x - 1)²
3) 8x²y - 8xy + 2x
= 2x(4xy - 4y + 1)
5) 2x² + 5x³ + x²y
= x²(2 + 5x + y)
6) 4x²y - 8xy² + 18x²y²
= 2xy(2x - 4y + 9xy)
\(A=5x^3-7x^2+3x^3-4x^2+x^2-x^3+5x-1=7x^3-10x^2+5x-1\)
\(B=5x^3+3x^2-7x^4-5x^3+4x^2-x^4+3=-8x^4+7x^2+3\)
Ta có: \(\frac{5.3+6}{8.3+9}=\frac{15+6}{24+9}=\frac{21}{33}=\frac{7}{11}\)
\(\frac{4.4-6}{7.4-8}=\frac{16-6}{28-8}=\frac{10}{20}=\frac{1}{2}\)
\(5\times3+9=24\)
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5 x 3 + 9 = 15 + 9 = 24