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(2007×20062006)−(2006×20072007)= ? ({2007} \times {20062006}) - ({2006} \times {20072007}) = \ ? (2007×20062006)−(2006×20072007)= ?
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1+a+2b+3c+2ab+3ac+6bc+6abc 1 + a + 2b + 3c + 2ab + 3ac + 6bc + 6abc1+a+2b+3c+2ab+3ac+6bc+6abc
We are given that a=999,b=666,c=333a = \color{#D61F06}{999} , b = \color{#3D99F6}{666} , c = \color{#20A900}{333}a=999,b=666,c=333. Find the value of the expression above.
Hint: Try to factorize the expression.
x2+x4+x6+x8+x10x2−x4+x6−x8+x10=x3+x5+x7+x9+x11x3−x5+x7−x9+x11\large\dfrac{x^2+x^4+x^6+x^8+x^{10}}{x^2-x^4+x^6-x^8+x^{10}}=\dfrac{x^3+x^5+x^7+x^9+x^{11}}{x^3-x^5+x^7-x^9+x^{11}}x2−x4+x6−x8+x10x2+x4+x6+x8+x10=x3−x5+x7−x9+x11x3+x5+x7+x9+x11
Does the equation above hold true for all real x≠0x\ne0x=0?
4(x2+2x+1)(x2+3x−2)+(x−3)2=(ax2+bx+c)24(x^2 + 2x + 1)(x^2 + 3x -2) + (x - 3)^2 \\ = (ax^2 + bx + c)^24(x2+2x+1)(x2+3x−2)+(x−3)2=(ax2+bx+c)2 If the above equation is always correct, then what is the value of a2+b2+c2 a^2 + b^2 +c^2a2+b2+c2?
(1−110)(1−111)(1−112)⋯(1−1100)= ?\displaystyle \left( 1 - \frac{1}{\color{teal}{10}}\right)\left(1 - \frac{1}{\color{teal}{11}}\right)\left(1 - \frac{1}{\color{teal}{12}}\right)\cdots\left(1 - \frac{1}{\color{teal}{100}}\right)= \ ? (1−101)(1−111)(1−121)⋯(1−1001)= ?
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