Câu hỏi tương tự cũng có

Giải:
Ta có bảng sau: ( \(x,y\in N\) và 2y - 5 là số lẻ )

Vậy cặp số ( x; y ) là ( 77; 3) ; ( 5; 21 )

1, \(x^2+x-6=x^2+3x-2x-6\)

\(=x\left(x+3\right)-2\left(x+3\right)=\left(x-2\right)\left(x+3\right)\)

2, \(x^2-3x+2=x^2-2x-x+2\)

\(=x\left(x-2\right)-\left(x-2\right)=\left(x-1\right)\left(x-2\right)\)

3, \(x^2-4x+3=x^2-3x-x+3\)

\(=x\left(x-3\right)-\left(x-3\right)=\left(x-1\right)\left(x-3\right)\)

\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)

\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)

\(\Rightarrowđpcm\)

a, \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{99}{100!}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)

\(\Rightarrowđpcm\)

d, \(D=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow3D=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)

\(\Rightarrow3D-D=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)

\(\Rightarrow2D=1-\dfrac{1}{3^{99}}\)

\(\Rightarrow D=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}< \dfrac{1}{2}\)

\(\Rightarrowđpcm\)

\(\left(a+b\right)^3+\left(a-b\right)^3\)

\(=\left(a+b+a-b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

\(=2a\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)\)

\(=2a\left(a^2+3b^2\right)\)

Ta có: \(VT=\left(x^2+y^2\right)^2-\left(2xy\right)^2\)

\(=\left(x^2+2xy+y^2\right)\left(x^2-2xy+y^2\right)\)

\(=\left(x-y\right)^2\left(x+y\right)^2=VP\)

\(\Rightarrowđpcm\)

\(x^4y^4+4=x^4y^4+4x^2y^2+4-4x^2y^2\)

\(=\left(x^2y^2+2\right)^2-4x^2y^2\)

\(=\left(x^2y^2+2-2xy\right)\left(x^2y^2+2+2xy\right)\)

Giải:

Ta có: IA = IB ( I thuộc trung trực của AB )

và SA = SB ( S thuộc trung trực của AB )

Mà CS + BS > BC

=> CS + AS > BI + IC

=> AS + CS > AI + IC

=> đpcm

\(x^2-\left(x-1\right)^2+2\left(x-1\right)-1\)

\(=x^2-x^2+2x-1+2x-2-1\)

\(=4x-4=4\left(x-1\right)\)