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30 tháng 7

\(\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(1.5+\dfrac{-3}{5}:x\right)=0\left(x\ne0\right)\\ TH1:\dfrac{3}{4}x-\dfrac{9}{16}=0\\ =>\dfrac{3}{4}x=\dfrac{9}{16}\\ =>x=\dfrac{9}{16}:\dfrac{3}{4}=\dfrac{3}{4}\left(tm\right)\\ TH2:1,5+\dfrac{-3}{5}:x=0\\ =>\dfrac{3}{5}:x=\dfrac{3}{2}\\ =>x=\dfrac{3}{5}:\dfrac{3}{2}=\dfrac{2}{5}\left(tm\right)\)

30 tháng 7

\(x=\dfrac{3}{4}\) và \(x=\dfrac{2}{5}\)

29 tháng 10 2021

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

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

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

\(\left[{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\left[{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\)

\(\left[{}\begin{matrix}x-\dfrac{4}{5}=\dfrac{3}{4}\\x-\dfrac{4}{5}=\dfrac{-3}{4}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{20}\\x=\dfrac{1}{20}\end{matrix}\right.\)

20 tháng 9 2021

sai rồi cậu thử ghép vào mà xem

nó có ra kết quả = 0 đâu

1:

a: =7/5(40+1/4-25-1/4)-1/2021

=21-1/2021=42440/2021

b: =5/9*9-1*16/25=5-16/25=109/25

28 tháng 9 2021

a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)

Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)

b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)

Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)

17 tháng 8 2017

\(\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(\dfrac{1}{3}-\dfrac{3}{5}:x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{16}=0\\\dfrac{1}{3}-\dfrac{3}{5}:x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{16}\\\dfrac{3}{5}:x=\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{9}{5}\end{matrix}\right.\)

17 tháng 8 2017

\(\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(\dfrac{1}{3}-\dfrac{3}{5}:x\right)=0\)

\(\Rightarrow\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(\dfrac{1}{3}-\dfrac{5}{3}x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{16}=0\\\dfrac{1}{3}-\dfrac{5}{3}x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{16}\Rightarrow x=\dfrac{3}{4}\\\dfrac{5}{3}x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{5}\end{matrix}\right.\)

20 tháng 8 2017

1.Tính

a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)

b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)

c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)

d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)

e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)

Bài 2

a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)

\(x=\dfrac{13}{49}\)

b.\(\left|x-1,5\right|=2\)

Xảy ra 2 trường hợp

TH1

\(x-1,5=2\)

\(x=3,5\)

TH2

\(x-1,5=-2\)

\(x=-0,5\)

Vậy \(x=3,5\) hoặc \(x=-0,5\) .

Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.

20 tháng 8 2017

Ths bn nhé

21 tháng 7 2021

Tham khảo:undefinedundefined

21 tháng 7 2021

ảnh sau mờ quá bạn ơi 

28 tháng 3 2018

a/ \(\dfrac{x}{9}=\dfrac{16}{x}\)

\(\Leftrightarrow x^2=9.16\)

\(\Leftrightarrow x^2=144\)

\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-12\end{matrix}\right.\)

Vậy ...

b/ \(x^3+27=0\)

\(\Leftrightarrow x^3=-27\)

\(\Leftrightarrow x^3=\left(-3\right)^3\)

\(\Leftrightarrow x=-3\)

Vậy ...

c/ \(\left|x\left(x^2-\dfrac{5}{4}\right)=x\right|\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^3-\left(\dfrac{5}{4}x+x\right)=0\\x^3-\left(\dfrac{5}{4}x-x\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^3-\dfrac{9}{4}x=0\\x^3-\dfrac{1}{4}x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\left(x^2-\dfrac{9}{4}\right)=0\\x\left(x^2-\dfrac{1}{4}\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x^2-\dfrac{9}{4}=0\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x^2-\dfrac{1}{4}=0\end{matrix}\right.\end{matrix}\right.\)

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

Vậy ...

c/ Với mọi x ta có :

\(\left|x-5\right|=\left|5-x\right|\)

\(\Leftrightarrow\left|x+3\right|+\left|x-5\right|=\left|x+3\right|+\left|5-x\right|\)

\(\Leftrightarrow\left|x+3\right|+\left|5-x\right|\ge\left|\left(x+3\right)+\left(5-x\right)\right|\)

\(\Leftrightarrow\left|x+3\right|+\left|5-x\right|\ge\left|8\right|\)

\(\Leftrightarrow\left|x+3\right|+\left|5-x\right|\ge8\)

Dấu "=" xảy ra khi :

\(\left(x+3\right)\left(5-x\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0\\5-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3\le0\\5-x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3\\5\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3\\5\le x\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-3\le x\le5\\x\in\varnothing\end{matrix}\right.\)

Vậy ...