\(\text{Ta có : }\dfrac{1989\cdot1990+3978}{1992\cdot1991-3984}=\dfrac{1989\cdot1990+1989\cdot2}{1992\cdot1991-1992\cdot2}\\ =\dfrac{1989\left(1990+2\right)}{1992\left(1991-2\right)}\\ =\dfrac{1989\cdot1992}{1992\cdot1989}\\ =1\)

\(\text{Ta có : }\)

\(\\ 3x^{n-2}\left(x^{n+2}-y^{n+2}\right)+y^{n+2}\left(3x^{n-2}-y^{n-5}\right)\)

\(\\ =3x^{n-2}\cdot x^{n+2}-3x^{n-2}\cdot y^{n+2}+y^{n+2}\cdot3x^{n-2}-y^{n+2}\cdot y^{n-5}\)

\(\\ =3x^{\left(n+2\right)+\left(n-2\right)}+\left(-3x^{n-2}\cdot y^{n+2}+y^{n+2}\cdot3x^{n-2}\right)-y^{\left(n+2\right)+\left(n-5\right)}\)

\(=3x^{n+2+n-2}-y^{n+2+n-5}\)

\(=3x^{\left(n+n\right)+\left(2-2\right)}-y^{\left(n+n\right)+\left(2-5\right)}\)

\(=3x^{2n}-y^{2n-3}\)

Gọi \(ƯCLN_{\left(2n+4;2n+3\right)}\) là \(d\)

\(\Rightarrow2n+4⋮d\)

\(\Rightarrow2n+3⋮d\)

\(\Leftrightarrow\left(2n+4\right)-\left(2n+3\right)⋮d\)

\(\Leftrightarrow2n+4-2n-3⋮d\)

\(\Leftrightarrow\left(2n-2n\right)+\left(4-3\right)⋮d\)

\(\Leftrightarrow1⋮d\)

\(\Leftrightarrow d\inƯ_{\left(1\right)}\)

\(\Leftrightarrow d=1\)

Mà \(d\) là \(ƯCLN_{\left(2x+4;2x+3\right)}\)

\(\Rightarrow\dfrac{2x+4}{2x+3}\) là phân số tối giản \(\left(ĐPCM\right)\)

Vậy \(\dfrac{2x+4}{2x+3}\) là phân số tối giản

\(\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=\dfrac{x^4-\left(x^2+4x^2\right)+4}{x^4-\left(x^2+9x^2\right)+9}\\ \)

\(=\dfrac{x^4-x^2-4x^2+4}{x^4-x^2-9x^2+9}\\ \)

\(=\dfrac{x^4-x^2-\left(4x^2-4\right)}{x^4-x^2-\left(9x^2-9\right)}\)

\(=\dfrac{x^2-\left(x^2-1\right)-4\left(x^2-1\right)}{x^2-\left(x^2-1\right)-9\left(x^2-1\right)}\\ \)

\(=\dfrac{\left(x^2-4\right)\left(x^2-1\right)}{\left(x^2-9\right)\left(x^2-1\right)}\\ \)

\(=\dfrac{x^2-4}{x^2-9}\)

Có rất nhiều là đằng khác.

Ví dụ : số 24 và số 25

\(\text{a) }\dfrac{17}{30}\text{ và }\dfrac{51}{92}\\ \text{Ta có : }\dfrac{17}{30}=\dfrac{17\cdot3}{30\cdot3}=\dfrac{51}{90}\\ \text{Mà }\dfrac{51}{90}>\dfrac{51}{92}\\ \Rightarrow\dfrac{17}{30}>\dfrac{51}{92}\\ \text{Vậy }\dfrac{17}{30}>\dfrac{51}{92}\\ \\ \)

\(\text{b) }-\dfrac{3}{5}\text{ và }-\dfrac{9}{23}\\ \text{Ta có : }-\dfrac{3}{5}=-\dfrac{3\cdot3}{5\cdot3}=-\dfrac{9}{15}\\ \text{ Mà }-\dfrac{9}{15}< -\dfrac{9}{23}\\ \Rightarrow-\dfrac{3}{5}< -\dfrac{9}{23}\\ \text{Vậy }-\dfrac{3}{5}< -\dfrac{9}{23}\\ \\ \)

\(\text{c) }-\dfrac{15}{23}\text{ và }\dfrac{-151515}{232323}\\ \text{Ta có : }-\dfrac{151515}{232323}=-\dfrac{15\cdot10101}{23\cdot10101}=\dfrac{-15}{23}\\ \text{Vậy }-\dfrac{15}{23}=\dfrac{-151515}{232323}\)

Theo bài ra ta có :

\(12a=72b\Rightarrow\dfrac{12a}{72}=\dfrac{72b}{72}\Rightarrow\dfrac{a}{6}=b\\ a-b=80\text{ }\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được :

\(\dfrac{a}{6}=b=\dfrac{a-b}{6-1}=\dfrac{80}{5}=16\)

\(\dfrac{a}{6}=16\Rightarrow a=96\\ b=16\)

\(\text{Vậy }a=96\\ b=16\)

Ta có : \(xx'\text{ // }yy'\)

\(\Rightarrow xAt=yBA\left(\text{2 góc đồng vị}\right)\)

\(\Rightarrow yBA=50^o\)

Ta lại có : \(yBA+ABy'=180^o\left(\text{2 góc kề bù}\right)\)

\(Hay:50^o+ABy'=180^o\)

\(\Rightarrow Aby'=180^o-50^o\)

\(\Rightarrow Aby'=130^o\)

Vậy \(Aby'=130^o\)

Bạn vào câu hỏi tương tự nha !!!