1 + 11 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 20 bn nhé

\(c,TH1:\frac{2}{3}x-\frac{4}{9}=0=>x=\frac{2}{3}\)

\(TH2:\frac{1}{2}+-\frac{3}{7}:x=0=>x=\frac{6}{7}\)

Vậy \(x\in\left\{\frac{2}{3};\frac{6}{7}\right\}\)

c) \(\left(\frac{2}{3}.x-\frac{4}{9}\right)\left(\frac{1}{2}+\frac{-3}{7}:x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}.x-\frac{4}{9}=0\\\frac{1}{2}+\frac{-3}{7}:x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}.x=\frac{4}{9}\\\frac{-3}{7}:x=-\frac{1}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}\)

người đó mua tất cả số mì sợi là:

80.175+50.100=5255(g)

5255g=5,255(kg)

\(x-x^2\)

\(=-\left(x^2-2x.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}\)

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

Ta có: \(-\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Dấu '' = '' xảy ra khi: \(x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

\(2x-2x^2-5\)

\(=-2[\left(x^2-2x.\frac{1}{2}+\frac{1}{4}\right)+\frac{9}{4}]\)

\(=-2[\left(x-\frac{1}{2}\right)^2+\frac{9}{4}]\)

\(=-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\)

Ta có: \(-\left(x-\frac{1}{2}\right)^2\le0\Rightarrow-2\left(x-\frac{1}{2}\right)^2\le0\Rightarrow-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\le\frac{9}{2}\)

Dấu '' = '' xảy ra khi: \(x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

`x-x^2`

`= - (x^2 - x)`

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

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

Dấu "=" xảy ra khi : `<=> (x-1/2)^2=0 <=>x=1/2`

Vậy GTLN của BT alf `1/4 <=> x=1/2`

`2x-2x^2-5`

`= -2x^2 +2x-5 = -2 (x^2 - x + 5/2)`

`= - 2 (x^2 - 2 . x . 1/2 + 1/4 +9/4)`

`= -2 (x-1/2)^2 -9/2 =< (-9)/2`

Dấu "=" xảy ra khi : `<=> (x-1/2)^2=0 <=>x=1/2`

Vậy GTLN của BT là `(-9)/2 <=>x=1/2`

trong ΔDEFΔDEF vuông tại D có

DK2=EK.KFDK2=EK.KF(đlý)⇒KF=DK2EK=628⇒KF=DK2EK=628=4,5

ta có:EF=EK+KF=8+4,5=12,5

DE2=EF.EK(đlý)DE2=EF.EK(đlý)=12,5.8=100⇒DE=10⇒DE=10

DF2=EF.KFDF2=EF.KF(đlý)=12,5.4,5=56,25⇒⇒DF=7,5