ngu

\(\frac{x}{y}=\frac{-1}{2}\Leftrightarrow2x=-y\Leftrightarrow y=-2x\)

Thay y=-2x vào biểu thức cần tính sẽ ra

bạn giải đi bạn

Theo đề ta có: \(x:y:z=3:4:5\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

Đặt: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\left(k\inℕ^∗\right)\)

Suy ra: \(x=3k;y=4k;z=5k\) Thay vào biểu thức P ta có:

\(P=\frac{3k+8k+15k}{6k+12k+20k}+\frac{6k+12k+20k}{9k+16k+25k}+\frac{9k+16k+25k}{12k+20k+30k}\)

\(P=\frac{26k}{38k}+\frac{38k}{50k}+\frac{50k}{62k}=\frac{13}{19}+\frac{19}{25}+\frac{25}{31}=\frac{33141}{14725}\)

1.Tính giá trị của biểu thức: A=\(\frac{5x^2+3y^2}{10x^2-3y^2}\left(1\right)biết\frac{x}{3}=\frac{y}{5}suyra:5x=3y;suyra:x=\frac{3y}{5};thayvào\left(1\right)taco:\frac{5\left(\frac{3y}{5}\right)^2+3y^2}{10\left(\frac{3y}{5}\right)^2-3y^2}=\frac{\frac{9y^2}{5}+3y^2}{\frac{18y^2}{5}-3y^2}=\frac{24y^2}{5}\cdot\frac{5}{3y^2}=8\)

2.\(\frac{x}{y}=\frac{7}{10}suyra;\frac{x}{7}=\frac{y}{10}\left(1\right)và\frac{y}{z}=\frac{5}{8}suyra;\frac{y}{5}=\frac{z}{8}suyra;\frac{y}{5}\cdot\frac{1}{2}=\frac{z}{8}\cdot\frac{1}{2}suyra;\frac{y}{10}=\frac{z}{16}\left(2\right)Tù\left(1\right)và\left(2\right)suyra\frac{x}{7}=\frac{y}{10}=\frac{z}{16}và2x+5y-2z=9;suyra:\frac{2x}{14}=\frac{5y}{50}=\frac{2z}{32}ápdụngtínhchấtcủadãytỉsốbằngnhautacó\frac{2x}{14}=\frac{5y}{50}=\frac{2z}{32}=\frac{2x+5y-2z}{14+50-32}=\frac{9}{32}suyra;x=\frac{63}{32};y=\frac{45}{16};z=\frac{9}{2}\)

B=3,406938828

xcănx=cănx mũ 3

y căn y = căn y mũ 3

\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)

\(=\frac{4y^2-\left(x-y\right)^2}{y^2\left(x-y\right)}.\frac{y^2-xy}{x-3y}+\frac{x\left(x-2y\right)-2\left(x^2-xy\right)}{2\left(x-2y\right)}.\frac{2x-4y}{xy+y^2}\)

\(=\frac{3y^2+2xy-x^2}{y^2\left(x-y\right)}.\frac{y^2-xy}{x-3y}+\frac{-x^2}{2\left(x-2y\right)}.\frac{2x-4y}{xy+y^2}\)

\(=\frac{\left(x+y\right)\left(3y-x\right)}{y^2\left(x-y\right)}.\frac{y\left(y-x\right)}{x-3y}-\frac{x^2}{2\left(x-2y\right)}.\frac{2\left(x-2y\right)}{y\left(x+y\right)}\)

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

\(=\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}=\frac{2xy+y^2}{y\left(x+y\right)}=\frac{2x+y}{x+y}\)

Giờ chỉ cần thế x, y vô nữa là xong nhé.

\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)

\(=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y\left(y-x\right)}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x\left(x-y\right)}{x-2y}\right):\frac{y\left(x+y\right)}{2\left(x-2y\right)}\)

\(=\frac{4y\left(y-x\right)}{\left(x-y\right)\left(x-3y\right)}-\frac{\left(x-y\right)y\left(y-x\right)}{y^2\left(x-3y\right)}\)\(+\frac{x.2\left(x-2y\right)}{2.y\left(x+y\right)}-\frac{x\left(x-y\right).2\left(x-2y\right)}{\left(x-2y\right).y\left(x+y\right)}\)

\(=\frac{-4y}{x-3y}+\frac{\left(x-y\right)^2}{y\left(x-3y\right)}+\frac{x\left(x-2y\right)}{y\left(x+y\right)}-\frac{2x\left(x-y\right)}{y\left(x+y\right)}\)

\(=\frac{-4y^2+x^2-2xy+y^2}{y\left(x-3y\right)}+\frac{x^2-2xy-2x^2+2xy}{y\left(x+y\right)}\)

\(=\frac{x^2-2xy-3y^2}{y\left(x-3y\right)}+\frac{-x^2}{y\left(x+y\right)}\)

\(=\frac{x^2+xy-3xy-3y^2}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)

\(=\frac{x\left(x+y\right)-3y\left(x+y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)

\(\frac{\left(x+y\right)\left(x-3y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)

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

\(=\frac{x^2-2xy+y^2-x^2}{y\left(x+y\right)}=\frac{-2xy+y^2}{y\left(x+y\right)}\)

\(=\frac{y\left(y-2x\right)}{y\left(x+y\right)}=\frac{y-2x}{x+y}\)

Thay \(x=\frac{1}{2};y=\frac{1}{3}\)vào A ta có :

\(A=\frac{\frac{1}{3}-2.\frac{1}{2}}{\frac{1}{2}+\frac{1}{3}}=\frac{\frac{1}{3}-1}{\frac{3}{6}+\frac{2}{6}}=\frac{2}{3}:\frac{5}{6}=\frac{2.6}{3.5}=\frac{4}{5}\)

Vậy \(A=\frac{4}{5}\)tại \(x=\frac{1}{2};y=\frac{1}{3}\)

Ừ nhở chị sai từ chỗ \(\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}=\frac{x^2+2xy+y^2-x^2}{y\left(x+y\right)}=\frac{y^2+2xy}{y\left(x+y\right)}\)em nhé