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ta tính riêng từng biểu thức:
\(A^2=\left(\frac{y}{z}+\frac{z}{y}\right)^2=\frac{y^2}{z^2}+\frac{z^2}{y^2}+2\)
\(B^2=\left(\frac{z}{x}+\frac{x}{z}\right)^2=\frac{z^2}{x^2}+\frac{x^2}{z^2}+2\)
\(C^2=\left(\frac{x}{y}+\frac{y}{x}\right)^2=\frac{x^2}{y^2}+\frac{y^2}{x^2}+2\)
cộng lại ta có:
\(A^2+B^2+C^2=\left(\frac{x^2}{y^2}+\frac{y^2}{x^2}\right)+\left(\frac{y^2}{z^2}+\frac{z^2}{y^2}\right)+\left(\frac{z^2}{x^2}+\frac{x^2}{z^2}\right)+6\)
\(A\cdot B\cdot C=\left(\frac{y}{z}+\frac{z}{y}\right)\left(\frac{z}{x}+\frac{x}{z}\right)\left(\frac{x}{y}+\frac{y}{x}\right)\)
\(A\cdot B\cdot C=\left(\frac{y}{x}+\frac{xy}{z^2}+\frac{z^2}{xy}+\frac{x}{y}\right)\left(\frac{x}{y}+\frac{y}{x}\right)\)
\(A\cdot B\cdot C=\left(\frac{x^2}{y^2}+\frac{y^2}{x^2}\right)+\left(\frac{y^2}{z^2}+\frac{z^2}{y^2}\right)+\left(\frac{z^2}{x^2}+\frac{x^2}{z^2}\right)+2\)
trừ \(A^2+B^2+C^2\) cho \(A\cdot B\cdot C\)
= 6-2
=4
b. Sử dụng các hằng đẳng thức
\(a^3+b^3+c^2-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=3\left(a^2+b^2+c^2-ab-bc-ca\right)\)
và \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
nên \(A=\frac{a^2+b^2+c^2-ab-bc-ca}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{1}{2}.\frac{\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Do (a - b) + (b - c) + (c - a) = 0 nên áp dụng hđt \(X^2+Y^2+Z^2=-2\left(XY+YZ+ZX\right)\)khi X + Y + Z = 0, ta có:
\(A=-2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right).\)
Bài 1 :
\(b,ax^2+3ax+9=a^2\)
\(\Leftrightarrow a^2x+3ax+9-a^2=0\)
\(\Leftrightarrow ax\left(a+3\right)+\left(a+3\right)\left(3-a\right)=0\)
\(\Leftrightarrow\left(a+3\right)\left(ax+3-a\right)=0\)
Vì \(a\ne3\Rightarrow\left(a+3\right)\ne0\Rightarrow ax+3-a=0\)
\(\Leftrightarrow ax=a-3\)
Vì \(a\ne0\Rightarrow x=\frac{a-3}{a}\)
\(P=\frac{x\left(x+5\right)+y\left(y+5\right)+2\left(xy-3\right)}{x\left(x+6\right)+y\left(y+6\right)+2xy}\)
\(=\frac{x^2+5x+y^2+5y+2xy-6}{x^2+6x+y^2+6y+2xy}\)
\(=\frac{\left(x+y\right)^2+5\left(x+y\right)-6}{\left(x+y\right)^2+6\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(x+y+5\right)-6}{\left(x+y\right)\left(x+y+6\right)}\)
\(=\frac{2005\times\left(2005+5\right)-6}{2005\times\left(2005+6\right)}\)
\(=\frac{2005\times2010-6}{2005\times2011}\)
\(=\frac{2004}{2005}\)
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=\(\frac12\left(1-\frac{1}{n^2+n+1}\right)\)
\(S=\frac{n\left(n+1\right)}{2\left(n^2+n+1\right)}\)
1.
a + b + c = 0 \(\Rightarrow\)a = - ( b + c ) \(\Rightarrow\)a2 = [ -( b + c ) ]2 \(\Rightarrow\)a2 = b2 + c2 + 2bc
Tương tự : b2 = a2 + c2 + 2ac ; c2 = a2 + b2 + 2ab
a + b + c = 0 \(\Rightarrow\)a3 + b3 + c3 = 3abc ( chứng minh )
Ta có : \(A=\frac{a^2}{b^2+c^2+2bc-b^2-c^2}+\frac{b^2}{a^2+c^2+2ac-a^2-c^2}+\frac{c^2}{a^2+b^2+2ab-a^2-b^2}\)
\(A=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}\)
\(A=\frac{a^3+b^3+c^3}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\)
2. quy đồng mà giải
tại sao a+b+c=0 lại suy ra đc \(a^3+b^3+c^3=3abc\)