\(\alpha=123456.123457-123455.123458\) và \(\beta=987654.987655-987653.987656\). Chứng minh \(\alpha=\beta\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta tính A = 123 456 . 123 457 - 123 455 . 123 458
= ( 123 455 + 1 ) . 123 457 - 123 455 . ( 123 457 + 1 )
= 123 457 . 123 455 + 123 457 - 123 455 . 123 457 + 123 455
= 0
Ta tính B = 987 654 . 987 655 - 987 653 . 987 656
= ( 987 653 + 1 ) . 987 655 - 987 653 . ( 987 655 + 1 )
= 987 655 . 987 653 + 987 655 - 987 653 . 987 655 + 987 653
= 0
Vậy A = B
![](https://rs.olm.vn/images/avt/0.png?1311)
A=123456.123457-123455.123458=(123455+1).123457-123455(123457+1)=123455.123457+123457-123455.123457-123455=2
B=987654.987655-987653.987656=(987653+1).987655-987653.(987655+1)=987653.987655+987655-987653.987655-987653=2
=> A=B
![](https://rs.olm.vn/images/avt/0.png?1311)
2.
ĐK: \(2x-y\ge0;y\ge0;y-x-1\ge0;y-3x+5\ge0\)
\(\left\{{}\begin{matrix}xy-2y-3=\sqrt{y-x-1}+\sqrt{y-3x+5}\left(1\right)\\\left(1-y\right)\sqrt{2x-y}+2\left(x-1\right)=\left(2x-y-1\right)\sqrt{y}\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left(1-y\right)\sqrt{2x-y}+y-1+2x-y-1-\left(2x-y-1\right)\sqrt{y}=0\)
\(\Leftrightarrow\left(1-y\right)\left(\sqrt{2x-y}-1\right)+\left(2x-y-1\right)\left(1-\sqrt{y}\right)=0\)
\(\Leftrightarrow\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}-1\right)\left(1+\sqrt{y}\right)+\left(\sqrt{2x-y}-1\right)\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}+1\right)=0\)
\(\Leftrightarrow\left(1-\sqrt{y}\right)\left(\sqrt{2x-y}-1\right)\left(\sqrt{y}+\sqrt{2x-y}+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=2x-1\end{matrix}\right.\) (Vì \(\sqrt{y}+\sqrt{2x-y}+2>0\))
Nếu \(y=1\), khi đó:
\(\left(1\right)\Leftrightarrow x-5=\sqrt{-x}+\sqrt{-3x+6}\)
Phương trình này vô nghiệm
Nếu \(y=2x-1\), khi đó:
\(\left(1\right)\Leftrightarrow2x^2-5x-1=\sqrt{x-2}+\sqrt{4-x}\) (Điều kiện: \(2\le x\le4\))
\(\Leftrightarrow2x\left(x-3\right)+x-3+1-\sqrt{x-2}+1-\sqrt{4-x}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{1+\sqrt{4-x}}-\dfrac{1}{1+\sqrt{x-2}}+2x+1\right)=0\)
Ta thấy: \(1+\sqrt{x-2}\ge1\Rightarrow-\dfrac{1}{1+\sqrt{x-2}}\ge-1\Rightarrow1-\dfrac{1}{1+\sqrt{x-2}}\ge0\)
Lại có: \(\dfrac{1}{1+\sqrt{4-x}}>0\); \(2x>0\)
\(\Rightarrow\dfrac{1}{1+\sqrt{4-x}}-\dfrac{1}{1+\sqrt{x-2}}+2x+1>0\)
Nên phương trình \(\left(1\right)\) tương đương \(x-3=0\Leftrightarrow x=3\Rightarrow y=5\)
Ta thấy \(\left(x;y\right)=\left(3;5\right)\) thỏa mãn điều kiện ban đầu.
Vậy hệ phương trình đã cho có nghiệm \(\left(x;y\right)=\left(3;5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{sin\left(a-b\right)}{sina.sinb}+\dfrac{sin\left(b-c\right)}{sinb.sinc}+\dfrac{sin\left(c-a\right)}{sinc.sina}\)
\(=\dfrac{sina.cosb-cosa.sinb}{sina.sinb}+\dfrac{sinb.cosc-cosb.sinc}{sinb.sinc}+\dfrac{sinc.cosa-cosc.sina}{sina.sinc}\)
\(=\dfrac{cosb}{sinb}-\dfrac{cosa}{sina}+\dfrac{cosc}{sincc}-\dfrac{cosb}{sinb}+\dfrac{cosa}{sina}-\dfrac{cosc}{sincc}\)
\(=0\)
Ta có A = 123456.123457 - 123455.123458
= (123455 + 1).123457 - 123455.123458
= 123455.123457 - 123455.123458 + 123457
= 123455(123457 - 123458) + 123457
= 123455.(-1) + 123457
= 2
Lại có B = 987654.987655 - 987653.987656
= (987653 + 1).987655 - 987653.987656
= 987653.987655 - 987653.987656 + 987655
= 987653.(987655 - 987656) + 987655
= 987653.(-1) + 987655 = 2
=> A = B (CÙNG = 2)