Xin lỗi mik viet sai chính tả ko phải là mắng mà là mng 

Ta có: \(\left(1-a\right)\left(1-b\right)=1-a-b+ab\)

-Vì \(a>0;b>0\) nên ab > 0

Suy ra: \(\left(1-a\right)\left(1-b\right)>1-a-b\) (*)

-Vì c < 1 nên 1-c > 0

Tương tự (*) => \(\left(1-a\right)\left(1-b\right)\left(1-c\right)>1-a-b-c\)

\(\Rightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)\left(1-d\right)>\left(1-a-b-c\right)\left(1-d\right)\)

\(d< 1\Rightarrow d-1>0\)

Vậy \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\left(1-d\right)>1-a-b-c-d\)

=> (đpcm)

Đặt \(A=\left(1-a\right)\left(1-b\right)\left(1-c\right)\left(1-d\right)\)

\(A=\left(1-a-b+ab\right)\left(1-c-d+cd\right)\)

\(A=1-c-d+cd-a+ac+ad-acd-b+bd-bcd+ab-abc-abd+abcd+bc\)

\(A=1-a-b-c-d+cd\left(1-a\right)+ac\left(1-b\right)+bc\left(1-d\right)+bd\left(1-c\right)+abcd\)

Có: 0<a,b,c,d<1

=> \(cd\left(1-a\right)>0;ac\left(1-b\right)>0;bc\left(1-d\right)>0;bd\left(1-c\right)>0;abcd>0\)

\(\Rightarrow A>A-cd\left(1-a\right)-ac\left(1-b\right)-bc\left(1-d\right)-bd\left(1-c\right)-abcd=1-a-b-c-d\)

                                                                                                                                        đpcm

a) ĐK: \(x\ne0;x\ne-1\)

\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{2}+x-2\right)\)

\(=\left(\frac{x+1-2+x}{\left(x^2+x\right)\left(x+1\right)}\right):\left(\frac{1+2x+4}{2}\right)\)

\(=\frac{2x-1}{\left(x^2+x\right)\left(x+1\right)}:\frac{2x+5}{2}\)\(=\frac{2\left(2x-1\right)}{\left(x^2+x\right)\left(x+1\right)\left(2x+5\right)}\)?? hình như hết tính tiếp được rồi :v

P/s: Có phải đề là tính giá trị biểu thức không?

x² -10x + 25 - y²

= (x-5)² - y²

= (x-5+y).(x-5-y)

\(x^2-10x+25-y^2\)

\(=\left(x-5\right)^2-y^2\)

\(=\left(x-5+y\right)\left(x-5-y\right)\)

\(\left(x^2-5x+8\right)^2-6x+8\)

\(=x^4+25x^2+64-10x^3-80x+16x^2-6x+8\)

\(=x^4-10x^3+41x^2-86x+72\)

\(=x^3\left(x-2\right)-8x^2\left(x-2\right)+25x\left(x-2\right)-36\left(x-2\right)\)

\(=\left(x-2\right)\left(x^3-8x^2+25x-36\right)\)

\(=\left(x-2\right)\left[x^2\left(x-4\right)-4x\left(x-4\right)+9\left(x-4\right)\right]\)

\(=\left(x-2\right)\left(x-4\right)\left(x^2-4x+9\right)\)

Phân thức đối của phân thức \(\frac{x-1}{x-y}\)là \(\frac{-\left(x-1\right)}{x-y}\)=\(\frac{1-x}{x-y}\)

=> Chọn C)