a: x-y-z=0

=>x=y+z; y=x-z; z=x-y

\(K=\dfrac{x-z}{x}\cdot\dfrac{y-x}{y}\cdot\dfrac{z+y}{z}=\dfrac{y\cdot\left(-z\right)\cdot x}{xyz}=-1\)

b: Tham khảo:

a) \(\frac{x+1}{2x+6}\)+\(\frac{2x+3}{x\left(x+3\right)}\)

= \(\frac{x+1}{2\left(x+3\right)}\)+ \(\frac{2x+3}{x\left(x+3\right)}\)

= \(\frac{x\left(x+1\right)}{2x\left(x+3\right)}\)+ \(\frac{2\left(2x+3\right)}{2x\left(x+3\right)}\)

= \(\frac{x^2+x+4x+6}{2x\left(x+3\right)}\)

= \(\frac{x^2+5x+6}{2x\left(x+3\right)}\)

= \(\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}\)

= \(\frac{x+2}{2x}\)

b) \(\frac{x-1}{x}\)+ \(\frac{x+2}{2}\)

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

= \(\frac{2x-2+x^2+2x}{2x}\)

= \(\frac{x^2+4x-2}{2x}\)

c) \(\frac{1}{x+y}\)+ \(\frac{-1}{x-y}\)+ \(\frac{2x}{x^2+y^2}\)

= \(\frac{\left(x-y\right)\left(x^2+y^2\right)}{\left(x^2+y^2\right)\left(x-y\right)\left(x+y\right)}\)+\(\frac{-\left(x+y\right)\left(x^2+y^2\right)}{\left(x^2+y^2\right)\left(x-y\right)\left(x+y\right)}\)+ \(\frac{2x\left(x-y\right)\left(x+y\right)}{\left(x^2+y^2\right)\left(x-y\right)\left(x+y\right)}\)

= \(\frac{x^3+xy^2-x^2y-y^3-x^3-xy^2-xy^2-y^3+2x^3+2x^2y-2x^2y+2xy^2}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)

= \(\frac{2x^3+xy^2-x^2y-2y^3}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)

= \(\frac{\left(2x^3-2y^3\right)-\left(x^2y-xy^2\right)}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)

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

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

= \(\frac{2x^2+xy+2y^2}{\left(x+y\right)\left(x^2+y^2\right)}\)

e) = \(\frac{3x^2-6xy+3y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

= \(\frac{3\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

=\(\frac{3x-3y}{x^2+xy+y^2}\)

( Mình bận rồi, lát làm câu d nhé)

giúp e vs các a cj soyeon_Tiểubàng giải

Phương An

Hoàng Lê Bảo Ngọc

Silver bullet

Nguyễn Huy Tú

Nguyễn Như Nam

Hoàng Tuấn Đăng

Nguyễn Trần Thành Đạt

Nguyễn Huy Thắng

Võ Đông Anh Tuấn

sửa đề CM biểu thức \(\le\frac{3}{16}\)

\(\frac{1}{2x+y+z}=\frac{1}{x+x+y+z}\le\frac{1}{16}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

bình phương hai vế:

\(\frac{1}{2x+y+z)^2}=\frac{1}{16^2}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)

áp dụng bđt phụ: \(\left(x_1+x_2+x_3+x_4\right)^2\le4\left(x_1+x_2+x_3+x_4\right)\)

áp dụng cho cụm \(\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)

=> \(\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\le4\left(\frac{2}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)

=> \(\frac{1}{\left(2x+y+z\right)^2}\le\frac{1}{64}\left(\frac{2}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)

áp dụng tương tự:

=> \(\frac{1}{\left(2y+x+z\right)}\le\frac{1}{64}\left(\frac{1}{x^2}+\frac{2}{y^2}+\frac{1}{z^2}\right)\)

\(\frac{1}{\left(2z+x+y\right)^2}\le\frac{1}{64}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{z^2}\right)\)

cộng cả ba biêu thức trên

=> \(VT\le\frac{1}{64}\left\lbrace\left(\frac{2}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+\left(\frac{1}{x^2}+\frac{2}{y^2}+\frac{1}{z^2}\right)+\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{z^2}\right)\right\rbrace\) \(VT\le\frac{1}{64}\left(\frac{4}{x^2}+\frac{4}{y^2}+\frac{4}{z^2}\right)\)

\(VT\le\frac{4}{64}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)=\frac{1}{16}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)

ta có \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=3\)

=> \(VT\le\frac{1}{16\cdot3}=\frac{3}{16}\)

dấu bằng xảy ra khi x=y=z=1

a) sửa đề: \(\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)

=\(\frac{-x^2\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{-y^2\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{-z^2\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

=\(-\frac{\left\lbrace x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\right\rbrace}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

xét tử số:

Tử=\(x^2y-x^2z+y^2z-y^2x+z^2x-z^2y\)

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

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

=\(\left(y-z\right)\left\lbrace x^2-x\left(y+z\right)+yz\right\rbrace\)

=\(\left(y-z\right)\left\lbrace x\left(x-y\right)-z\left(x-y\right)\right\rbrace\)

=\(\left(y-z\right)\left(x-y\right)\left(x-z\right)\)

=\(-\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

thay lại vào biểu thức cũ:

\(\Rightarrow-\frac{\left\lbrace-\left(x-y\right)\left(y-z\right)\left(z-x\right)\right\rbrace}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

=\(1\)

b) \(\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(c-a\right)\left(a-b\right)}\)

=\(\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

Bài 1:

\(\frac{A}{x-1}+\frac{B}{x-2}=\frac{A\left(x-2\right)+B\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}\)

\(=\frac{Ax-2A+Bx-B}{x^2-3x+2}=\frac{\left(A+B\right)x-\left(2A+B\right)}{x^2-3x+2}\)

so sách với tử số vừa tìm dc với đề bài:

=> A+B=1

2A+B=-2

=>(2A+B)-(A+B)=-2-1

A=-3

=> B=1+3=4

b) sửa đề \(\frac{A}{x-1}+\frac{\left(Bx+C\right)}{x^2+1}=\frac{A}{x-1}+\frac{\left(Bx+C\right)}{x^2+1}\)

=> \(\frac{A}{x-1}+\frac{\left(Bx+C\right)}{x^2+1}=\frac{A\left(x^2+1\right)+\left(Bx+C\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+1\right)}\)

\(=\frac{Ax^2+A+Bx^2-Bx+Cx-C}{\left(x-1\right)\left(x^2+1\right)}=\frac{\left(A+B\right)x^2+\left(C-B\right)x+\left(A-C\right)}{\left(x-1\right)\left(x^2+1\right)}\)

so sánh với tử số bên cạnh là \(x^2+2x-1\)

=>\(A+B=1\)

\(C-B=2\)

\(A-C=-1\)

=> \(A=1,B=0,C=2\)

bài 2:

quy đồng hai hạng tử đầu tiên:

=> \(\frac{x}{1-x^2}+\frac{y}{1-y^2}=\frac{x\left(1-y^2\right)+y\left(1-x^2\right)}{\left(1-x^2\right)\left(1-y^2\right)}=\frac{\left(x+y\right)\left(1-xy\right)}{\left(1-x^2\right)\left(1-y^2\right)}\)

từ xy+yz+xz=1=> 1-xy=z(x+y) thay vào biểu thức vừa tìm dc ta có:

\(\frac{\left(x+y\right)z\left(x+y\right)}{\left(1-x^2\right)\left(1-y^2\right)}=\frac{z\left(x+y\right)^2}{\left(1-x^2\right)\left(1-y^2\right)}\)

\(VT=\frac{z\left(x+y\right)^2}{\left(1-x^2\right)\left(1-y^2\right)}+\frac{z}{1-z^2}=z\left\lbrace\frac{\left(x+y\right)^2\left(1-z^2\right)+\left(1-x^2\right)\left(1-y^2\right)}{\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)}\right)\)

ta có:

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

=\(\left(x^2+2xy+y^2\right)-z^2\left(x+y\right)^2+1-x^2-y^2+x^2y^2\)

=\(\left(1+xy\right)^2-z^2\left(x+y\right)^2=\left(1+xy-xz-yz\right)\left(1+xy+xz+yz\right)\)

=\(4xy\)

thay vào biểu thức ban đầu:

\(z\cdot\frac{4xy}{\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)}=\frac{4xyz}{\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)}\left(đpcm\right)\)

bài 3:

xếp hạng tổng k của dãy số:

\(a_{k}=\frac{k}{k^4+k+1}\)

=> \(a_{k}=\frac12\left\lbrace\frac{\left(k^2+k+1\right)-\left(k^2-k+1\right)}{\left(k^2-k+1\right)\left(k^2+k+1\right)}\right\rbrace=\frac12\left(\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right)\)

thay k=1,2,3,4,...,n)

=> \(S=\frac12\left\lbrace\left(\frac11-\frac13\right)+\left(\frac13-\frac17\right)+\cdots+\left(\frac{1}{n^2-n+1}-\right.\frac{1}{n^2+n+1}\right)\) S=\(\frac12\left(1-\frac{1}{n^2+n+1}\right)\)

\(S=\frac{n\left(n+1\right)}{2\left(n^2+n+1\right)}\)