với n thuộc N* hãy chứng tỏ rằng :
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}\left[\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\right]\)
\(=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
chứng tỏ rằng với mọi n thuộc N* ta có :
\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}=\frac{n}{2\left(3n+2\right)}\)
\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\frac{3n}{2.\left(3n+2\right)}\)
\(=\frac{n}{2\left(3n+2\right)}\)
Với mọi số tự nhiên n > 2 . Chứng minh rằng \(\frac{1}{\left(n-1\right).n.\left(n+1\right)}=\frac{1}{2}\left[\frac{1}{\left(n-1\right).n}-\frac{1}{n.\left(n+1\right)}\right]\)
\(\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\frac{2}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}\)
\(=\frac{1}{2}\left[\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]\)
Ta có đpcm.
Câu 8 : Cho biểu thức :
\(N=\left(\frac{x-1}{\left(x-1\right)^2+x}-\frac{2}{x-2}\right):\left(\frac{\left(x-1\right)^4+2}{\left(x-1\right)^3-1}-x+1\right)\)
Chứng minh rằng với mọi giá trị thích hợp của x thì giá trị N luôn là số nguyên
Câu 8 :
\(N=\left(\frac{x-1}{\left(x-1\right)^2+x}-\frac{2}{x-2}\right):\left(\frac{\left(x-1\right)^4+2}{\left(x-1\right)^3-1}-x+1\right)\)
Đặt \(x-1=a\)
\(N=\left(\frac{a}{a^2+x}-\frac{2}{a-1}\right):\left(\frac{a^4+2}{a^3-1}-a\right)\)
\(N=\frac{a\left(a-1\right)-2\left(a^2+x\right)}{\left(a^2+x\right)\left(a-1\right)}:\frac{a^4+2-a\left(a^3-1\right)}{a^3-1}\)
\(N=\frac{a^2-a-2a^2-2x}{\left(a^2+x\right)\left(a-1\right)}:\frac{a^4+2-a^4+a}{a^3-1}\)
\(N=\frac{-a^2-a-2x}{\left(a^2+x\right)\left(a-1\right)}\cdot\frac{\left(a-1\right)\left(a^2+a+1\right)}{2+a}\)
\(N=\frac{-\left(a^2+a+2x\right)\left(a^2+a+1\right)}{\left(a^2+x\right)\left(2+a\right)}\)
\(N=\frac{-\left[\left(x-1\right)^2+x-1+2x\right]\left[\left(x-1\right)^2+x-1+1\right]}{\left[\left(x-1\right)^2+x\right]\left(2+x-1\right)}\)
\(N=\frac{-\left(x^2+x\right)\left(x^2-x+1\right)}{\left(x^2-x+1\right)\left(x+1\right)}\)
\(N=\frac{-x\left(x+1\right)}{x+1}\)
\(N=-x\)( đpcm )
Câu 9 : Tìm giá trị nhỏ nhất của biểu thức :
\(P=\frac{x^2}{x+4}\cdot\left(\frac{x^2+16}{x}+8\right)+9\)
Bài làm :
\(P=\frac{x^2}{x+4}\cdot\frac{x^2+8x+16}{x}+9\)
\(P=\frac{x^2\left(x+4\right)^2}{x\left(x+4\right)}+9\)
\(P=x\left(x+4\right)+9\)
\(P=x^2+4x+9\)
\(P=\left(x+2\right)^2+5\ge5\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-2\)
Bài 10 : Tìm GTLN
\(Q=\left(\frac{x^3+8}{x^3-8}\cdot\frac{4x^2+8x+16}{x^2-4}-\frac{4x}{x-2}\right):\frac{-16}{x^4-6x^3+12x^2-8x}\)
\(Q=\left[\frac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\cdot\frac{4\left(x^2+2x+4\right)}{\left(x-2\right)\left(x+2\right)}-\frac{4x}{x-2}\right]:\frac{-16}{x\left(x^3-6x^2+12x-8\right)}\)
\(Q=\left(\frac{4\left(x^2-2x+4\right)}{\left(x-2\right)^2}-\frac{4x\left(x-2\right)}{\left(x-2\right)^2}\right):\frac{-16}{x\left[x^2\left(x-2\right)-4x\left(x-2\right)+4\left(x-2\right)\right]}\)
\(Q=\frac{4x^2-8x+16-4x^2+8x}{\left(x-2\right)^2}:\frac{-16}{x\left(x-2\right)\left(x^2-4x+4\right)}\)
\(Q=\frac{16}{\left(x-2\right)^2}\cdot\frac{-x\left(x-2\right)\left(x-2\right)^2}{16}\)
\(Q=-x\left(x-2\right)\)
\(Q=-x^2+2x\)
\(Q=-x^2+2x-1+1\)
\(Q=1-\left(x-1\right)^2\le1\forall x\)
Dấu "=" \(\Leftrightarrow x=1\)
Vậy....
a, Chứng tỏ rằng với mọi \(n\inℕ^∗\) ta luôn có: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
b, Áp dụng tính nhanh tổng sau: \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+....+\frac{1}{90}\)
a ) Ta có : \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n.\left(n+1\right)}-\frac{n}{n.\left(n+1\right)}\) \(=\frac{1}{n.\left(n+1\right)}\)
b ) Áp dụng công thức trên tính tổng này như sau :
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
Chúc học giỏi !!!
a, \(VP=\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}\)
\(=\frac{1}{n\left(n+1\right)}=VT\RightarrowĐPCM\)
a) Ta có: \(VP:\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}=VT\)
v) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
Chứng tỏ \(A=\frac{1}{n\times\left(n+1\right)\times\left(n+2\right)}=\frac{\frac{1}{ }}{2}\times\left(\frac{1}{n\times\left(n+1\right)}-\frac{1}{\left(n+1\right)\times\left(n+2\right)}\right)\)với n\(\in\)N*
a, Cm công thức
\(\forall n\ge1\) ta có \(\frac{2}{\left(2n+1\right)\left(\sqrt{n}-\sqrt{n+1}\right)}< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b, áp dụng tính
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{4023\cdot\left(\sqrt{2011}+\sqrt{2012}\right)}< \frac{2011}{2013}\)
chỗ \(\sqrt{n}-\sqrt{n+1}\)phải là \(\sqrt{n}+\sqrt{n+1}\)
a, Ta có
\(\frac{2}{\left(2n+1\right)\left(\sqrt{n}-\sqrt{n+1}\right)}=\frac{2\cdot\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n}-\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
\(=\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{2n+1}=\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{4n^2+4n+1}}< \frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{4n^2+4n}}\)
mà \(\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{4n^2+4n}}=\frac{2\cdot\left(\sqrt{n+1}-\sqrt{n}\right)}{2\sqrt{n\left(n+1\right)}}=\frac{\sqrt{n+1}}{\sqrt{n}\cdot\sqrt{n+1}}-\frac{\sqrt{n}}{\sqrt{n}\cdot\sqrt{n+1}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b, áp dụng bđt ta có
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{4023\cdot\left(\sqrt{2011}+\sqrt{2012}\right)}< \frac{2011}{2013}\)
\(=\frac{1}{\left(2\cdot1+1\right)\left(1+\sqrt{2}\right)}+\frac{1}{\left(2\cdot2+1\right)\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{\left(2\cdot2011+1\right)\left(\sqrt{2011}-\sqrt{2012}\right)}\)
\(< 1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}\)..
\(=1-\frac{1}{\sqrt{2012}}=\frac{\sqrt{2012}-1}{\sqrt{2012}}=\frac{2011}{\sqrt{2012}\cdot\left(\sqrt{2012}+1\right)}\)
\(=\frac{2011}{2012+\sqrt{2012}}< \frac{2011}{2013}\)
Bạn Nhật sai đề bài
Câu. a. Dòng thứ nhất xuống dòng thứ 2. Em chú ý mẫu số sai rồi.
b. Công thức có số 2 trên tử số. Mà em ko đưa số 2 vào thì sao áp dụng dc công thức?
Chứng minh với mọi số nguyên dương n, n>=2 ta có :
\(\left(1-\frac{2}{6}\right)\left(1-\frac{2}{12}\right)\left(1-\frac{2}{20}\right)...\left(1-\frac{2}{n\left(n+1\right)}\right)>\frac{1}{3}\)
a/Chứng minh rằng \(\frac{2}{\left(2n+1\right)\sqrt{n}+\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b/Áp dụng chứng minh
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{4003\left(\sqrt{2001}+\sqrt{2002}\right)}<\frac{2001}{2003}\)
