a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\)( do \(x^2\ge0,\left(y-\dfrac{1}{10}\right)^4\ge0\))

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)

b) \(\left(\dfrac{1}{2}.x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\)( do \(\left(\dfrac{1}{2}x-5\right)^{20}\ge0,\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\))

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)

\(a,\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\\ b,\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\end{matrix}\right.\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)

Mà \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)

\(\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)

a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)

Mà \(x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\forall x;y\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=0\\\left(y-\dfrac{1}{10}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(0;\dfrac{1}{10}\right)\)

b) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)

Mà \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\forall x;y\)

\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}=0\\\left(y^2-\dfrac{1}{4}\right)^{10}=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=10\\\left[{}\begin{matrix}y=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(10;\dfrac{1}{2}\right);\left(10;-\dfrac{1}{2}\right)\right\}\)

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)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: \(\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)

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 \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)

Bài 1:

a)\(\left(5x+1\right)^2=\dfrac{36}{49}\)

\(\Leftrightarrow\left(5x+1\right)^2=\left(\dfrac{6}{7}\right)^2=\left(-\dfrac{6}{7}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5x=-\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\)

Bài 2:

a)\(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)

Dễ thấy: \(\left\{{}\begin{matrix}x^2\ge0\\\left(y-\dfrac{1}{10}\right)^4\ge0\end{matrix}\right.\)

\(\Rightarrow x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\)

Xảy ra khi \(\left\{{}\begin{matrix}x^2=0\\\left(y-\dfrac{1}{10}\right)^4=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)

b)\(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{40}\le0\)

Dễ thấy: \(\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\\left(y^2-\dfrac{1}{4}\right)^{40}\ge0\end{matrix}\right.\)

\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{40}\ge0\)

Mà \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{40}\le0\)

Xảy ra khi \(\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}=0\\\left(y^2-\dfrac{1}{4}\right)^{40}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)

\(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\) \(\left(1\right)\)

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 ...

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!

1: =>3x+1=4

=>3x=3

hay x=1

2: \(\Leftrightarrow172\cdot x^2=\dfrac{1}{2^3}+\dfrac{7^9}{98^3}=\dfrac{1}{2^3}+\dfrac{7^9}{7^6\cdot2^3}\)

\(\Leftrightarrow172\cdot x^2=\dfrac{1}{2^3}+\dfrac{7^3}{2^3}=\dfrac{344}{2^3}\)

\(\Leftrightarrow x^2=\dfrac{1}{4}\)

=>x=1/2 hoặc x=-1/2

3: \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{9}=\dfrac{4}{9}\\x-\dfrac{2}{9}=-\dfrac{4}{9}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{9}\end{matrix}\right.\)

4: =>x+2=0 và y-1/10=0

=>x=-2 và y=1/10