TĐB, ta có: (chép lại đề bài)

=> 3x = x + 5 + 1

=> 3x = x + 6

=> 3x - x = 6

=> x(3-1) = 6

=> 2x = 6

=> x = 6 : 2

=> x = 3

Vậy x = 3

a:

Sửa đề: Chiều dài là 25m; chiều rộng là 20m

Diện tích thửa ruộng là 25x20=500(m²)

b: Khối lượng thóc thu được là:

500:100x65=5x65=325(kg)

\(\dfrac{-15}{8}\cdot\left(\dfrac{8}{-15}+\dfrac{32}{27}\right)\cdot\dfrac{15}{-7}\)

\(=\left(\dfrac{-15}{8}\cdot\dfrac{8}{-15}-\dfrac{15}{8}\cdot\dfrac{32}{27}\right)\cdot\dfrac{15}{-7}\)

\(=\left(1-4\cdot\dfrac{5}{9}\right)\cdot\dfrac{15}{-7}\)

\(=\dfrac{-11}{9}\cdot\dfrac{15}{-7}=\dfrac{165}{63}=\dfrac{55}{21}\)

\(\Leftrightarrow\left(x^2+xy\right)-\left(x+y\right)-\left(2xy^2-2y^2\right)=1\)

\(\Leftrightarrow x\left(x+y\right)-\left(x+y\right)-2y^2\left(x-1\right)=1\)

\(\Leftrightarrow\left(x-1\right)\left(x+y\right)-2y^2\left(x-1\right)=1\)

\(\Leftrightarrow\left(x-1\right)\left(x+y-2y^2\right)=1=1.1=\left(-1\right).\left(-1\right)\)

Th1: \(\left\{{}\begin{matrix}x-1=1\\x+y-2y^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2+y-2y^2=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y^2-y-1=0\end{matrix}\right.\)

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

Th2: \(\left\{{}\begin{matrix}x-1=-1\\x+y-2y^2=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\0+y-2y^2=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2y^2-y-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\\left[{}\begin{matrix}y=1\\y=-\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy pt có 2 cặp nghiệm nguyên \(\left(x;y\right)=\left(2;1\right);\left(0;1\right)\)

Thời gian kể từ khi hai xe xuất phát đến lúc gặp nhau là:

8h-7h15p=45p=0,75(giờ)

Tổng vận tốc hai xe là 50+40=90(km/h)

Độ dài quãng đường từ bến A đến bến B là:

90x0,75=67,5(km)

\(Q\left(2\right)=3.2-6=0\)

\(P\left(-1\right)=4.\left(-1\right)+2=-2\)

a: \(-\dfrac{1}{2}+x=\dfrac{3}{4}\)

=>\(x=\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{3}{4}+\dfrac{2}{4}=\dfrac{5}{4}\)

b: \(-\dfrac{1}{3}+x=\dfrac{4}{5}\)

=>\(x=\dfrac{4}{5}+\dfrac{1}{3}=\dfrac{12}{15}+\dfrac{5}{15}=\dfrac{17}{15}\)

c: \(\dfrac{x}{5}=\dfrac{-5}{6}+\dfrac{19}{30}\)

=>\(\dfrac{x}{5}=-\dfrac{25}{30}+\dfrac{19}{30}=-\dfrac{6}{30}=-\dfrac{1}{5}\)

=>x=-1

d: \(\dfrac{x}{5}=\dfrac{5}{6}+\dfrac{-19}{30}\)

=>\(\dfrac{x}{5}=\dfrac{25}{30}-\dfrac{19}{30}=\dfrac{6}{30}=\dfrac{1}{5}\)

=>x=1

e: \(\dfrac{5}{7}-\dfrac{2}{3}x=-\dfrac{3}{4}\)

=>\(\dfrac{2}{3}x=\dfrac{5}{7}+\dfrac{3}{4}=\dfrac{20}{28}+\dfrac{21}{28}=\dfrac{41}{28}\)

=>\(x=\dfrac{41}{28}:\dfrac{2}{3}=\dfrac{41}{28}\cdot\dfrac{3}{2}=\dfrac{123}{56}\)

g: \(2\dfrac{4}{5}x-x-=-1\dfrac{1}{2}\)

=>\(x\left(\dfrac{14}{5}-1\right)=-\dfrac{3}{2}\)

=>\(x\cdot\dfrac{9}{5}=-\dfrac{3}{2}\)

=>\(x=-\dfrac{3}{2}:\dfrac{9}{5}=-\dfrac{3}{2}\cdot\dfrac{5}{9}=\dfrac{-15}{18}=-\dfrac{5}{6}\)

???

\(\dfrac{7}{8}+x=\dfrac{7}{20}\)

=>\(x=\dfrac{7}{20}-\dfrac{7}{8}=\dfrac{14}{40}-\dfrac{35}{40}=-\dfrac{21}{40}\)

=)x=7/20-7/8

=)x=-21/40

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