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a: Ta có: \(\left(x-2\right)\left(x+1\right)< 0\)

\(\Leftrightarrow-1< x< 2\)

b: Ta có: \(\left(x+\dfrac{1}{3}\right)\left(x-1\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le-\dfrac{1}{3}\\x\ge1\end{matrix}\right.\)

22 tháng 12 2017

a)

\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)

b)

\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)

1 tháng 1 2018

tiếp đi bạn

24 tháng 7 2017

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

24 tháng 7 2017

b) \(\left|x-3\right|=-4\)

t/h1:\(x-3=-4\)

\(x=3-\left(-4\right)\)

\(x=7\)

t/h2:\(x-3=4\)

\(x=3-4\)

\(x=-1\)

11 tháng 9 2017

\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)

=>Trong 2 số phải có 1 số âm và 1 số dương

\(2-x>\dfrac{4}{5}-x\)

=>\(\dfrac{4}{5}< x< 2\)

Vậy...

a: \(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}\left(x+1\right)\)

=>\(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}x+\dfrac{1}{2}\)

=>\(-\dfrac{3}{2}x-\dfrac{1}{2}x=\dfrac{1}{2}-\dfrac{1}{4}\)

=>\(-2x=\dfrac{1}{4}\)

=>\(2x=-\dfrac{1}{4}\)

=>\(x=-\dfrac{1}{4}:2=-\dfrac{1}{8}\)

b: ĐKXĐ: x>=0

\(\left(6-3\sqrt{x}\right)\left(\left|x\right|-7\right)=0\)

=>\(\left\{{}\begin{matrix}6-3\sqrt{x}=0\\\left|x\right|-7=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3\sqrt{x}=6\\\left|x\right|=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-7\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)

4 tháng 12 2023

bài nào cũng thấy Phước Thịnh :)

21 tháng 1 2018

đề sai

29 tháng 10 2021

3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

27 tháng 11 2017

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 \(\left(x;y\right)=\left(-\dfrac{9}{25};-\dfrac{9}{25}\right)\)

27 tháng 11 2017

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 \(0;\dfrac{1}{10}\)

20 tháng 6 2017

a/dễ --> tự lm

b/ \(\left(x-\dfrac{4}{7}\right)\left(1\dfrac{3}{5}+2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\1\dfrac{3}{5}+2x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\2x=\dfrac{8}{5}\Rightarrow x=\dfrac{4}{5}\end{matrix}\right.\)

Vậy...............

c/ \(\left(x-\dfrac{4}{7}\right):\left(x+\dfrac{1}{2}\right)>0\)

TH1: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{4}{7}\\x>-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{4}{7}\)

TH2: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x< -\dfrac{1}{2}\)

Vậy \(x>\dfrac{4}{7}\) hoặc \(x< -\dfrac{1}{2}\) thì thỏa mãn đề

d/ \(\left(2x-3\right):\left(x+1\dfrac{3}{4}\right)< 0\)

TH1: \(\left\{{}\begin{matrix}2x-3>0\\x+1\dfrac{3}{4}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>1,5\\x< -\dfrac{7}{4}\end{matrix}\right.\)(vô lý)

TH2: \(\left\{{}\begin{matrix}2x-3< 0\\x+1\dfrac{3}{4}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< 1,5\\x>-\dfrac{7}{4}\end{matrix}\right.\)\(\Rightarrow-\dfrac{7}{4}< x< 1,5\)

Vậy...................

11 tháng 9 2017

x< -7/4(vô lí ) vì sao bạn