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ĐKXĐ: \(x\ge0;x\ne4\)
\(N=\frac{x+12}{x-4}+\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(N=\frac{x+12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(N=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(A=\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}+.......+\frac{\sqrt{n}-\sqrt{n-1}}{\left(\sqrt{n}-\sqrt{n-1}\right)\left(\sqrt{n}+\sqrt{n}-1\right)}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{2-1}+........+\frac{\sqrt{n}-\sqrt{n-1}}{n-\left(n-1\right)}\)
\(=\sqrt{2}-\sqrt{1}+...........+\sqrt{n}-\sqrt{n-1}\)
\(=\sqrt{n}-\sqrt{1}=\sqrt{n}-1\)
bài B tương tự
Với mọi \(k\in N\)Ta có :
\(1-\frac{1}{k^2}=\frac{k^2-1}{k^2}=\frac{\left(k-1\right)\left(k+1\right)}{k^2}\)
Áp dụng ta có :
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{n^2}\right)\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.......\frac{\left(n-1\right)\left(n+1\right)}{n^2}\)
\(=\frac{\left[1.2.3.....\left(n-1\right)\right]\left[3.4.5.....\left(n+1\right)\right]}{\left(2.3.4.....n\right)\left(2.3.4.....n\right)}\)
\(=\frac{n+1}{2n}\)
x khác 1
\(N=\frac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2+4}{\left(x+1\right)\left(x^2+x+1\right)}\)
\(N=\frac{x^2+2x-x-2-2x^2-2x-2+2x^2+4}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x}{x^2+x+1}\)
Xét hiệu 1/3-N=\(\frac{1}{3}-\frac{x}{x^2+x+1}=\frac{x^2+x+1-3x}{3\left(x^2+x+1\right)}=\frac{x^2-2x+1}{3\left(x^2+x+1\right)}=\frac{\left(x-1\right)^2}{3\left(x^2+x+1\right)}>0\)với mọi x khác 1
=> 1/3 >N
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=\(\frac12\left(1-\frac{1}{n^2+n+1}\right)\)
\(S=\frac{n\left(n+1\right)}{2\left(n^2+n+1\right)}\)
Dấu "....." ở giữa \(\frac{2^2-1}{2^2}\) và \(\frac{3^2-1}{3^2}\) là gì vậy?
\(A=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.....\frac{\left(n-1\right)\left(n+1\right)}{n^2}=\frac{\left(1.2.3.....\left(n-1\right)\right)\left(3.4.5......\left(n+1\right)\right)}{\left(2.3.4.....n\right)\left(2.3.4......n\right)}=\frac{1.\left(n+1\right)}{n.2}=\frac{n+1}{2n}\)