Áp dụng t/c dãy tỉ số bằng nhau:

a.

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x}{6}=\dfrac{4y}{20}=\dfrac{2x+4y}{6+20}=\dfrac{28}{26}=\dfrac{14}{13}\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.\dfrac{14}{13}=\dfrac{52}{13}\\y=5.\dfrac{14}{13}=\dfrac{70}{13}\end{matrix}\right.\)

(Em có nhầm đề 26 thành 28 ko nhỉ, số xấu quá)

b.

\(4x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{3x}{15}=\dfrac{-2y}{-8}=\dfrac{3x-2y}{15-8}=\dfrac{35}{7}=5\)

\(\Rightarrow\left\{{}\begin{matrix}x=5.5=25\\y=4.2=20\end{matrix}\right.\)

c.

\(\dfrac{x}{-3}=\dfrac{y}{-7}=\dfrac{2x}{-6}=\dfrac{4y}{-28}=\dfrac{2x+4y}{-6-28}=\dfrac{68}{-34}=-2\)

\(\Rightarrow\left\{{}\begin{matrix}x=-3.\left(-2\right)=6\\y=-7.\left(-2\right)=14\end{matrix}\right.\)

d.

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{-3y}{9}=\dfrac{-2z}{-8}=\dfrac{4x-3y-2z}{8+9-8}=\dfrac{16}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\dfrac{16}{9}=\dfrac{32}{9}\\y=-3.\dfrac{16}{9}=-\dfrac{48}{9}\\z=4.\dfrac{16}{9}=\dfrac{64}{9}\end{matrix}\right.\)

a, Ta có : \(-\dfrac{3}{2}-2x+\dfrac{3}{4}=-2\)

\(\Rightarrow-2x=-2+\dfrac{3}{2}-\dfrac{3}{4}=-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{5}{8}\)

Vậy ...

b, Ta có : \(\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)\left(-\dfrac{3}{2}-\dfrac{10}{3}\right)=\dfrac{2}{5}\)

\(\Rightarrow-\dfrac{29}{6}\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)=\dfrac{2}{5}\)

\(\Rightarrow-\dfrac{2}{3}x-\dfrac{3}{5}=-\dfrac{12}{145}\)

\(\Rightarrow-\dfrac{2}{3}x=-\dfrac{12}{145}+\dfrac{3}{5}=\dfrac{15}{29}\)

\(\Rightarrow x=-\dfrac{45}{58}\)

Vậy...

- Tham khảo thanh này để đánh nha bạn ;-;

Ta có: \(\left(x+2\right)^4+\left(x+4\right)^4\)

\(=\left(x^2+4x+4\right)^2+\left(x^2+8x+16\right)^2\)

\(=x^4+16x^2+16+8x^3+8x^2+32x+x^4+64x^2+256+16x^3+32x^2+256x\)

\(=2x^4+24x^3+120x^2+288x+272\)

x+3/4=3/6

x=3/6-3/4

x=-1/4

\(x=\dfrac{3}{6}-\dfrac{3}{4}=\dfrac{6}{12}-\dfrac{9}{12}=\dfrac{-3}{12}\)

\(2xy-x^2-y^2+16\)

\(=-\left(x^2-2xy+y^2-16\right)\)

\(=-\left(x-y-4\right)\left(x-y+4\right)\)

Bạn tham khảo nha. Chúc bạn học tốt

Ta có \(-2< -\dfrac{4}{3}< -1\) nên \(\left[-\dfrac{4}{3}\right]=-2\).

\(0< \dfrac{1}{2}< 1\) nên \(\left[\dfrac{1}{2}\right]=0\).

a) \(\left(4\frac{1}{2}-2x\right)\cdot3\frac{2}{3}=\frac{11}{5}\)

\(\left(\frac{9}{2}-2x\right)=\frac{11}{5}\cdot\frac{3}{11}\)

\(2x=\frac{45-6}{10}\)

\(2x=\frac{39}{10}\)

\(x=\frac{39}{10\cdot2}=\frac{39}{20}\)

b) \(\frac{3}{4}\cdot x+\frac{4}{7}\cdot x=-\frac{15}{8}\)

\(x\cdot\left(\frac{21+16}{28}\right)=-\frac{15}{8}\)

\(x=-\frac{15}{8}\cdot\frac{28}{37}\)

\(x=-\frac{105}{74}\)

Bài 4: 

b: Ta có: \(2x\left(x-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)