2 + 2 + 2 + 4 +....+ 2+ 1996
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câu 1:
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\)
\(\hept{\begin{cases}15=3\cdot5\\35=5\cdot7\end{cases}}\\ \hept{\begin{cases}3=3\\63=3^2\cdot7\\99=3^2\cdot11\end{cases}}\)
=>\(\frac{2310}{3465}+\frac{462}{3465}+\frac{198}{3465}+\frac{110}{3465}+\frac{70}{3465}\)
=>\(\frac{2310+462+198+110+70}{3465}\)
=>\(\frac{3150}{3465}\)=\(\frac{10}{11}\)
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(X -10/1994 -1) + (X-8/1996 - 1) + (X-6/1998 - 1)+ (X-4/2000 - 1) + (X-2/2002 - 1) = (X-2002/2 - 1) + (X-2000/4 - 1) + (X-1998/6 - 1) + (X-1996/8 - 1) + (X-1994/10 - 1)
=> x-2004/1994 + x-2004/1996 + x-2004/1998 + x-2004/2000 + x-2004/2002 = x-2004/2 + x-2004/4 + x-2004/6 + x-2004/8 + x-2004/1994
=> x-2004/1994 + x-2004/1996 + x-2004/1998 + x-2004/2000 + x-2004/2002 - x-2004/2 - x-2004/4 - x-2004/6 - x-2004/8 - x-2004/1994 = 0
=> (x - 2004)(1/994 + 1/1996 + 1/1998 + 1/2000 + 1/2002 + 1/2 + 1/4 + 1/6 + 1/8) = 0
Mà (1/994 + 1/1996 + 1/1998 + 1/2000 + 1/2002 + 1/2 + 1/4 + 1/6 + 1/8) \(\ne\)0
=> x - 2004 = 0
=> x = 2004
Vậy x = 2004
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a)
\(x^4+1996x^2+1995x+1996\)
\(=\left(x^4-x\right)+\left(1996x^2+1996x+1996\right)\)
\(=x\left(x^3-1\right)+1996\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+1996\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x\left(x-1\right)+1996\right]\)
\(=\left(x^2+x+1\right)\left(x^2-x+1996\right)\)
b)
\(x^4+1997x^2+1996x+1997\)
\(=\left(x^4-x\right)+\left(1997x^2+1997x+1997\right)\)
\(=x\left(x^3-1\right)+1997\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+1997\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x\left(x-1\right)+1997\right]\)
\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)
x4+1996x2+1995x+1996
=(x4_x)+(1996x2+1996x+1996)
=x(x3-1)+1996(x2+x+1)
=x(x-1)(x2+x+1)+1996(x2+x+1)
=(x2+x+1)((x2-1)+1996)
=(x2+x+1)((x+1)(x-1)+1996)
Câu 2 tương tự bạn nhé!
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= có x +y+z=a=>x2+y2+z2+2(xy+yz+xz)=a2
Thay vào a2=b+3992=>xy+zy+xz=1996
thay vào P ta có
P=x\(\sqrt{\dfrac{\left(xy+yz+zx+z^2\right)\left(zx+xy+yz+x^2\right)}{xy+yz+zx+x^2}}\)
+y\(\sqrt{\dfrac{\left(zx+zy+xy+z^2\right)\left(zx+zy+xy+x^2\right)}{xy+yz+xz+y^2}}\)
+\(\sqrt{\dfrac{\left(zx+xy+zy+x^2\right)\left(xz+xy+zy+y^2\right)}{xz+xy+zy+z^2}}\)
=x\(\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+z\right)\left(x+y\right)}}\)
+y\(\sqrt{\dfrac{\left(x+z\right)\left(z+y\right)\left(x+y\right)\left(z+x\right)}{\left(y+z\right)\left(x+y\right)}}\)
+z\(\sqrt{\dfrac{\left(y+z\right)\left(x+y\right)\left(z+x\right)\left(x+y\right)}{\left(z+x\right)\left(z+y\right)}}\)
=x\(\sqrt{\left(y+z\right)^2}\)+y\(\sqrt{\left(x+z\right)^2}\)+z\(\sqrt{\left(x+y\right)^2}\)=x(z+y)+y(x+z)+z(x+y)
=2(xy+zx+zy)=3992
*có gì ko hiểu thì hỏi
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Trung bình cộng của dãy số trên là:
( 1996+2): 2 = 999
Dãy số trên có số số hạng là:
(1996-2):2 + 1=998
Tổng của dãy số trên là:
999 x 998=997002
Đáp số:997002
Số số hạng là
(1996-2):2+1=998(số)
tổng các số chẵn từ 2 đến 1996 là:
(1996+2).998:2=997002