5/8

3/4

\(a)\dfrac{3}{4}.\dfrac{5}{6}=\dfrac{5}{8}\)

\(b)\dfrac{2}{7}:\dfrac{4}{9}=\dfrac{9}{14}\)

\(\dfrac{3}{4}x\dfrac{5}{6}=\dfrac{3x5}{4x6}=\dfrac{15}{24}=\dfrac{5}{8}\)

b) \(\dfrac{2}{7}:\dfrac{4}{9}=\dfrac{2}{7}x\dfrac{9}{4}=\dfrac{18}{28}=\dfrac{9}{14}\)

\(\dfrac{3}{4}\cdot\dfrac{2}{5}+\dfrac{3}{4}\cdot\dfrac{2}{5}\)

\(=\dfrac{3}{4}\cdot\dfrac{2}{5}\left(1+1\right)\)

\(=\dfrac{6}{20}\cdot2\)

\(=\dfrac{12}{20}=\dfrac{3}{5}\)

3/5 

giúp mình với, đi mà T.T

Sao đến bài này là ko ai giúp mình vậy T.T

\(\dfrac{5}{4}\times x=\dfrac{3}{8}+\dfrac{5}{12}\)

\(\dfrac{5}{4}\times x=\dfrac{19}{24}\)

\(x=\dfrac{19}{24}:\dfrac{5}{4}\)

\(x=\dfrac{19}{30}\)

`a, 3/4 - 5/4 :(x-1) =1/2`

`=> 5/4:(x-1)= 3/4 -1/2`

`=> 5/4:(x-1)= 3/4 - 2/4`

`=> 5/4:(x-1)= 1/4`

`=> x-1= 5/4 : 1/4`

`=> x-1=5`

`=>x=5+1`

`=>x=6`

__

`(1/2-x)^2 -2^2 =12`

`=> (1/2-x)^2 = 12+4`

`=> (1/2-x)^2= 16`

`=> (1/2-x)^2 =4^2`

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-x=4\\\dfrac{1}{2}-x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)

__

`(1/2)^(2x-1) =1/16`

`=> (1/2)^(2x-1) = (1/2)^4`

`=> 2x-1=4`

`=> 2x=4+1`

`=>2x=5`

`=>x=5/2`

\(a,\dfrac{3}{4}-\dfrac{5}{4}:\left(x-1\right)=\dfrac{1}{2}\)

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

\(\dfrac{5}{4}:\left(x-1\right)=\dfrac{1}{4}\)

\(x-1=\dfrac{5}{4}:\dfrac{1}{4}\)

\(x-1=5\)

\(x=6\)

\(\left(\dfrac{1}{2}-x\right)^2-2^2=12\)

\(\left(\dfrac{1}{2}-x\right)^2-4=12\)

\(\left(\dfrac{1}{2}-x\right)^2=16\)

\(\left(\dfrac{1}{2}-x\right)^2=4^2hoặc\left(\dfrac{1}{2}-x\right)^2=\left(-4\right)^2\)

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

=>1/2 -x =4      1/2 -x= -4

=> x=1/2-4              x=1/2-(-4)

=>x=-7/2                 x=9/2

vậy x∈{-7/2 ; 9/2}

\(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{16}\)

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

\(=>2x-1=4\)

\(=>2x=5\)

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

a,

\(\Leftrightarrow\left(\left(2x^2-4\right)-2\left(x+1\right)^2\right)< 0\)

\(\Leftrightarrow2x^2-4-2\left(x^2+2x+1\right)< 0\)

\(\Leftrightarrow2x^2-4-2x^2-4x-2< 0\)

\(\Leftrightarrow-4x-6< 0\)

\(\Rightarrow x+\dfrac{3}{2}>0\)

\(\Rightarrow x>-\dfrac{3}{2}\)

\(x\in\left\{-\dfrac{3}{2};\infty\right\}\)

b/

\(\Leftrightarrow\left(x-3\right)^2-5+6x< 0\)

\(\Leftrightarrow x^2-6x+9-5+6x< 0\)

\(\Leftrightarrow x^2+4< 0\) ( điều này vô lý vì không có giá trị nào của x khiến x^2+4<0)

từ trên suy ra:

không có giá trị nào của x để pt này đúng .

c, đưa các hệ số vào công thức bậc 2 ( áp dụng ct bậc 2):

có: \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2.a}\)

Với : a=1 ; b=-4 ; c =3

Ta có:

\(x=\dfrac{-1.-4\pm\sqrt{-4^2-4.1.3}}{2.1}\)

\(\Leftrightarrow x=\dfrac{4\pm\sqrt{4}}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{4+2}{2}\\x_2=\dfrac{4-2}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=3\\x_2=1\end{matrix}\right.\)

tìm khoảng bđt bằng parabol :

 có một dấu bất đẳng thức

\(\Rightarrow1< x< 3\)

`@` `\text {Ans}`

`\downarrow`

\(\left(x-\dfrac{1}{5}\right)^2+1=3,5\div7\%\)

`=> (x-1/5)^2 + 1 = 3,5 \div 0,07`

`=> (x-1/5)^2 +1=50`

`=> (x-1/5)^2 = 49`

`=> (x-1/5)^2 = (+-7)^2`

`=>`\(\left[{}\begin{matrix}x-\dfrac{1}{5}=7\\x-\dfrac{1}{5}=-7\end{matrix}\right.\)

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

`=>`\(\left[{}\begin{matrix}x=\dfrac{36}{5}\\x=-\dfrac{34}{5}\end{matrix}\right.\)

Vậy, `x={36/5; -34/5}.`

\(\dfrac{1}{3}+\dfrac{5}{6}\cdot\left(x-\dfrac{11}{5}\right)=\dfrac{3}{4}\)

\(\dfrac{5}{6}\cdot\left(x-\dfrac{11}{5}\right)=\dfrac{3}{4}-\dfrac{1}{3}\)

\(\dfrac{5}{6}\cdot\left(x-\dfrac{11}{5}\right)=\dfrac{5}{12}\)

\(x-\dfrac{11}{5}=\dfrac{5}{12}\cdot\dfrac{6}{5}\)

\(x-\dfrac{11}{5}=\dfrac{1}{2}\)

\(x=\dfrac{1}{2}+\dfrac{11}{5}\)

\(x=\dfrac{27}{10}\)

\(\dfrac{5}{6}\left(x-\dfrac{11}{5}\right)=\dfrac{3}{4}-\dfrac{1}{3}\)

\(\dfrac{5}{6}\left(x-\dfrac{11}{5}\right)=\dfrac{5}{12}\)

\(x-\dfrac{11}{5}=\dfrac{5}{12}:\dfrac{5}{6}\)

\(x-\dfrac{11}{5}=\dfrac{1}{2}\)

\(x=\dfrac{1}{2}+\dfrac{11}{5}=\dfrac{27}{10}\)

`@` `\text {Ans}`

`\downarrow`

\(\dfrac{1}{3}+\dfrac{5}{6}\times\left(x-\dfrac{11}{5}\right)=\dfrac{3}{4}\)

\(\dfrac{5}{6}\times\left(x-\dfrac{11}{5}\right)=\dfrac{3}{4}-\dfrac{1}{3}\)

\(\dfrac{5}{6}\times\left(x-\dfrac{11}{5}\right)=\dfrac{5}{12}\)

\(x-\dfrac{11}{5}=\dfrac{5}{12}\div\dfrac{5}{6}\)

\(x-\dfrac{11}{5}=\dfrac{5}{12}\times\dfrac{6}{5}\)

\(x-\dfrac{11}{5}=\dfrac{6}{12}\)

\(x=\dfrac{6}{12}+\dfrac{11}{5}\)

\(x=\dfrac{27}{10}\)

\(\left(x-2\right)\left(2x+1\right)-5\left(x+3\right)=2x\left(x-3\right)+4\left(1+2x\right)-2\left(1+x\right)\)

\(2x^2+x-4x-2-5x-15=2x^2-6x+4+8x-2-2x\)

\(x-4x-2-5x-15=-6x+4+8x-2-2x\)

\(\Rightarrow-8x-17=2\)

\(-8x=19\Rightarrow x=-\dfrac{19}{8}\)

Vậy \(x=-\dfrac{19}{8}\)

\(\left(x-2\right)\left(2x+1\right)-5\left(x+3\right)=2x\left(x-3\right)+4\left(1+2x\right)-2\left(1+x\right)\)

\(\Rightarrow2x^2+x-4x-2-5x-15-2x^2+6x-4-8x+2+2x=0\\ \Leftrightarrow-8x-19=0\\ \Leftrightarrow-8x=19\\ \Leftrightarrow x=-\dfrac{19}{8}\)