a: \(\dfrac{2.75}{x}=\dfrac{0.4}{1.5}=\dfrac{4}{15}\)

\(\Leftrightarrow x=\dfrac{11}{4}\cdot\dfrac{15}{4}=\dfrac{165}{16}\)

b: \(3\dfrac{1}{2}:\left(2x-3\right)=\dfrac{-3}{4}:0.2\)

\(\Leftrightarrow\dfrac{7}{2}:\left(2x-3\right)=\dfrac{-3}{4}:\dfrac{1}{5}=\dfrac{-15}{4}\)

\(\Leftrightarrow2x-3=\dfrac{7}{2}:\dfrac{-15}{4}=\dfrac{-7}{2}\cdot\dfrac{4}{15}=\dfrac{-28}{30}=\dfrac{-14}{15}\)

=>2x=-14/15+3=45/45-14/15=31/45

=>x=31/90

c: \(\dfrac{3x+2}{27}=\dfrac{3}{3x+2}\)

\(\Leftrightarrow\left(3x+2\right)^2=81\)

=>3x+2=9 hoặc 3x+2=-9

=>3x=7 hoặc 3x=-11

=>x=7/3 hoặc x=-11/3

d: \(\dfrac{5-x}{4}=\dfrac{2x+3}{2}\)

=>10-2x=8x+12

=>-10x=2

hay x=-1/5

a) Vì

)

A=\(\dfrac{9}{4}-\left(\dfrac{-5}{2}\right)+\left(\dfrac{-25}{6}\right)\)

A=\(\dfrac{19}{4}\)-\(\dfrac{25}{6}\)

A=\(\dfrac{14}{24}\)=\(\dfrac{7}{12}\)

b, Thay vào, ta sẽ có:

A=3.\(\left(\dfrac{-2}{3}\right)-4.\left(\dfrac{-2}{3}\right).\dfrac{4}{5}+5.\dfrac{4}{5}\)

A=-2-\(\left(\dfrac{-32}{15}\right)\)+4

A=\(\dfrac{2}{15}\)+4

A=\(\dfrac{62}{15}\)

a. Ta có : ( a + b )( c - d ) = ac-ad+bc-bd (1)

( a - b )( c + d ) = ac+ad-bc+bd (2)

Từ giả thuyết : \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow ad=bc\) (3)

Từ (1) , ( 2) và ( 3) \(\Rightarrow\)( a + b )( c - d) = ( a - b)( c + d )

\(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\left(đpcm\right)\)

a/ \(\left|2x-1,6\right|-2,3=1,4\)

\(\Leftrightarrow\left|2x-1,6\right|=3,7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1,6=3,7\\2x-1,6=-3,7\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2,65\\x=-1,05\end{matrix}\right.\)

Vậy ....

b/ \(5,4-\left|3x-1,2\right|=5,5\)

\(\Leftrightarrow\left|3x-1,2\right|=-0,1\)

Mà \(\left|3x-1,2\right|\ge0\)

\(\Leftrightarrow x\in\varnothing\)

c/ \(\left|x+1,3\right|+\left|x+2,4\right|=4x\)

Mà \(\left\{{}\begin{matrix}\left|x+1,3\right|\ge0\\\left|x+2,4\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow4x\ge0\)

\(\Leftrightarrow x+1,3+x+2,4=4x\)

\(\Leftrightarrow2x+3,7=4x\)

\(\Leftrightarrow3,7=4x-2x\)

\(\Leftrightarrow2x=3,7\)

\(\Leftrightarrow x=1,85\)

Vậy ....

d/ \(\left|x-1,2\right|+\left|2,5-x\right|=0\)

Mà \(\left\{{}\begin{matrix}\left|x-1,2\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)

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

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

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

Vậy ..

a, \(\left|2x-1,6\right|-2,3=1,4\)

\(\Rightarrow\left|2x-1,6\right|=3,7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1,6=3,7\\2x-1,6=-3,7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2,65\\x=-1,05\end{matrix}\right.\)

b,\(5,4-\left|3x-1,2\right|=5,5\)

\(\Rightarrow\left|3x-1,2\right|=-0,1\) (vô lí)

Vì \(\left|x\right|\ge0\) mà \(\left|3x-1,2\right|< 0\)

Vậy, không có giá trị của x thỏa mãn.

c, \(\left|x+1,3\right|+\left|x+2,4\right|=4x\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x+1,3\right|\ge0\\\left|x+2,4\right|\ge0\end{matrix}\right.\Leftrightarrow4x\ge0\)

\(\Leftrightarrow x+1,3+x+2,4=4x\)

\(\Leftrightarrow x+x+1,3+2,4=4x\)

\(\Leftrightarrow2x+3,7=4x\)

\(\Leftrightarrow2x-4x=-3,7\)

\(\Leftrightarrow-2x=-3,7\)

\(\Leftrightarrow x=\dfrac{3,7}{2}\)

d, \(\left|x-1,2\right|+\left|2,5-x\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x-1,2\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)

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

tick đi giải cho

đáp án là a=4.............