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\(\frac{3x-y}{x+y}=\frac{3}{4}\)
\(=>4\left(3x-y\right)=3\left(x+y\right)\)
\(12x-4y=3x+3y\)
\(12x-3x=3y+4y\)
\(9x=7y\)
\(=>\frac{x}{y}=\frac{7}{9}\)
Có : a/ab+a+1 = a/ab+a+abc = 1/b+1+bc = 1/bc+b+1
c/ca+c+1 = bc/abc+bc+b = b/1+bc+b = b/bc+b+1
=> A = 1+bc+b/bc+b+1 = 1
Tk mk nha
BÀI 1:
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{a\left(bc+b+1\right)}+\frac{abc}{ab\left(ca+c+1\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a} +\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\) (thay abc = 1)
\(=\frac{a+ab+1}{a+ab+1}=1\)
Ta có: \(\frac{3x-y}{x+y}\)=\(\frac{3}{4}\)
\(\Leftrightarrow\)4(3x-y)=3(x+y)
\(\Leftrightarrow\)12x-4y=3x+3y
\(\Leftrightarrow\)12x-3x=4x+3y
\(\Leftrightarrow\)9x=7y
\(\Leftrightarrow\)\(\frac{x}{y}\)=\(\frac{7}{9}\)
\(\frac{3x-y}{x+y}=\frac{3}{4}\Leftrightarrow\frac{3x-y}{x+y}+1=\frac{3}{4}+1\Leftrightarrow\frac{4x}{x+y}=\frac{7}{4}.\) Ở vế trái chia cả tử và mẫu cho y , được:
\(\frac{4.\frac{x}{y}}{\frac{x}{y}+1}=\frac{7}{4}\) Suy ra : \(16.\frac{x}{y}=7\left(\frac{x}{y}+1\right)\) Vậy \(9.\frac{x}{y}=7\Leftrightarrow\frac{x}{y}=\frac{7}{9}\)
Ta có: \(\frac{3x-y}{x+y}=\frac{3}{4}\)
\(\Rightarrow\frac{3x+3y-4y}{x+y}=\frac{3}{4}\)
\(\Rightarrow3-\frac{4y}{x+y}=\frac{3}{4}\)
\(\Rightarrow\frac{4y}{x+y}=3-\frac{3}{4}=\frac{9}{4}\)
\(\Rightarrow4.4y=9.\left(x+y\right)\)
\(\Rightarrow16y=9y+9x\)
\(\Rightarrow9x=16y-9y=7y\)
\(\Rightarrow\frac{x}{y}=\frac{7}{9}\)
Vậy tỉ số \(\frac{x}{y}=\frac{7}{9}\)
=> 2(3x - y) = x + y (ở đây nhân chéo)
=> 6x - 2y = x + y
=> 5x = 3y (ở đây áp dụng qt chuyển vế )
=> x/y =3/5
Chuẩn luôn bạn ạ, k nhoa mn
Ta có: \(\frac{3x-y}{x+y}=\frac{3}{4}\Rightarrow\left(3x-y\right)4=\left(x+y\right)3\)
\(\Rightarrow13x-4y=3x+3y\)
\(\Rightarrow13x-3x=3y+4y\)
\(\Rightarrow10x=7y\)
\(\Rightarrow\frac{x}{7}=\frac{7}{10}\)
Tại sao : \(\frac{3x-y}{x+y}=\frac{3}{4}\)hả Phạm Ngọc Thạch
\(\frac{5x-2y}{x+3y}=\frac{7}{4}\)
=> (5x - 2y).4 = 7.(x + 3y)
=> 20x - 8y = 7x + 21y
=>> 20x - 7x = 21y + 8y
=> 13x = 29y
\(\Rightarrow\frac{x}{y}=\frac{29}{13}\)
\(\frac{5x-2y}{x+3y}=\frac{7}{4}\)
\(\Rightarrow4\left(5x-2y\right)=7\left(x+3y\right)\)
\(\Rightarrow20x-8y=7x+21y\)
\(\Rightarrow20x-7x=8y+21y\)
\(\Rightarrow13x=29y\)
\(\Rightarrow\frac{x}{y}=\frac{29}{13}\)
Vậy \(\frac{x}{y}=\frac{29}{13}\)
\(\frac{3x-y}{x+y}=\frac{3}{4}\Rightarrow12x-4y=3x+3y\Rightarrow9x=7y\Rightarrow\frac{x}{y}=\frac{7}{9}\)
4(3x-y) = 3(x+y)
12x -4y = 3x+3y
9x = 7y
x/y = 7/9