Let \(a,b,c,d\) be positive real numbers. Prove that \((a+b)\times(c+d) = ac+ad+bc+bd\). Find both algebraic solution and geometric interpretation.
Let \(a,b,c,d\) be positive real numbers such that \(a\geq b\) and \(c\geq d\). Prove that \((a-b)\times(c-d) = ac-ad-bc+bd\). Find both algebraic solution and geometric interpretation.
Using the area of a rectangle prove that \(a\times b=b\times a\).
Do there exist two numbers such that their sum, quotient and product would be all equal to each other?
Find the mistake in the sequence of equalities: \(-1=(-1)^{\frac{2}{2}}=((-1)^2)^{\frac{1}{2}}=1^{\frac{1}{2}}=1\).
Recall that \((n+1)^2=n^2+2n+1\) and after expansion we get \((n+1)^2-(2n+1)=n^2\). Subtract \(n(2n+1)\) from both sides \((n+1)^2-(2n+1)-n(2n+1)=n^2-n(2n+1)\) and rewrite it as \((n+1)^2-(n+1)(2n+1)=n^2-n(2n+1)\).
Now we add \(\frac{(2n+1)^2}{4}\) to both sides: \((n+1)^2-(n+1)(2n+1)+\frac{(2n+1)^2}{4}=n^2-n(2n+1)+\frac{(2n+1)^2}{4}\).
Factor both sides into square: \(((n+1)-\frac{2n+1}{2})^2=(n-\frac{2n+1}{2})^2\).
Now take the square root: \((n+1)-\frac{2n+1}{2}=n-\frac{2n+1}{2}\).
Add \(\frac{2n+1}{2}\) to both sides and we get \(n+1=n\) which is equivalent to \(1=0\).
Let’s prove that \(1\) is the smallest positive real number: Assume the contrary and let \(x\) be the smallest positive real number. If \(x>1\) then \(1\) is smaller, thus \(x\) is not the smallest. If \(x<1,\) then \(\frac{x}{2}<x\) so \(x\) can not be the smallest either. Then \(x\) can only be equal to \(1\).
Find the last two digits of the number \[33333333333333333347^4 - 11111111111111111147^4\]
Replace all stars with ”+” or ”\(\times\)” signs so the equation holds: \[1*2*3*4*5*6=100\] Extra brackets may be added if necessary. Please write down the expression into the answer box.
Is it true that if \(a\) is a positive number, then \(a^2 \ge a\)? What about \(a^2 +1 \ge a\)?