Bài 1: Cho \(\dfrac{3a+b+2c}{2a+c}=\dfrac{a+3b+c}{2b}=\dfrac{a+2b+2c}{b+c}\). Tính giá trị biểu thức A=\(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Bài 2: Cho x; y; z ≠ 0 và \(\dfrac{x+3y-z}{z}=\dfrac{y+3x-x}{x}=\dfrac{z+3x-y}{y}\). Tính P=\(\left(\dfrac{x}{y}+3\right)\left(\dfrac{y}{z}+3\right)\left(\dfrac{z}{x}+3\right)\)
Cứu tui với :<
1.
\(\dfrac{3a+b+2c}{2a+c}=\dfrac{a+3b+c}{2b}=\dfrac{a+2b+2c}{b+c}\)
\(\Leftrightarrow\dfrac{a+b+c+2a+c}{2a+c}=\dfrac{a+b+c+2b}{2b}=\dfrac{a+b+c+b+c}{b+c}\)
\(\Leftrightarrow\dfrac{a+b+c}{2a+c}+1=\dfrac{a+b+c}{2b}+1=\dfrac{a+b+c}{b+c}+1\)
\(\Leftrightarrow\dfrac{a+b+c}{2a+c}=\dfrac{a+b+c}{2b}=\dfrac{a+b+c}{b+c}\)
TH1: \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=-1\)
TH2: \(a+b+c\ne0\)
\(\Rightarrow\dfrac{1}{2a+c}=\dfrac{1}{2b}=\dfrac{1}{b+c}\)
\(\Rightarrow\left\{{}\begin{matrix}2a+c=b+c\\2b=b+c\\\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a=b\\b=c\end{matrix}\right.\) \(\Rightarrow2a=b=c\)
\(\Rightarrow P=\dfrac{\left(a+2a\right)\left(2a+2a\right)\left(2a+a\right)}{a.2a.2a}=9\)
Bài 2 đề sai
Ở phân thức thứ 2 không thể là \(\dfrac{y+3x-x}{x}\)
Bài 2:
\(P=\dfrac{x+3y}{y}\cdot\dfrac{y+3z}{z}\cdot\dfrac{z+3x}{x}=\dfrac{\left(x+3y\right)\left(y+3z\right)\left(z+3x\right)}{xyz}\)
Với \(x+y+z=0\)
\(\dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}\\ \Leftrightarrow\dfrac{x+3y+x+y}{z}=\dfrac{y+3z+y+z}{x}=\dfrac{z+3x+x+z}{y}\\ \Leftrightarrow\dfrac{2\left(x+2y\right)}{z}=\dfrac{2\left(y+2z\right)}{x}=\dfrac{2\left(z+2x\right)}{y}\\ \Leftrightarrow\dfrac{2\left(y-z\right)}{z}=\dfrac{2\left(z-x\right)}{x}=\dfrac{2\left(x-y\right)}{y}\\ \Leftrightarrow\dfrac{2y-2z}{z}=\dfrac{2z-2x}{x}=\dfrac{2x-2y}{y}\\ \Leftrightarrow\dfrac{2y}{z}-2=\dfrac{2z}{x}-2=\dfrac{2x}{y}-2\\ \Leftrightarrow\dfrac{2y}{z}=\dfrac{2z}{x}=\dfrac{2x}{y}\\ \Leftrightarrow\dfrac{y}{z}=\dfrac{z}{x}=\dfrac{x}{y}\Leftrightarrow x=y=z=0\left(\text{trái với GT}\right)\)
Với \(x+y+z\ne0\)
\(\Leftrightarrow\dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}=\dfrac{3\left(x+y+z\right)}{x+y+z}=3\\ \Leftrightarrow\left\{{}\begin{matrix}x+3y-z=3z\\y+3z-x=3x\\z+3x-y=3y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y=4z\\y+3z=4x\\z+3x=4y\end{matrix}\right.\\ \Leftrightarrow P=\dfrac{4x\cdot4y\cdot4z}{xyz}=64\)
Bài 1: Cho \(\dfrac{3a+b+2c}{2a+c}=\dfrac{a+3b+c}{2b}=\dfrac{a+2b+2c}{b+c}\). Tính giá trị biểu thức A=\(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Bài 2: Cho x; y; z ≠ 0 và \(\dfrac{x+3y-z}{z}=\dfrac{y+3x-x}{x}=\dfrac{z+3x-y}{y}\). Tính P=\(\left(\dfrac{x}{y}+3\right)\left(\dfrac{y}{z}+3\right)\left(\dfrac{z}{x}+3\right)\)
Bài 1: Cho \(\dfrac{3a+b+2c}{2a+c}=\dfrac{a+3b+c}{2b}=\dfrac{a+2b+2c}{b+c}\). Tính giá trị biểu thức A=\(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Bài 2: Cho x; y; z ≠ 0 và \(\dfrac{x+3y-z}{z}=\dfrac{y+3x-x}{x}=\dfrac{z+3x-y}{y}\). Tính P=\(\left(\dfrac{x}{y}+3\right)\left(\dfrac{y}{z}+3\right)\left(\dfrac{z}{x}+3\right)\)
Bài 1: Cho a, b, c > 1. CMR: \(a^{\log_bc}+b^{\log_ca}+c^{\log_ab}\ge3\sqrt[3]{abc}\)
Bài 2: Cho các số x, y, z > 0 thoả mãn: \(\dfrac{x\left(y+z-x\right)}{logx}=\dfrac{y\left(z+x-y\right)}{logy}=\dfrac{z\left(x+y-z\right)}{logz}\). CMR: xy.yx = yz.zy = xz.zx
Cho \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\)
C/m rằng: \(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\) với các mẫu số ≠ 0
Áp dụng t/c của DTSBN , ta có :
+, \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\\ =\dfrac{x+2y+z}{a+2b+c+2\left(2a+b-c\right)+4a-4b+c}\\ =\dfrac{x+2y+z}{a+2b+c+4a+2b-2a-2c+4a-4b+c}\\ =\dfrac{x+2y+z}{\left(a+4a+4a\right)+\left(2b+2b-4b\right)+\left(c-2c+c\right)}\\ =\dfrac{x+2y+z}{9a}\left(1\right)\)
+, \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\\ =\dfrac{2x+y-z}{2\left(a+2b+c\right)+2a+b-c-4a+4b+c}\\ =\dfrac{2x+y-z}{2a+4b+2c+2a+b-c-4a+4b+c}\\ =\dfrac{2x+y-z}{\left(2a+2a-4a\right)+\left(4b+b+4b\right)+\left(2c-c+c\right)}\\ =\dfrac{2x+y-z}{9b}\left(2\right)\)
+, \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\\ =\dfrac{4x-4y+z}{4\left(a+2b+c\right)-4\left(2a+b-c\right)++4a-4b+c}\\ =\dfrac{4x-4y+z}{4a+8b+4c-8a-4b+4c+4a-4b+c}\\ =\dfrac{4x-4y+z}{\left(4a-8a+4a\right)+\left(8b-4b-4b\right)+\left(4c+4c+c\right)}\\ =\dfrac{4x-4y+z}{9c}\left(3\right)\)
Từ (1);(2) và (3)
\(\Rightarrow\dfrac{x+2y+z}{9a}=\dfrac{2a+y-z}{9b}=\dfrac{4x-4y+z}{9c}\\ \Rightarrow\dfrac{x+2y+z}{9a}\cdot9=\dfrac{2a+y-z}{9b}\cdot9=\dfrac{4x-4y+z}{9c}\cdot9\\ \Rightarrow\dfrac{x+2y+z}{a}=\dfrac{2a+y-z}{b}=\dfrac{4x-4y+z}{c}\\ \Rightarrow\dfrac{a}{a+2y+z}=\dfrac{b}{2a+y-z}=\dfrac{c}{4x-4y+z}\left(đpcm\right)\)
Đặt \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=k\left(a+2b+c\right)\\y=k\left(2a+b-c\right)\\z=k\left(4a-4b+c\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{x+2y+z}=\dfrac{a}{k\left(a+2b+c\right)+2k\left(2a+b-c\right)+k\left(4a-4b+c\right)}=\dfrac{a}{k.9a}=\dfrac{1}{9k}\)
Tượng tự:
\(\dfrac{b}{2x+y-z}=\dfrac{b}{9bk}=\dfrac{1}{9k}\) ; \(\dfrac{c}{4x-4y+z}=\dfrac{c}{9k.c}=\dfrac{1}{9k}\)
\(\Rightarrow\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\)
Cho\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\)
C/m rằng:\(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\) với các mẫu số ≠ 0
Bài 1: a;b;c > 0
Chứng minh : \(\dfrac{a}{3a+b+c}+\dfrac{b}{3b+a+c}+\dfrac{c}{3c+a+b}\le\dfrac{3}{5}\)
Bài 2: x;y;z \(\ne\) 1 và xyz = 1
Chứng minh : \(\dfrac{x^2}{\left(x-1\right)^2}+\dfrac{y^2}{\left(y-1\right)^2}+\dfrac{z^2}{\left(z-1\right)^2}\ge1\)
1.
Áp dụng BĐT Cauchy-Schwarz:
\(\dfrac{a}{2a+a+b+c}=\dfrac{a}{25}.\dfrac{\left(2+3\right)^2}{2a+a+b+c}\le\dfrac{a}{25}\left(\dfrac{2^2}{2a}+\dfrac{3^2}{a+b+c}\right)=\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{a}{a+b+c}\)
Tương tự:
\(\dfrac{b}{3b+a+c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{b}{a+b+c}\)
\(\dfrac{c}{a+b+3c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{c}{a+b+c}\)
Cộng vế:
\(VT\le\dfrac{6}{25}+\dfrac{9}{25}.\dfrac{a+b+c}{a+b+c}=\dfrac{3}{5}\)
Dấu "=" xảy ra khi \(a=b=c\)
2.
Đặt \(\dfrac{x}{x-1}=a;\dfrac{y}{y-1}=b;\dfrac{z}{z-1}=c\)
Ta có: \(\dfrac{x}{x-1}=a\Rightarrow x=ax-a\Rightarrow a=x\left(a-1\right)\Rightarrow x=\dfrac{a}{a-1}\)
Tương tự ta có: \(y=\dfrac{b}{b-1}\) ; \(z=\dfrac{c}{c-1}\)
Biến đổi giả thiết:
\(xyz=1\Rightarrow\dfrac{abc}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}=1\)
\(\Rightarrow abc=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
\(\Rightarrow ab+bc+ca=a+b+c-1\)
BĐT cần chứng minh trở thành:
\(a^2+b^2+c^2\ge1\)
\(\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\ge1\)
\(\Leftrightarrow\left(a+b+c\right)^2-2\left(a+b+c-1\right)\ge1\)
\(\Leftrightarrow\left(a+b+c-1\right)^2\ge0\) (luôn đúng)
Bài 1 : Cho các số thực a,b,c khác 0 thỏa mãn \(a+b+c=2;a^2+b^2+c^2=4\) và \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)
Chứng minh rằng : xy+yz+zx=0
Bài 2 : Cho x khác -1;0;1 thỏa mãn \(\dfrac{a}{x-1}=\dfrac{b}{x}=\dfrac{c}{x+1}\) Chứng minh rằng : \(4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\)
Bài 3 : Cho các số thực a,b,c khác 0 thỏa mãn \(\dfrac{x}{a+2b-c}=\dfrac{y}{2a+b+c}=\dfrac{z}{4b+c-4a}\) . Chứng minh rằng : \(\dfrac{a}{x+2y-z}=\dfrac{b}{2x+b+c}=\dfrac{c}{4y+z-4x}\)
GIÚP MÌNH ĐI CHIỀU 1 GIỜ ĐI HOK RỒI !!!
Bài 1 : Thực hiện phép tính: P=\(\sqrt{\dfrac{9}{25}}+2018^0+\left[-0.4\right]\)
Tìm x thỏa mãn :\((\sqrt{x}-4)-(\left[x+2\right]-1).\left(x^2-3\right)=0\)
Bài 2 :
a, Tìm x;y biết : \(\dfrac{x+y}{2017}=\dfrac{xy}{2018}=\dfrac{x-y}{2019}\)
b.Cho x; y; z;a;b;c thỏa mãn: \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\). CMR:\(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\) ( với các điều kiện các mẫu thức khác 0)
