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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!
a)
\(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=-\dfrac{1}{4}-y\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}-\dfrac{1}{3}+x=-\dfrac{1}{4}-y\\\dfrac{1}{2}-\dfrac{1}{3}+x=\dfrac{1}{4}+y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=-\dfrac{5}{12}\\x-y=\dfrac{1}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{6}\\y=-\dfrac{1}{4}\end{matrix}\right.\)
b)\(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)
ta thấy : \(\left|x-y\right|\ge0\\ \left|y+\dfrac{9}{25}\right|\ge0\)\(\Rightarrow\left|x-y\right|+\left|y+\dfrac{9}{25}\right|\ge0\)
đẳng thửc xảy ra khi : \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Rightarrow x=y=-\dfrac{9}{25}\)
vậy \(\left(x;y\right)=\left(-\dfrac{9}{25};-\dfrac{9}{25}\right)\)
c) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)
ta thấy \(\left(\dfrac{1}{2}x-5\right)^{20}\:và\:\left(y^2-\dfrac{1}{4}\right)^{10}\) là các lũy thừa có số mũ chẵn
\(\Rightarrow\:\)\(\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\ \left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)
đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=10\\\left[{}\begin{matrix}y=-\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
vậy cặp số x,y cần tìm là \(\left(10;\dfrac{1}{2}\right)\:hoặc\:\left(10;-\dfrac{1}{2}\right)\)
d)
\(\left|x\left(x^2-\dfrac{5}{4}\right)\right|=x\\ \Leftrightarrow x\left(x^2-\dfrac{5}{4}\right)=x\left(vì\:x\ge0\right)\\ \Leftrightarrow x\left(x^2-\dfrac{9}{4}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{9}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
vậy x cần tìm là \(-\dfrac{3}{2};0;\dfrac{3}{2}\)
e)\(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
ta thấy: \(x^2\ge0;\left(y-\dfrac{1}{10}\right)^4\ge0\)
\(\Rightarrow x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\)
đẳng thức xảy ra khi: \(\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
vậy cặp số cần tìm là \(0;\dfrac{1}{10}\)
a) \(\left|x-\dfrac{4}{11}\right|+\left|5+y\right|=0\)
<=>\(\left[{}\begin{matrix}x-\dfrac{4}{11}=0\\5+y=0\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}x=\dfrac{4}{11}\\y=-5\end{matrix}\right.\)
phần b, c tương tự
a)
Ta thấy \(\left\{\begin{matrix} |x+\frac{19}{5}|\geq 0\\ |y+\frac{1890}{1975}|\geq 0\\ |z-2005|\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{Z}\)
\(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|\geq 0\)
Do đó, để \(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|=0\) thì :
\(\left\{\begin{matrix} |x+\frac{19}{5}|= 0\\ |y+\frac{1890}{1975}|= 0\\ |z-2005|=0\end{matrix}\right.\Rightarrow x=\frac{-19}{5}; y=\frac{-1890}{1975}; z=2005\)
b) Giống phần a, vì trị tuyệt đối của một số luôn không âm nên để tổng các trị tuyệt đối bằng $0$ thì:
\(\left\{\begin{matrix} |x+\frac{3}{4}|=0\\ |y-\frac{1}{5}|=0\\ |x+y+z|=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=-\frac{3}{4}\\ y=\frac{1}{5}\\ z=-(x+y)=\frac{11}{20}\end{matrix}\right.\)
c) \(\frac{16}{2^x}=1\Rightarrow 16=2^x\)
\(\Leftrightarrow 2^4=2^x\Rightarrow x=4\)
d) \((2x-1)^3=-27=(-3)^3\)
\(\Rightarrow 2x-1=-3\)
\(\Rightarrow 2x=-2\Rightarrow x=-1\)
e) \((x-2)^2=1=1^2=(-1)^2\)
\(\Rightarrow \left[\begin{matrix} x-2=1\\ x-2=-1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=1\end{matrix}\right.\)
f) \((x+\frac{1}{2})^2=\frac{4}{25}=(\frac{2}{5})^2=(\frac{-2}{5})^2\)
\(\Rightarrow \left[\begin{matrix} x+\frac{1}{2}=\frac{2}{5}\\ x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-1}{10}\\ x=\frac{-9}{10}\end{matrix}\right.\)
g) \((x-1)^2=(x-1)^6\)
\(\Leftrightarrow (x-1)^6-(x-1)^2=0\)
\(\Leftrightarrow (x-1)^2[(x-1)^4-1]=0\)
\(\Rightarrow \left[\begin{matrix} (x-1)^2=0\\ (x-1)^4=1=(-1)^4=1^4\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x-1=-1\\ x-1=1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x=0\\ x=2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x=\left\{0;1;2\right\}\)
a)\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1980}{1975}\right|+\left|z-2004\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{1980}{1975}\right|=0\\\left|z-2004\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{1980}{1975}\\z=2004\end{matrix}\right.\)
b) \(\left|\dfrac{3}{4}+x\right|+\left|-\dfrac{1}{5}+y\right|+\left|x+y+z\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|\dfrac{3}{4}+x\right|=0\\\left|-\dfrac{1}{5}+y\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\z=\dfrac{11}{20}\end{matrix}\right.\)
\(1)\)
\(VT=\left(\left|x-6\right|+\left|2022-x\right|\right)+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)
\(\ge\left|x-6+2022-x\right|+\left|0\right|+\left|0\right|+\left|0\right|=2016\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-6\right)\left(2022-x\right)\ge0\left(1\right)\\x-10=y-2014=z-2015=0\left(2\right)\end{cases}}\)
\(\left(2\right)\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=10\\y=2014\\z=2015\end{cases}}\)
\(\left(1\right)\)
TH1 : \(\hept{\begin{cases}x-6\ge0\\2022-x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge6\\x\le2022\end{cases}\Leftrightarrow}6\le x\le2022}\) ( nhận )
TH2 : \(\hept{\begin{cases}x-6\le0\\2022-x\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le6\\x\ge2022\end{cases}}}\) ( loại )
Vậy \(x=10\)\(;\)\(y=2014\) và \(z=2015\)
\(2)\)
\(VT=\left|x-5\right|+\left|1-x\right|\ge\left|x-5+1-x\right|=\left|-4\right|=4\)
\(VP=\frac{12}{\left|y+1\right|+3}\le\frac{12}{3}=4\)
\(\Rightarrow\)\(VT\ge VP\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-5\right)\left(1-x\right)\ge0\left(1\right)\\\left|y+1\right|=0\left(2\right)\end{cases}}\)
\(\left(1\right)\)
TH1 : \(\hept{\begin{cases}x-5\ge0\\1-x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge5\\x\le1\end{cases}}}\) ( loại )
TH2 : \(\hept{\begin{cases}x-5\le0\\1-x\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le5\\x\ge1\end{cases}\Leftrightarrow}1\le x\le5}\) ( nhận )
\(\left(2\right)\)\(\Leftrightarrow\)\(y=-1\)
Vậy \(1\le x\le5\) và \(y=-1\)
a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=1.6=\dfrac{8}{5}\)
=>x+4/15=8/5 hoặc x+4/15=-8/5
=>x=4/3 hoặc x=-28/15
c: =>x-y=0 và y+9/25=0
=>x=y=-9/25
d: =>-1/3<x-3/5<1/3
=>4/15<x<14/15
e: =>|x+5,5|>5,5
=>x+5,5>5,5 hoặc x+5,5<-5,5
=>x>0 hoặc x<-11
2: |x-y|+|y+9/25|=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)
\(\dfrac{x-2}{2}=\dfrac{y-4}{3}=\dfrac{z-8}{5}\)
\(\Rightarrow\dfrac{x-2}{2}+2=\dfrac{y-4}{3}+2=\dfrac{z-8}{5}+2\)
\(\Rightarrow\dfrac{x+2}{2}=\dfrac{y+2}{3}=\dfrac{z+2}{5}\)
\(\Rightarrow\left(\dfrac{x+2}{2}\right)^2=\left(\dfrac{y+2}{3}\right)^2=\left(\dfrac{z+2}{5}\right)^2\)
\(\Rightarrow\dfrac{\left(x+2\right)^2}{4}=\dfrac{\left(y+2\right)^2}{9}=\dfrac{\left(z+2\right)^2}{25}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{\left(x+2\right)^2}{4}=\dfrac{\left(y+2\right)^2}{9}=\dfrac{\left(z+2\right)^2}{25}=\dfrac{3.\left(y+2\right)^2}{27}\dfrac{\left(x+2\right)^2+3\left(y+2\right)^2-\left(z+2\right)^2}{4+27-25}=\dfrac{24}{6}=4\)\(\Rightarrow\left\{{}\begin{matrix}\left(x+2\right)^2=16\\\left(y+2\right)^2=36\\\left(z+2\right)^2=100\end{matrix}\right.\)
Bạn chia trường hợp rồi tìm x,y,z nhé
vì \(\left(4x^2-4x+1\right)^{2022}\ge0\left(\forall x\right)\),\(\left(y^2-\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}\ge0\left(\forall y\right)\),\(\left|x+y+z\right|\ge0\)
mà \(\left(4x^2-4x+1\right)^{2022}+\left(y^2+\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}+\left|x+y-z\right|=0\)
=>\(\left\{{}\begin{matrix}4x^2-4x+1=0\\y^2+\dfrac{4}{5}y+\dfrac{4}{25}=0\\x+y-z=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-1=0\\y+\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\\dfrac{1}{2}-\dfrac{2}{5}-z=0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)
KL: vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)