\(1,3x-7=19\\ \Rightarrow3x=26\\ \Rightarrow x=\dfrac{26}{3}\\ 2,\left(2x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\\ 3,3x+\dfrac{2}{4}+1=5x-\dfrac{1}{3}\\ \Rightarrow5x-\dfrac{1}{3}-3x-\dfrac{2}{4}-1=0\\ \Rightarrow2x-\dfrac{11}{6}=0\\ \Rightarrow2x=\dfrac{11}{6}\\ \Rightarrow x=\dfrac{11}{12}\)

\(4,\dfrac{x}{15}+\dfrac{1}{2}-\dfrac{x}{50}=\dfrac{5}{6}\\ \Rightarrow\dfrac{x}{15}-\dfrac{x}{50}=\dfrac{5}{6}-\dfrac{1}{2}\\ \Rightarrow x\left(\dfrac{1}{15}-\dfrac{1}{50}\right)=\dfrac{1}{3}\\ \Rightarrow\dfrac{7}{150}x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{50}{7}\)

Bài 4 là:

\(\dfrac{1}{2.5}\) + \(\dfrac{1}{5.8}\) +....+\(\dfrac{1}{92.95}\) + \(\dfrac{1}{95.98}\)

                  Đúng không bạn

Bạn chỉ nên đưa câu khó lên vì làm cả đề thì ko làm kịp đc

a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)

\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)

b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)

\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)

c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)

\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)

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

a) \(\dfrac{4}{5}\left(x-\dfrac{1}{3}\right)-3\dfrac{1}{2}=50\%\)

\(\dfrac{4}{5}x-\dfrac{4}{15}-\dfrac{7}{2}=\dfrac{1}{2}\)

\(\dfrac{4}{5}x=\dfrac{1}{2}+\dfrac{7}{2}+\dfrac{4}{15}\)

\(\dfrac{4}{5}x=\dfrac{64}{15}\)

\(x=\dfrac{64}{15}:\dfrac{4}{5}\\ x=\dfrac{16}{3}\)

b) \(8\dfrac{3}{5}.x-2\dfrac{1}{5}.x=16\%\)

\(\dfrac{43}{5}x-\dfrac{11}{5}x=\dfrac{4}{25}\)

\(\dfrac{32}{5}x=\dfrac{4}{25}\)

\(x=\dfrac{4}{25}:\dfrac{32}{5}\)

\(x=\dfrac{1}{40}\)

a: =>10|x-2|=50

=>|x-2|=5

=>x-2=5 hoặc x-2=-5

=>x=7 hoặc x=-3

b: =>|3x-12|+|5x-20|=56

=>8|x-4|=56

=>|x-4|=7

=>x-4=7 hoặc x-4=-7

=>x=11 hoặc x=-3

2.(5x-8)-(4x-5)=4.(3x-4)+11

2(5x−8)−3(4x−5)=4(3x−4)+11

⇔10x−16−12x+15=12x+16+11

⇔−2x−1=12x+27⇔−14x−28=0⇔x=−2

(8x+2).(1-3x)+(6x-1).(4x-1)=-50

(8x+2)(1−3x)+(6x−1)(4x−10)=−50

⇔8x−24\(x^2\)+2−6x+24\(x^2\)−60x−4x+10=−50

⇔−62x+12=−50

⇔−62x=−62

⇔x=1

bài 2

Xét ΔABC có 

\(\frac{AE}{AB}\)=\(\frac{AD}{AC}\)

Do đó: DE//CB

Xét tứ giác BEDC có DE//BC

nên BEDC là hình thang

mà ˆEBC=ˆDCB

nên BEDC là hình thang cân