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\(2^3+3.\left(\dfrac{1}{9}\right)^0-2^{-2}.4+\left[\left(-2\right)^2:\dfrac{1}{2}\right].8\)
\(=8+3.1-\dfrac{1}{2^2}.4+\left(4.2\right).8\\ =8+3-1+64\\ =74\)
\(2^3+3\cdot\left(\dfrac{1}{9}\right)^0-2^{-2}\cdot4+\left[\left(-2\right)^2:\dfrac{1}{2}\right]\cdot8\)
\(=8+3-\dfrac{1}{4}\cdot4+\left(4\cdot2\right)\cdot8\)
\(=11-1+64\)
=74
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1:
a: M=4/9x^2y^2*4x^2y^2=16/9x^4y^4
b: bậc là 8
hệ số là 16/9
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1.
\(f\left(x\right)=2x^4+6x^3+8x^2+12x+1\)
2.
\(h\left(x\right)=\left(2x^4+6x^3+8x^2+12x+1\right)-\left(2x^4+6x^3+17x^2+12x-26\right)\)
\(=-9x^2+27\)
3.
\(h\left(x\right)=0\Leftrightarrow-9x^2+27=0\)
\(\Leftrightarrow x^2=3\Rightarrow x=\pm\sqrt{3}\)
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\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{39}{19^2.20^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{19^2}-\frac{1}{20^2}\)
\(=1-\frac{1}{20^2}\)
\(=\frac{399}{400}\)
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a: \(A\left(x\right)=5x^3-10x^2+11x-14\)
b: \(A\left(1\right)=5-10+11-14=-8\)
\(A\left(-1\right)=5\cdot\left(-1\right)-10\cdot1+11\cdot\left(-1\right)-14=-40\)
A(0)=-14