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* Với \(a=1\) ta thấy BĐT đúng.
* Ta xét khi \(a>1\)
Hàm nghi số \(y=\) \(y=\frac{1}{a^1}=\left(\frac{1}{a}\right)^1\) nghịch biến với \(\forall t\in R,\) khi \(a>1\).
Khi đó ta có
Ta có: \(\left(x-y\right)\left(\frac{1}{a^x}-\frac{1}{a^y}\right)\le0,\forall x,y\in R\Rightarrow\frac{x}{a^x}+\frac{y}{a^y}\le\frac{x}{a^y}+\frac{y}{a^x}\) (1)
Chứng minh tương tự \(\frac{y}{a^y}+\frac{z}{a^z}\le\frac{z}{a^y}+\frac{y}{a^z}\) (2) \(\frac{z}{a^z}+\frac{x}{a^x}\le\frac{x}{a^z}+\frac{z}{a^x}\) (3)
Cộng vế với vế (1), (2) và (3) ta được \(2\left(\frac{x}{a^x}+\frac{y}{a^y}+\frac{z}{a^z}\right)\le\frac{y+z}{a^x}+\frac{z+x}{a^y}+\frac{x+y}{a^z}\) (4)
Cộng 2 vế của (4) với biểu thức \(\frac{x}{a^x}+\frac{y}{a^y}+\frac{z}{a^z}\) ta được
\(3\left(\frac{x}{a^x}+\frac{y}{a^y}+\frac{z}{a^z}\right)\le\frac{x+y+z}{a^x}+\frac{x+y+z}{a^y}+\frac{x+y+z}{a^z}=\left(x+y+z\right)\left(\frac{1}{a^x}+\frac{1}{a^y}+\frac{1}{a^z}\right)\)
Ta có a.(a+b+c)+b.(a+b+c)+c.(a+b+c)=1/144
=>ta sử dụng phép phân phối có a+b+c chung
=>(a+b+c)(a+b+c)=1/144
=>a+b+c=1/12
từ đó tính a,b,c lần lượt là -1/2;3/4;-1/6
cậu toàn chép sai đề bài à nếu là c.(a+b+c)=-1/72 mới tính được
a) \(\frac{2}{\left(x+2\right).\left(x+4\right)}+\frac{4}{\left(x+4\right).\left(x+8\right)}+\frac{6}{\left(x+8\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{x+14}{\left(x+2\right).\left(x+14\right)}-\frac{x+2}{\left(x+2\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{x+14-x+2}{\left(x+2\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{16}{\left(x+2\right).\left(x+4\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow x=16\)
Vậy x = 16
\(b,\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(vì\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
\(\text{Vậy }x=-1\)
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)
\(\Rightarrow x=2k\)
\(y=3k\)
\(z=5k\)
Thay \(x=2k;y=3k;z=5k\) vào \(x.y.z=810\) ta được:
\(2k.3k.5k=810\)
\(30k^3=810\)
\(k^3=27\)
\(k^3=3^3\)
\(\Rightarrow k=3\)
\(\Rightarrow x=2k=2.3=6\)
\(y=3k=3.3=9\)
\(z=5k=5.3=15\)
Vậy \(x=6;y=9;z=15\)
b. (x+1)(1/10+1/11+1/12-1/13-1/14)=0
x+1=0 (vì : 1/10+1/11+1/12-1/13-1/14>0)
x=-1
\(\left(-2\frac{3}{4}+\frac{1}{2}\right)^2\)
\(=\left(-\frac{11}{4}+\frac{1}{2}\right)^2\)
\(=\left(-\frac{11}{4}+\frac{2}{4}\right)^2\)
\(=\left(-\frac{9}{4}\right)^2\)
\(=\frac{81}{16}\)
\(\left(-2\frac{3}{4}+\frac{1}{2}\right)^2\)
\(=\left(\frac{-11}{4}+\frac{1}{2}\right)^2\)
\(=\left(\frac{-11}{4}+\frac{2}{4}\right)^2\)
\(=\left(\frac{-9}{4}\right)^2\)
\(=\frac{81}{16}\)
Đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\)
=> \(x=2k+1\)
\(y=3k+2\)
\(z=4k+3\)
Thay \(x=2k+1;y=3k+2;z=4k+3\) vào \(2x+3y-z=50\) ta được:
\(2\left(2k+1\right)+3\left(3k+2\right)-4\left(4k+3\right)=50\)
\(4k+2+9k+6-4k-3=50\)
\(9k+5=50\)
\(9k=45\)
\(k=5\)
\(\Rightarrow x=2k+1=2.5+1=11\)
\(y=3k+2=3.5+2=17\)
\(z=4k+3=4.5+3=23\)
Vậy \(x=11;y=17;z=23\)
\(B=\left[\frac{x^2-y^2}{xy}-\frac{1}{x+y}\left(\frac{x^2}{y}-\frac{y^2}{x}\right)\right]:\frac{x-y}{x}\)
=>\(B=\left[\frac{x^2-y^2}{xy}-\frac{1}{x+y}\left(\frac{x^3}{xy}-\frac{y^3}{xy}\right)\right].\frac{x}{x-y}\)
=>\(B=\left(\frac{x^2-y^2}{xy}-\frac{1}{x+y}.\frac{x^3-y^3}{xy}\right).\frac{x}{x-y}\)
=>\(B=\left(\frac{x^2-y^2}{xy}-\frac{1}{x+y}.\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{xy}\right).\frac{x}{x-y}\)
=>\(B=\left(\frac{x^2-y^2}{xy}-\frac{x^2-xy+y^2}{xy}\right).\frac{x}{x-y}\)
=>\(B=\frac{x^2-y^2-x^2+xy-y^2}{xy}.\frac{x}{x-y}\)
=>\(B=\frac{xy}{xy}.\frac{x}{x-y}\)
=>\(B=1.\frac{x}{x-y}\)
=>\(B=\frac{x}{x-y}\)