Cho \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\) và \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\). CMR: \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
1, Cho x; y; z ≠0 và \(\dfrac{1}{x}\) + \(\dfrac{1}{y}\)+ \(\dfrac{1}{z}\)=\(\dfrac{2}{2x+y+2z}\). Cmr: (2x+y)(y+2z)(z+x)= 0
2, Cho \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\). Cmr: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\)
Gấp ạ, ai giúp mình với!!!!
2: Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{c+a}+\dfrac{c\left(a+b+c\right)}{a+b}-a-b-c=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c-a-b-c=0\)
1: Sửa đề: Cho \(x,y,z\ne0\) và \(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}=\dfrac{2}{2x+y+2z}\).
CM:....
Đặt 2x = x', 2z = z'.
Ta có: \(\dfrac{2}{x'}+\dfrac{2}{y}+\dfrac{2}{z'}=\dfrac{2}{x'+y+z'}\)
\(\Leftrightarrow\dfrac{1}{x'}+\dfrac{1}{y}+\dfrac{1}{z'}=\dfrac{1}{x'+y+z'}\)
\(\Leftrightarrow\dfrac{1}{x'}-\dfrac{1}{x'+y+z'}+\dfrac{1}{y}+\dfrac{1}{z'}=0\)
\(\Leftrightarrow\dfrac{y+z'}{x'\left(x'+y+z'\right)}+\dfrac{y+z'}{yz'}=0\)
\(\Leftrightarrow\dfrac{\left(y+z'\right)\left(yz'+x'^2+x'y+x'z'\right)}{x'yz'\left(x'+y+z'\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x'+y\right)\left(y+z'\right)\left(z'+x'\right)}{x'yz'\left(x'+y+z'\right)}=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(2z+2x\right)=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(z+x\right)=0\left(đpcm\right)\)
1) Rút gọn bt:
(x+y+z)3+(x-y-z)3+(y-x-z)3+(z-y-x)3
2)Tìm x,y,z t/m: 9x2+y2+2z2-18x+4z-6y+20=0
3)Cho \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\)=1 và \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}\)=0 . CMR:
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)=1
1)cho Q=\(\dfrac{a^4+a^3-a^2-2a-2}{a^4+2a^3-a^2-4a-2}\)
Tìm GTNN của Q
2)cho \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\) và \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
CMR: \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
\(1,Q=\dfrac{a^4-2a^2+a^3-2a+a^2-2}{a^4-2a^2+2a^3-4a+a^2-2}\\ Q=\dfrac{\left(a^2-2\right)\left(a^2+a+1\right)}{\left(a^2-2\right)\left(a^2+2a+1\right)}=\dfrac{a^2+a+1}{a^2+2a+1}\)
\(Q=\dfrac{x^2+x+1}{\left(x+1\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{x^2+x+1-\dfrac{3}{4}x^2-\dfrac{3}{2}x-\dfrac{3}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}\\ Q=\dfrac{\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}=\dfrac{\dfrac{1}{4}\left(x-1\right)^2}{\left(x+1\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\\ Q_{min}=\dfrac{3}{4}\Leftrightarrow x=1\)
\(2,\text{Từ GT }\Leftrightarrow\dfrac{ayz+bxz+czy}{xyz}=0\\ \Leftrightarrow ayz+bxz+czy=0\\ \text{Ta có }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ca}\right)=0\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{cxy+ayz+bzx}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{0}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
cho \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\) và \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\), tính A \(=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)
Từ giả thiết : \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ca}\right)=1\)
\(\Rightarrow A+2.\left(\dfrac{xyc+yza+xzb}{abc}\right)=1\left(1\right)\)
Mà theo gt : \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Rightarrow ayz+bzx+cxy=0\)
Do đó : \(\left(1\right)=A=1\)
Đặt $ X = a - b; Y = b - c; Z = c - a \Rightarrow X + Y + Z = 0$
Với X + Y + Z = 0, ta chứng minh được :
$ ( \dfrac{1}{X} + \dfrac{1}{Y} + \dfrac{1}{Z} )^2 = \dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2}$
Thật vậy, ta có :
$ ( \dfrac{1}{X} + \dfrac{1}{Y} + \dfrac{1}{Z} )^2 = \dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2} + \dfrac{2}{XY} + \dfrac{2}{YZ} + \dfrac{2}{ZX}$
$ = \dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2} + 2.\dfrac{X + Y + Z}{XYZ}$
$ = \dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2}$ ( do X + Y + Z = 0)
$ \Rightarrow \sqrt{\dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2}} = \sqrt{( \dfrac{1}{X} + \dfrac{1}{Y} + \dfrac{1}{Z} )^2} = |\dfrac{1}{X} + \dfrac{1}{Y} + \dfrac{1}{Z}|$
Suy ra : $ \sqrt{\dfrac{1}{(a - b)^2} + \dfrac{1}{(b - c)^2} +\dfrac{1}{( c - a)^2}} = |\dfrac{1}{a - b} + \dfrac{1}{b - c} + \dfrac{1}{c - a}|$
Do a, b, c là số hữu tỷ nên $|\dfrac{1}{a - b} + \dfrac{1}{b - c} + \dfrac{1}{c - a}|$ cũng là số hữu tỷ. Ta có điều phải chứng minh.
1.
a) CMR: Nếu a+b+c=0 thì \(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{c^2+a^2-b^2}=0\)
b) Nếu \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\) thì:
\(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+2y-z}=\dfrac{c}{4x-4y+z}\)
2. Cho \(\dfrac{x}{x^2+x+1}=a\) .Tính \(M=\dfrac{x^2}{x^4-x^2+1}\)
Cho a,b,c và x,y,z khác nhau và khác 0
CMR: \(\text{Nếu }\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0,\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\text{Thì }\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
+) \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Rightarrow\dfrac{ayz}{xyz}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\)
\(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Rightarrow ayz+bxz+cxy=0\)
+) \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\dfrac{xy}{ab}+2\dfrac{xz}{ac}+2\dfrac{yz}{bc}=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy}{abc}+\dfrac{bxz}{abc}+\dfrac{ayz}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{ayz+bxz+cxy}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{0}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)Bài 1: a;b;c > 0 và abc = 1
Chứng minh : \(\dfrac{a}{b^4+c^4+a}+\dfrac{b}{a^4+c^4+b}+\dfrac{c}{a^4+b^4+c}\le1\)
Bài 2: x;y;z > 0 và x + y + z = 2
Chứng minh : \(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)
1.
Ta có:
\(x^4+y^4\ge\dfrac{1}{2}\left(x^2+y^2\right)^2=\dfrac{1}{2}\left(x^2+y^2\right)\left(x^2+y^2\right)\ge\left(x^2+y^2\right)xy\)
Đặt vế trái của BĐT cần chứng minh là P, áp dụng bồ đề vừa chứng minh ta có:
\(P\le\dfrac{a.abc}{bc\left(b^2+c^2\right)+a.abc}+\dfrac{b.abc}{ca\left(c^2+a^2\right)+b.abc}+\dfrac{c.abc}{ab\left(a^2+b^2\right)+c.abc}\)
\(P\le\dfrac{a^2.bc}{bc\left(a^2+b^2+c^2\right)}+\dfrac{b^2.ac}{ca\left(a^2+b^2+c^2\right)}+\dfrac{c^2.ab}{ab\left(a^2+b^2+c^2\right)}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
2.
\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}=1\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{2}{3}\)
Cho: \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=0\) và \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2\). Tính giá trị của biểu thức: \(\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}\)
Bài này dễ thôi:vv
Theo đề ta có: \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=0\Leftrightarrow\dfrac{xbc+yac+zab}{abc}=0\Leftrightarrow xbc+yac+zab=0\)
Lại có:\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2\Rightarrow\left(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}\right)^2=4\)
=>\(\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{ab}{xy}+\dfrac{bc}{yz}+\dfrac{ca}{xz}\right)=4\)
=>\(\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{abz+bcx+cay}{xyz}\right)=4\)
=>\(\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2.0=4\Rightarrow\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}=2\)
Vậy...