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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 \(0;\dfrac{1}{10}\)
Giải:
a) \(x+\left(-\dfrac{31}{12}\right)^2=\left(\dfrac{49}{12}\right)^2-x=y\)
\(\Leftrightarrow x+\left(-\dfrac{31}{12}\right)^2=\left(\dfrac{49}{12}\right)^2-x\)
\(\Leftrightarrow x+\left(-\dfrac{31}{12}\right)^2-\left(\dfrac{49}{12}\right)^2+x=0\)
\(\Leftrightarrow2x+\left(-\dfrac{31}{12}\right)^2-\left(\dfrac{49}{12}\right)^2=0\)
\(\Leftrightarrow2x+\dfrac{\left(-31\right)^2}{12^2}-\dfrac{49^2}{12^2}=0\)
\(\Leftrightarrow2x+\dfrac{\left(-31\right)^2-49^2}{144}=0\)
\(\Leftrightarrow2x+\dfrac{961-2401}{144}=0\)
\(\Leftrightarrow2x+\dfrac{-1440}{144}=0\)
\(\Leftrightarrow2x+\left(-10\right)=0\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=5\)
Mà \(x+\left(-\dfrac{31}{12}\right)^2=y^2\)
\(\Leftrightarrow5+\dfrac{961}{144}=y^2\)
\(\Leftrightarrow y^2=\dfrac{1681}{144}\)
\(\Leftrightarrow y=\pm\dfrac{41}{12}\)
Vậy ...
b) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
Vì \(\left(\dfrac{1}{2}x-5\right)^{20}\ge0;\forall x\)
và \(\left(y^2-\dfrac{1}{4}\right)^{10}\ge0;\forall y\)
\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
\(\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
Chúc bạn học tốt!
Với mọi x,y ta có :
\(\left\{{}\begin{matrix}\left(12x-5\right)^{20}\ge0\\\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(12x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)
Mà \(\left(12x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(12x-5\right)^{20}=0\\\left(y^2-\dfrac{1}{4}\right)^{10}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{12}\\\left[{}\begin{matrix}y=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy..
\(x^2+\left(y-\dfrac{1}{10}\right)^{2018}=0\\ \Leftrightarrow x^2+\left[\left(y-\dfrac{1}{10}\right)^{1009}\right]^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=0\\\left(y-\dfrac{1}{10}\right)^{1009}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
a)Với mọi x thuộc R: \(x^2\ge0;\left(y-\dfrac{1}{10}\right)^4\ge0\Leftrightarrow x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\)
\("="\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
b) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\forall x\in R\)
mà: \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
Xảy ra khi: \(\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
a: =>|3x-5|=|x+2|
=>3x-5=x+2 hoặc 3x-5=-x-2
=>2x=7 hoặc 4x=3
=>x=7/2 hoặc x=3/4
b: \(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
c: \(\Leftrightarrow\left|3x-5\right|=x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(3x-5-x+2\right)\left(3x-5+x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(2x-3\right)\left(4x-7\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
d: \(\dfrac{11}{2}\le\left|x\right|< \dfrac{17}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{11}{2}< =x< \dfrac{17}{2}\\-\dfrac{17}{2}< x< =-\dfrac{11}{2}\end{matrix}\right.\)
câu 1 \(A=\dfrac{3^2}{5^2}.5^2-\dfrac{9^3}{4^3}:\dfrac{3^3}{4^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{\left(3^2\right)^3}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{3^6}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}=3^2-3^3+\dfrac{1}{2}=-18+\dfrac{1}{2}=-\dfrac{35}{2}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{22}.2\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{4^4}{8^2}\right)^{2009}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{11}\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{\left(2^2\right)^4}{\left(2^3\right)^2}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{1}{2^2}.\dfrac{2^8}{2^6}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{2^8}{2^8}\right)^{2009}\)
\(B=1^{2010}-1^{2009}=1-1=0\)
câu 2
a) \(2x-\dfrac{5}{4}=\dfrac{20}{15}\)
\(\Leftrightarrow2x=\dfrac{4}{3}+\dfrac{5}{4}\)
\(\Leftrightarrow2x=\dfrac{31}{12}\)
\(\Leftrightarrow x=\dfrac{31}{24}\)
b) \(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
mình làm lại câu b) nha
b) |x-3|=-4
th1: x-3=-4
x=3+(-4)
x=-1
th2: x-3=4
x=3+4
x=7
b) \(\left|x-3\right|=-4\)
t/h1:\(x-3=-4\)
\(x=3-\left(-4\right)\)
\(x=7\)
t/h2:\(x-3=4\)
\(x=3-4\)
\(x=-1\)
a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
\(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\) \(\left(1\right)\)
Ta có :
\(\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\) \(\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)
Mặt khác :
\(\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=10\\\left[{}\begin{matrix}y=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy ...