Tính tích Q= (3+1)(3^2+1)(3^4+1)...(3^2^1997+1)
cac ban jup minh lam voi
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1+2+1+2+3+1+2+3+4+1+2+3+4+5
=(1+2)x4+3x3+4x2+5
=3x4+9+8+5
=12+9+8+5
=34
hinh nhu trong sach phat trien lop 6 co thi phai,lau roi quen
a, \(343\text{ : }\left(2^3-2^2+3^2\cdot x\right)=7\)
\(343\text{ : }\left(8-4+9\cdot x\right)=7\)
\(343\text{ : }\left(4+9\cdot x\right)=7\)
\(4+9\cdot x=343\text{ : }7\)
\(4+9\cdot x=49\)
\(9\cdot x=49-4\)
\(9\cdot x=45\)
\(x=45\text{ : }9\)
\(x=5\)
\(M=1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{3}+\dfrac{3}{15}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)
(x + 1) + (x + 2) + (x + 3 )+ ... + (x + 20) = 250 ( có 20 nhóm )
=> ( x + x + x +...+ x ) + ( 1 + 2 + 3 +...+ 20) = 250 ( có 20 x và 20 số hạng )
=> x . 20 + 20 . 21 : 2 = 250
=> x . 20 + 210 = 250
=> x . 20 = 250 - 210
=> x . 20 = 40
=> x = 40 : 20
x = 2
a/ \(\left|5x+\frac{3}{4}\right|-\frac{5}{4}=2\)
\(\left|5x+\frac{3}{4}\right|=\frac{13}{4}\)
\(\Rightarrow x=\left\{\frac{1}{2};-\frac{4}{5}\right\}\)
b/\(\frac{3}{2}-\left|\frac{1}{2}x+1\right|=\frac{1}{4}\)
\(\left|\frac{1}{2}x+1\right|=\frac{5}{4}\)
1/\(\frac{1}{2}x+1=\frac{5}{4}\)
\(\frac{1}{2}x=\frac{1}{4}\)
\(x=\frac{1}{2}\)
2/\(\frac{1}{2}x+1=-\frac{5}{4}\)
\(\frac{1}{2}x=-\frac{9}{4}\)
\(x=-\frac{9}{2}\)
\(\Rightarrow x=\left\{\frac{1}{2};-\frac{9}{2}\right\}\)