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Đặt \(\dfrac{x}{2015}=\dfrac{y}{2016}=\dfrac{z}{2017}=k\)
\(\Rightarrow x=2015k;y=2016k;z=2017k\) \(\left(1\right)\)
Thay (1) vào đề bài ta được:
\(\left(2015k-2017k\right)^3:\left[\left(2015k-2016k^2\right)\left(2016k-2017k\right)\right]\)
\(=\left(-2k\right)^3:\left[-k^2\left(-k\right)\right]\)
\(=-8k^3:\left(-k\right)^3\)
\(=8\)
Vậy \(\left(x-z\right)^3:\left[\left(x-y\right)^2\left(y-z\right)\right]=8.\)
a) Tính chất dãy tỉ số bằng nhau: \(\dfrac{x+y}{2014}=\dfrac{x-y}{2016}=\dfrac{x+y+x-y}{2014+2016}=\dfrac{2x}{4030}=\dfrac{x}{2015}\)
\(\dfrac{x+y}{2014}=\dfrac{x-y}{2016}=\dfrac{x+y-x+y}{2014-2016}=\dfrac{2y}{-2}=\dfrac{y}{-1}\)
Nên: \(\dfrac{x}{2015}=\dfrac{y}{-1}=\dfrac{xy}{2015}\)
Xét: \(\left\{{}\begin{matrix}\dfrac{x}{2015}=\dfrac{xy}{2015}\Leftrightarrow2015x=2015xy\Leftrightarrow y=1\\\dfrac{y}{-1}=\dfrac{xy}{2015}\Leftrightarrow2015y=-1xy\Leftrightarrow2015=-1x\Leftrightarrow x=-2015\end{matrix}\right.\)
2) \(VT=\left|x-6\right|+\left|x-10\right|+\left|x-2022\right|+\left|y-2014\right|+\left|z-2015\right|\)
\(VT=\left|x-6\right|+\left|2022-x\right|+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)
\(VT\ge\left|x-6+2022-x\right|+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)
\(VT\ge2016+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\ge2016=VP\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}6\le x\le2022\\x=10\\y=2014\\z=2015\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=2014\\z=2015\end{matrix}\right.\)
a, \(\left|3x-4\right|+\left|3y+5\right|=0\)
Ta có :
\(\left|3x-4\right|\ge0\forall x;\left|3y+5\right|\ge0\forall x\\ \)
\(\Rightarrow\left|3x-4\right|+\left|3y+5\right|\ge0\forall x\\ \Rightarrow\left\{{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-\dfrac{5}{3}\end{matrix}\right.\\ Vậy.........\)
b, \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)
Ta có :
\(\left|x+\dfrac{19}{5}\right|\ge0\forall x;\left|y+\dfrac{1890}{1975}\right|\ge0\forall y;\left|z-2004\right|\ge0\forall z \)
\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{1890}{1975}\\z=2004\end{matrix}\right.\\ Vậy............\)
c, \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)
Ta có : \(\left|x+\dfrac{9}{2}\right|\ge0\forall x;\left|y+\dfrac{4}{3}\right|\ge0\forall y;\left|z+\dfrac{7}{2}\right|\ge0\forall z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\\ Vậy............\)
d, \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)
Ta có :
\(\left|x+\dfrac{3}{4}\right|\ge0\forall x;\left|y-\dfrac{1}{5}\right|\ge0\forall y;\left|x+y+z\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\z=0-\dfrac{1}{5}+\dfrac{3}{4}=\dfrac{11}{20}\end{matrix}\right.\\ Vậy.......\)
e, Câu cuối bn làm tương tự như câu a, b, c nhé!
@ Mashiro Shiina
@Akai Haruma
@Nguyễn Thanh Hằng
@Đẹp Trai Không Bao Giờ Sai
Ta có :
\(\dfrac{x+y-z}{z}=\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}\\ \Leftrightarrow\dfrac{x+y+z}{z}=\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}\left(cùngcộngthêm2\right)\)
TH1: \(x+y+z\ne0\)
\(\Rightarrow x=y=z\)
\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)\\ =2\cdot2\cdot2=8\)
TH2: \(x+y+z=0\Rightarrow\left\{{}\begin{matrix}x=-\left(y+z\right)\\y=-\left(x+z\right)\\z=-\left(y+x\right)\end{matrix}\right.\)(*)
\(\Rightarrow P=\left(1+\dfrac{-\left(y+z\right)}{y}\right)\left(1+\dfrac{-\left(z+x\right)}{z}\right)\left(1+\dfrac{-\left(x+y\right)}{z}\right)\\ =\left(1-1-\dfrac{z}{y}\right)\left(1-1-\dfrac{x}{z}\right)\left(1-1-\dfrac{y}{z}\right)\\ =\left(-\dfrac{z}{y}\right)\left(-\dfrac{x}{z}\right)\left(-\dfrac{y}{z}\right)\\ =-1\)
Vậy P=8 hoặc P=-1
1) Phân số đầu nhân 2.
_ Phân số thứ 2 nhân 3, p/s thứ 3 giữ nguyên.
_ Lấy phân số đầu + p/s thứ 2 - p/s thứ 3.
_ Dựa vào dãy tỉ số bằng nhau tìm x, y, z.
2) \(x-y-z=0\Rightarrow x=y+z\)
Khi đó thay vào B được:
\(B=\left(1-\dfrac{z}{y+z}\right)\left(1-\dfrac{y+z}{y}\right)\left(1+\dfrac{y}{z}\right)\)
\(=\dfrac{y}{y+z}.\dfrac{z}{y}.\dfrac{y+z}{z}\)
\(=1\)
Vậy B = 1.
TH1: \(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(y+z=-x\)
\(x+z=-y\)
\(\Rightarrow M=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8xyz}=\dfrac{-xyz}{8xyz}=\dfrac{-1}{8}\)
TH2: \(x+y+z\ne0\)
\(\Rightarrow2x+2y-z=3\)
\(\Rightarrow2x+2y=4z\)
\(\Rightarrow x+y=2z\)
\(x+z=2y\)
\(y+z=2x\)
\(\Rightarrow M=\dfrac{2z.2y.2x}{8xyz}=1\)
Vậy: \(M=\dfrac{-1}{8}\) hoặc \(1\)
Ta có \(\dfrac{2x+2y-z}{z}=\dfrac{2x+2z-y}{y}=\dfrac{2y+2z-x}{x}\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\Rightarrow\dfrac{2x+2y-z}{z}=\dfrac{2x+2z-y}{y}=\dfrac{2y+2z-x}{x}=\dfrac{3\left(x+y+z\right)}{x+y+z}=3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x+2y-z}{z}=3\\\dfrac{2x+2z-y}{y}=3\\\dfrac{2y+2z-x}{x}=3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x+2y-z=3z\\2x+2z-y=3y\\2y+2z-x=3x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x+2y=4z\\2x+2z=4y\\2y+2z=4x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=2z\\x+z=2y\\y+z=2x\end{matrix}\right.\)
Ta có \(M=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8xyz}\)
\(\Rightarrow M=\dfrac{2x.2y.2z}{8xyz}=\dfrac{8xyz}{8xyz}=1\)
Vậy \(M=1\)
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\(\dfrac{x}{2015}=\dfrac{y}{2016}=\dfrac{z}{2017}=\dfrac{x-y}{-1}=\dfrac{y-z}{-1}=\dfrac{x-z}{-1}\Leftrightarrow x-z=x-y=y-z\Rightarrow x=y=z.\)