Bài 7: Phép nhân các phân thức đại số

Nguyễn Ngọc Thanh Trúc đề là gì

thực hiện phép tính

a) \(\dfrac{25x^3y}{7z}.\dfrac{28z}{15x^2y^5}\)

\(=\dfrac{25x^3y.28z}{7z.15x^2y^5}\)

\(=\dfrac{700x^3yz}{105x^2y^5z}\)

\(=\dfrac{20x}{3y^4}\)

b) \(\dfrac{x^2+3x+9}{2x+10}.\dfrac{x+5}{x^3+27}\)

\(=\dfrac{\left(x^2+3x+9\right)\left(x+5\right)}{\left(2x+10\right)\left(x^3+27\right)}\)

\(=\dfrac{x^3+3x^2+9x+5x^2+15x+45}{2x^4+54x+10x^3+270}\)

\(=\dfrac{x^3+8x^2+24x+45}{2x^4+10x^3+54x+270}\)

a)Nhận xét

\(\dfrac{n^3+1}{n^3-1}=\dfrac{\left(n+1\right)\left(n^2-n+1\right)}{\left(n-1\right)\left(n^2+n+1\right)}=\dfrac{\left(n+1\right)\left[\left(n-0,5\right)^2+0;75\right]}{\left(n-1\right)\left[\left(n+0,5\right)^2+0,75\right]}\)

Áp dụng công thức trên:

\(A=\dfrac{2^3+1}{2^3-1}.\dfrac{3^3+1}{3^3-1}....\dfrac{9^3+1}{9^3-1}\)

\(=\dfrac{\left(2+1\right)\left[\left(2-0,5\right)^2+0,75\right]}{\left(2-1\right)\left[\left(2+0,5\right)^2+0,75\right]}.\dfrac{\left(3+1\right)\left[\left(3-0,5\right)^2+0,75\right]}{\left(3-1\right)\left[\left(3+0,5\right)^2+0,75\right]}...\dfrac{\left(9+1\right)\left[\left(9-0,5\right)^2+0,75\right]}{\left(9-1\right)\left[\left(9+0,5\right)^2+0,75\right]}\)

\(=\dfrac{3\left(1,5^2+0,75\right)}{\left(2,5^2+0,75\right)}.\dfrac{4\left(2,5^2+0,75\right)}{2\left(3,5^2+0,75\right)}...\dfrac{10\left(8,5^2+0,75\right)}{8\left(9,5^2+0,75\right)}\)

\(=\dfrac{3.4....10}{1.2.....8}.\dfrac{1,5^2+0,75}{9,5^2+0,75}\)

\(=\dfrac{9.10}{2}.\dfrac{3}{91}\)

\(=\dfrac{3}{2}.\dfrac{90}{91}< \dfrac{3}{2}\)

\(\Rightarrowđpcm\)

b) Làm tương tự

ta có : \(xy+yz+xz=0\Rightarrow\dfrac{xy+yz+xz}{xyz}=0\)

\(\Leftrightarrow\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{y}=0\Rightarrow\dfrac{1}{z}=-\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

\(\Rightarrow\dfrac{1}{z^3}=-\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3\)

\(\Rightarrow\dfrac{1}{z^3}=-\left(\dfrac{1}{x^3}+3.\dfrac{1}{x^2}.\dfrac{1}{y}+3.\dfrac{1}{x}.\dfrac{1}{y^2}+\dfrac{1}{y^3}\right)\)

\(\Rightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-3.\dfrac{1}{x}.\dfrac{1}{y}.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

\(\Rightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=3.\dfrac{1}{xyz}\)

Do đó : \(xyz.\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=3\)

\(\Leftrightarrow\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}=3\)

\(\Leftrightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)

Vậy giá trị của biểu thức \(\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)

a,\(A=\left[\dfrac{\left(x-2\right)\left(x+1\right)}{x-1}-\left(x+2\right)\right].\dfrac{x^2-2x+1}{2}\)

A xác định \(\Leftrightarrow x-1\ne0\)

\(\Leftrightarrow x\ne1\)

Rút gọn:

\(A=\left[\dfrac{\left(x-2\right)\left(x+1\right)}{x-1}-\left(x+2\right)\right].\dfrac{\left(x-1\right)^2}{2}\)

\(A=\left[\dfrac{\left(x-2\right)\left(x+1\right)}{x-1}-\dfrac{x+2}{1}\right].\dfrac{\left(x-1\right)^2}{2}\)

\(A=\left[\dfrac{\left(x-2\right)\left(x+1\right)}{x-1}-\dfrac{\left(x+2\right)\left(x-1\right)}{x-1}\right].\dfrac{\left(x-1\right)^2}{2}\)

\(A=\left(\dfrac{x^2-x-2-x^2-x+2}{x-1}\right).\dfrac{\left(x-1\right)^2}{2}\)

\(A=\dfrac{-2}{x-1}.\dfrac{\left(x-1\right)^2}{2}\)

\(A=\dfrac{-2.\left(x-1\right)^2}{\left(x-1\right).2}\)

\(A=-x+1\)

b,\(A=-2\Leftrightarrow-x+1=-2\)

\(\Leftrightarrow-x=-3\)

\(\Leftrightarrow x=3\)

Thêm điều kiện đề: a, b, c > 0

Giải:

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\left[\left(a+b\right)^3+c^3\right]-\left[3ab\left(a+b\right)+3abc\right]=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow\dfrac{\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{2}=0\)

Ta có a, b, c > 0 => a + b + c > 0

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\Leftrightarrow a=b=c\)

Vậy ...

Chúc bạn học tốt!

a) Tìm điều kiện xác định A và rút gọc A

b) Tìm x để A = -2

c) Tìm giá trị nhỏ nhất của A