Prove that $(a^3+b^3+c^3+d^3)^2=9(ab-cd)(bc-ad)(ca-bd)$.
My solution:
First, we draw some very useful and helpful identities from the given equality, $a+b+c+d=0$:
Prove that $(a^3+b^3+c^3+d^3)^2=9(ab-cd)(bc-ad)(ca-bd)$.
Prove that $(a^3+b^3+c^3+d^3)^2=9(ab-cd)(bc-ad)(ca-bd)$.
My solution:
First, we draw some very useful and helpful identities from the given equality, $a+b+c+d=0$:
Analysis Quiz 20: Multiple-Choice Algebra Test
Answer the following questions based on the evaluation of a sum (without the help of calculator) below:
Evaluate [MATH] \frac{(-\sqrt{6}+\sqrt{7}+\sqrt{8})^4}{4(\sqrt{7}-\sqrt{6})(\sqrt{8}-\sqrt{6})}+\frac{(\sqrt{6}-\sqrt{7}+\sqrt{8})^4}{4(\sqrt{6}-\sqrt{7})(\sqrt{8}-\sqrt{7})}+\frac{(\sqrt{6}+\sqrt{7}-\sqrt{8})^4}{4(\sqrt{6}-\sqrt{8})(\sqrt{7}-\sqrt{8})}[/MATH].
Evaluate [MATH] \frac{(-\sqrt{6}+\sqrt{7}+\sqrt{8})^4}{4(\sqrt{7}-\sqrt{6})(\sqrt{8}-\sqrt{6})}+\frac{(\sqrt{6}-\sqrt{7}+\sqrt{8})^4}{4(\sqrt{6}-\sqrt{7})(\sqrt{8}-\sqrt{7})}+\frac{(\sqrt{6}+\sqrt{7}-\sqrt{8})^4}{4(\sqrt{6}-\sqrt{8})(\sqrt{7}-\sqrt{8})}[/MATH].
Evaluate [MATH]\small \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^4-12x^3-x+2}{x+2}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)[/MATH].
Evaluate [MATH]\large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^4-12x^3-x+2}{x+2}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)[/MATH].
My solution:
First, note that $6x^4-12x^3-x+2$ can be factorized as $6(2)^4-12(2)^3-(2)+2=96-96-2+2=0$, factorize it using the long polynomial division by the factor $x-2$, we get $6x^4-12x^3-x+2=(x-2)(6x^3-1)$.
My solution:
First, note that $6x^4-12x^3-x+2$ can be factorized as $6(2)^4-12(2)^3-(2)+2=96-96-2+2=0$, factorize it using the long polynomial division by the factor $x-2$, we get $6x^4-12x^3-x+2=(x-2)(6x^3-1)$.
Find all real solutions for the system [MATH]4x^2-40\left\lfloor{x}\right\rfloor+51=0[/MATH] where [MATH]\left\lfloor{x}\right\rfloor[/MATH] represents the floor of $x$.
Find all real solutions for the system [MATH]4x^2-40\left\lfloor{x}\right\rfloor+51=0[/MATH] where [MATH]\left\lfloor{x}\right\rfloor[/MATH] represents the floor of $x$.
My solution:
First, notice that if we rewrite the equality as [MATH]4x^2+51=40\left\lfloor{x}\right\rfloor[/MATH], we can tell [MATH]\left\lfloor{x}\right\rfloor[/MATH] must be a positive figure.
My solution:
First, notice that if we rewrite the equality as [MATH]4x^2+51=40\left\lfloor{x}\right\rfloor[/MATH], we can tell [MATH]\left\lfloor{x}\right\rfloor[/MATH] must be a positive figure.
Find the number of real solutions for the system $x^4-x^3+x^2-4x-12=0$.
Find the number of real solutions for the system $x^4-x^3+x^2-4x-12=0$.
First, we let $f(x)=x^4-x^3+x^2-4x-12$. $f(x)$ clearly is a quartic function and it has at most 4 real roots.
If we can factorize $f(x)$ as $f(x)=(x-a)(x-b)(x-c)(x-d)$, then $f(x)$ has 4 real roots.
First, we let $f(x)=x^4-x^3+x^2-4x-12$. $f(x)$ clearly is a quartic function and it has at most 4 real roots.
If we can factorize $f(x)$ as $f(x)=(x-a)(x-b)(x-c)(x-d)$, then $f(x)$ has 4 real roots.
Analysis Quiz 19: Multiple-Choice Test (Improve Analytical Skill)
Question 1: If you're asked to simplify [MATH]\left(1+\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\right)^2[/MATH], do you think by turning the $1$ as [MATH]\left(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\right)[/MATH] is feasible in order to simplify the expression?
A. Yes.
B. No.
A. Yes.
B. No.
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