ta có : x+3/5+x+4/5=x+5/3+x+6/2

   => (x+x)+(3/5+4/4)=(x+x)+(5/3+6/2)

   => 2x+8/5=2x+14/3  ( vô lí ) 

Ta có: \(x+\frac{3}{5}+x+\frac{4}{4}=x+\frac{5}{3}+x+\frac{6}{2}\)

=> \(x+\frac{3}{5}+x+1=x+\frac{5}{3}+x+3\)

=> \(x+1+x+3=x+\frac{5}{3}-x-\frac{3}{5}\)

=> \(2x+4=\frac{25-9}{15}\)

=> \(2x+4=\frac{16}{15}\)

=> \(2x+4=1+\frac{1}{15}\)

=> \(2x+4-1=\frac{1}{15}\)

=> \(2x+3=\frac{1}{15}\)

=> \(2x=\frac{1}{15}-3\)

=> \(2x=\frac{1-45}{15}\)

=> \(2x=-\frac{44}{15}\)

=> \(x=\frac{\left(-\frac{44}{15}\right)}{2}\)

=> \(x=-\frac{44}{15}.\frac{1}{2}\)

=> \(x=-\frac{22}{15}\)

\(A=3y^2+6y+5\)

\(\Leftrightarrow A=3\left(y^2+2y+1\right)+2\)

\(\Leftrightarrow A=3\left(y+1\right)^2+2\ge2\) Với \(\forall y\in R\)

Dấu "=" xảy ra khi y = -1

Vậy GTNN của A là 2 khi y = -1

\(B=\left(x+1\right)\left(x^2+4x+5\right)\left(x+5\right)\)

\(\Leftrightarrow B=\left(x^2+6x+5\right)\left(x^2+4x+5\right)\)

\(\Leftrightarrow B=\left(t+x\right)\left(t-x\right)=t^2-x^2\)

\(\Leftrightarrow B=x^4+10x^2+25-x^2=x^4+9x^2+25\)

\(\Leftrightarrow B=\left(x^2+\dfrac{9}{2}\right)^2+\dfrac{19}{4}\ge\left(\dfrac{9}{2}\right)^2+\dfrac{19}{4}=25\) Với \(\forall x\in R\)

Dấu "=" xảy ra khi x = 0

Vậy GTNN Của B là 25 khi x = 0 .

\(5^x+5^{x+1}=150\)

\(\Leftrightarrow5^x=25\)

hay x=2

\(A=-x^2+4x+7=-\left(x^2-4x+4\right)+11=-\left(x-2\right)^2+11\)

Ta thấy : \(-\left(x-2\right)^2+11\le11\)\(\Leftrightarrow maxA=11\)khi    \(x=2\)

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

Ta thấy : \(-\left(2x-1\right)^2-4\le-4\)\(\Leftrightarrow maxB=-4\)khi   \(x=\frac{1}{2}\)

\(C=-x^2+x+5=-\left(x^2-2\cdot\frac{1}{2}\cdot x+\frac{1}{4}\right)+\frac{21}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{21}{4}\)

Ta thấy : \(-\left(x-\frac{1}{2}\right)^2+\frac{21}{4}\le\frac{21}{4}\)\(\Leftrightarrow maxC=\frac{21}{4}\)khi    \(x=\frac{1}{2}\)

tk mk nka !!! 

thanks

mình câng gấp ạ

đề khó hiểu quá, bạn ghi lại đi

\(x-\dfrac{3}{4}=\dfrac{1}{7}\\ \Rightarrow x=\dfrac{25}{28}\)

\(\left(-x+\dfrac{3}{5}\right)+\dfrac{13}{20}=\dfrac{5}{6}\\ \Leftrightarrow\left(-x+\dfrac{3}{5}\right)=\dfrac{11}{60}\\ \Leftrightarrow-x=\dfrac{11}{60}-\dfrac{3}{5}\\ \Rightarrow-x=-\dfrac{5}{12}\\ \Rightarrow x=\dfrac{5}{12}\)