a) \(\dfrac{3}{5}\times x=\dfrac{4}{7}\)

            \(x=\dfrac{4}{7}\div\dfrac{3}{5}\)

            \(x=\dfrac{20}{21}\)

b) \(\dfrac{1}{8}\div x=\dfrac{1}{5}\)

           \(x=\dfrac{1}{8}\div\dfrac{1}{5}\)

          \(x=\dfrac{5}{8}\)

20/21

5/8

20/21

5/8

Để A là số nguyên thì x-5+7 chia hết cho x-5

=>x-5 thuộc {1;-1;7;-7}

=>x thuộc {6;4;12;-2}

Quy đồng : \(\dfrac{20}{80}>x>\dfrac{16}{80}\)

Giá trị của x là : \(\dfrac{19}{80};\dfrac{18}{80};\dfrac{17}{80}\)

24

24

24 nhé

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

a) \(\left(x+1\right)\left(x^2+1\right)=0\)

Vì \(\left(x^2+1\right)>0\forall x\)

\(\Rightarrow x=-1\)

b) \(5y^2-20=0\)

\(y^2-4=0\)

\(\left(y-2\right)\left(y+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\)

a, Ta có : \(\left(x+1\right)\left(x^2+1>0\right)=0\Leftrightarrow x=-1\)

b, \(5y^2=20\Leftrightarrow y^2=4\Leftrightarrow\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\)

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

d, \(\left|y-2\right|+5=0\)( vô lí ) 

Vậy ko có gtr y để bth bằng 0 

c) \(\left|x-2\right|-1=0\)

\(\left|x-2\right|=1\)

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

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

d) \(\left|y-2\right|+5=0\)

\(\left|y-2\right|=-5\)

Vì \(\left|y-2\right|\ge0\forall y\)

⇒ pt vô nghiệm

`A=-10/(sqrtx+5)(x>=0)`

`x>=0=>sqrtx>=0`

`=>sqrtx+5>=5>0`

`=>10/(sqrtx+5)<=10/5=2`

`=>A>=-2`

Dấu "=" xảy ra khi `x=0`

Vậy GTNN `A=-2<=>x=0`

giúp mình với

a. x + 3/10= 25/8

x = 25/8-3/10

x = 113/40

b. x/35 = 4/5-5/7

x/35 = 3/35

vậy x=3

\(x+\dfrac{3}{10}=4-\dfrac{7}{8}\)

\(x+\dfrac{3}{10}=\dfrac{25}{8}\)

        \(x=\dfrac{25}{8}-\dfrac{3}{10}\)

       \(x=\dfrac{113}{40}\)

\(\dfrac{5}{7}+\dfrac{x}{35}=\dfrac{4}{5}\)

        \(\dfrac{x}{35}=\dfrac{4}{5}-\dfrac{5}{7}\)

        \(\dfrac{x}{35}=\dfrac{3}{35}\)

    =>x=3