Giải phương trình: (x + 1)4 + (x - 1)4 = 16
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I __played____ the piano while my father played the guitar and my mother played the violin.
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-6 < x < 11 và x là số nguyên
⇒ x ∈ {-5; -4; -3; ...; 8; 9; 10}
S = (-5) + (-4) + (-3) + ... + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
= 6 + 7 + 8 + 9 + 10
= 40
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a,Với a = 1945; b = 1980; c = 1930 thì (a+b)+c = (1945 +1980) + 1930 = 3925 + 1930 = 5855
b,Với a = 2023; b+c = 1977 thì (a+b)+c = 2023 + 1977(ko bít b vs c là j) = 4000
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\(a)n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ 0,2\leftarrow-0,3\leftarrow-0,1\leftarrow---0,3\)
\(a=m_{Al}=0,2.27=5,4g\\ b)m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\\ c)C_{\%H_2SO_4}=\dfrac{0,3.98}{100}\cdot100=29,4\%\)
a)nH2=22,46,72=0,3mol2Al+3H2SO4→Al2(SO4)3+3H20,2←−0,3←−0,1←−−−0,3
�=���=0,2.27=5,4��)���2(��4)3=0,1.342=34,2��)�%�2��4=0,3.98100⋅100=29,4%a=mAl=0,2.27=5,4gb)mAl2(SO4)3=0,1.342=34,2gc)C%H2SO4=1000,3.98⋅100=29,4%
\(\Leftrightarrow\left[\left(x+1\right)^2\right]^2+\left[\left(x-1\right)^2\right]^2=16\)
\(\Leftrightarrow\left(x^2+2x+1\right)^2+\left(x^2-2x+1^2\right)=16\)
\(\Leftrightarrow x^4+4x^2+1+4x^3+4x+2x^2+x^4+4x^2+1-4x^3-4x+2x^2=16\)
\(\Leftrightarrow2x^4+12x^2+2=16\)
\(\Leftrightarrow x^4+6x^2-7=0\)
Đặt \(x^2=t\ge0\)
\(\Rightarrow t^2+6t-7=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-7\left(loai\right)\end{matrix}\right.\)
\(t=1\Rightarrow x^2=1\Rightarrow x=\pm1\)