chứng tỏ rằng \(\frac{1}{a}=\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}\) với a\(\in Z;a\ne0;a\ne-1\)
áp dụng: viết phân số \(\frac{1}{5}\)thành tổng của 3 p/s ai cập khác nhau
ai giải đc cho 3 tk. thề.hứa.đảm bảo.
Chứng tỏ rằng \(\frac{1}{a}=\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}\)với a thuộc Z;a không bằng 0;a không bằng -1
a) Cho \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
Chứng minh rằng: \(x^2+y^2+z^2=\left(x+y+z\right)^2\)
b) Cho a, b, c khác nhau đôi một. Chứng minh rằng:
\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}=\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)^2\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\frac{yz}{xyz}+\frac{xz}{xyz}+\frac{xy}{xyz}=0\)
\(\frac{yz+xz+xy}{xyz}=0\)
yz + xz + xy = 0
\(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2xz+2yz=x^2+y^2+z^2+2\times\left(xy+xz+yz\right)=x^2+y^2+z^2+2\times0=x^2+y^2+z^2\left(\text{đ}pcm\right)\)
a) Từ giả thiết suy ra: xy + yz + zx = 0
Do đó:
\(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)=x^2+y^2+z^2\)
b) Đặt \(\frac{1}{a-b}=x\); \(\frac{1}{b-c}=y\); \(\frac{1}{c-a}=z\)
Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=a-b+b-c+c-a=0\)
Theo câu a ta có: \(x^2+y^2+z^2=\left(x+y+z\right)^2\)
Suy ra điều phải chứng minh
a)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{xy+yz+xz}{xyz}=0\)
\(\Rightarrow xy+yz+xz=0\)
\(x^2+y^2+z^2=\left(x+y+z\right)^2\)
\(\Rightarrow x^2+y^2+z^2=x^2+y^2+z^2+2xy+2yz+2xz\)
\(\Rightarrow x^2+y^2+z^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)
Do \(xy+yz+xz=0\)
\(\Rightarrow x^2+y^2+z^2=x^2+y^2+z^2\) ( đpcm )
b)
\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}=\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)^2\)
\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}=\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}+\frac{2}{\left(a-b\right)\left(b-c\right)}+\frac{2}{\left(b-c\right)\left(c-a\right)}+\frac{2}{\left(a-b\right)\left(c-a\right)}\)
\(\Rightarrow\frac{2}{\left(a-b\right)\left(b-c\right)}+\frac{2}{\left(b-c\right)\left(c-a\right)}+\frac{2}{\left(a-b\right)\left(c-a\right)}=0\)
\(\Rightarrow2\left(\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)}\right)=0\)
\(\Rightarrow\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)}=0\)
\(\Rightarrow\frac{\left(c-a\right)^2\left(b-c\right)\left(a-b\right)+\left(a-b\right)^2\left(b-c\right)\left(c-a\right)+\left(b-c\right)^2\left(a-b\right)\left(c-a\right)}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}=0\)
\(\Rightarrow\frac{\left(c-a\right)\left(b-c\right)\left(a-b\right)\left[\left(a-b\right)+\left(b-c\right)+\left(c-a\right)\right]}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}=0\)
\(\Rightarrow\frac{\left(c-a\right)\left(b-c\right)\left(a-b\right)\left[\left(-a+a\right)+\left(-b+b\right)+\left(-c+c\right)\right]}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}=0\)
\(\Rightarrow\frac{\left(c-a\right)\left(b-c\right)\left(a-b\right).0}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}=0\)
\(\Rightarrow0=0\) ( đpcm )
Cho $A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\left(n\in Z;n\ge2\right)$A=142 +162 +182 +...+1(2n)2 (n∈Z;n≥2)
Chứng tỏ A$\notin$∉ N
Chứng tỏ rằng:
\(\frac{1}{a}\) = \(\frac{1}{a+1}\)+ \(\frac{1}{a\left(a+1\right)}\) với a thuộc Z; a không bằng 0; a không bằng -1
Ta có : \(\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}=\frac{a}{a\left(a+1\right)}+\frac{1}{a\left(a+1\right)}=\frac{a+1}{a\left(a+1\right)}=\frac{1}{a}\) Với mọi a
=> \(\frac{1}{a}=\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}\)
\(\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}=\frac{a}{a\left(a+1\right)}+\frac{1}{a\left(a+1\right)}=\frac{a+1}{a\left(a+1\right)}=\frac{1}{a}\)
Vậy....
chứng tỏ
\(y=\frac{1}{a}=\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}\)với a thuộc Z; a khác 0; a khác -1
giải giúp
Ta có:
\(\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}=\frac{a}{a\left(a+1\right)}+\frac{1}{a\left(a+1\right)}=\frac{a+1}{a\left(a+1\right)}=\frac{1}{a}=y\)
Đúng 100%
Cho \(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\left(n\in Z;n\ge2\right)\)
Chứng tỏ A\(\notin\) N
1, cho a,b,c là các số thực dương chứng minh rằng \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{2a+b}{a\left(a+2b\right)}+\frac{2b+c}{b\left(b+2c\right)}+\frac{2c+a}{c\left(a+2c\right)}\)
2,cho x,y,z thỏa mãn x+y+z=5 và xy+yz+xz=8 chứng minh rằng \(1\le x\le\frac{7}{3}\)
3, cho a,b,c>0 chứng minh rằng\(\frac{a^2}{2a^2+\left(b+c-a\right)^2}+\frac{b^2}{2b^2+\left(b+c-a\right)^2}+\frac{c^2}{2c^2+\left(b+a-c\right)^2}\le1\)
4,cho a,b,c là các số thực bất kỳ chứng minh rằng \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\left(ab+bc+ac-1\right)^2\)
5, cho a,b,c > 1 và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)chứng minh rằng \(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\le\sqrt{a+b+c}\)
Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)
\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)
\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)
\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)
\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)
Dấu "=" xảy ra khi x=y=z
chứng tỏ rằng\(\frac{1}{a.\left(a+1\right)}\)-\(\frac{1}{\left(a+1\right).\left(a+2\right)}\)=\(\frac{2}{a.\left(a+1\right).\left(a+2\right)}\)
Biến đổi vế trái ta có :
\(\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}=\frac{1}{a}-\frac{1}{a+1}-\left(\frac{1}{a+1}-\frac{1}{a+2}\right)=\frac{1}{a}-\frac{1}{a+1}-\frac{1}{a+1}+\frac{1}{a+2}\)
\(\frac{1}{a}-\frac{2}{a+1}+\frac{1}{a+2}=\frac{\left(a+1\right)\left(a+2\right)-2a\left(a+2\right)+a\left(a+1\right)}{a\left(a+1\right)\left(a+2\right)}\)
\(=\frac{a^2+3a+2-2a^2-4a+a^2+a}{a\left(a+1\right)\left(a+2\right)}=\frac{2}{a\left(a+1\right)\left(a+2\right)}\)
Vậy Vế trái = Vế phải
Bài 1: Cho a,b,c đôi một khác nhau. Chứng minh rằng:
\(\frac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}+\frac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}=1\)
Bài 2: CMR: nếu \(\frac{1}{x}-\frac{1}{y}-\frac{1}{z}=1\) và x=y+z thì:
\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1\)
Mọi người làm nhanh giúp em với ạ!
2) 1/x - 1/y - 1/z = 1
=> (1/x - 1/y - 1/z)^2 = 1
<=> 1/x^2 + 1/y^2 + 1/z^2 - 2/xy - 2/xz + 2/yz = 1
<=> 1/x^2 + 1/y^2 + 1/z^2 - 2.(1/xy + 1/xz - 1/yz) = 1
<=> 1/x^2 + 1/y^2 + 1/z^2 - 2.(z+y-x/xyz) = 1
<=> 1/x^2 + 1/y^2 + 1/z^2 - 2.0 = 1
<=> 1/x^2 + 1/y^2 + 1/z^2 = 1 (đpcm)