Tìm x,y
\(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)
Tìm x,y, biết
\(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)
Ta có : \(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)
\(\Rightarrow\frac{1+3y}{12}=\frac{1+9y}{4x}=\frac{1+3y+1+9y}{12+4x}=\frac{2+12y}{12+4x}\)
\(\Rightarrow\frac{1+6y}{16}=\frac{2.\left(1+6y\right)}{12+4x}\)
Do đó : \(16=\frac{12+4x}{2}\)
Từ đó suy ra : x = 5
tìm x
\(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)
Tìm x:
\(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)
tìm x thỏa mãn: \(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)
Ta có \(\frac{1+3y}{12}=\frac{1+6y}{16}\)
\(=>\frac{2\left(1+3y\right)}{24}=\frac{1+6y}{16}\)
\(=>\frac{2+6y}{24}=\frac{1+6y}{16}\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(=>\frac{2+6y}{24}=\frac{1+6y}{16}=\frac{2+6y-\left(1+6y\right)}{8}=\frac{2+6y-1-6y}{8}=\frac{1}{8}\)
\(=>\frac{1+6y}{16}=\frac{1}{8}\)
\(=>8\left(1+6y\right)=16\)
\(=>8+48y=16\)
\(=>48y=8\)
\(=>y=\frac{1}{8}\)
Ta có
\(\frac{2+6y}{24}=\frac{1+6y}{16}=\frac{1}{8}\)
\(=>\frac{1+9y}{4x}=\frac{1}{8}\)
Thế \(y=\frac{1}{6}\) vào biểu thức ta có
\(\frac{1+9y}{4x}=\frac{1}{8}\)
\(=>\frac{1+9.\frac{1}{6}}{4x}=\frac{1}{8}\)
\(=>\frac{\frac{5}{2}}{4x}=\frac{1}{8}\)
\(=>20=4x\)
\(=>x=5\)
Tìm x thỏa mãn: \(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)
\(\frac{1+3y}{12}=\frac{1+6y}{16}\)
<=> (1 + 3y).16 = (1 + 6y).12
<=> 16 + 48y = 12 + 72y
<=> 16 - 12 = 72y - 48y
<=> 24y = 4
<=> y = 1/6
Thay y = 1/6 vào đề bài ta được:
\(\frac{1+3.\frac{1}{6}}{12}=\frac{1+9.\frac{1}{6}}{4x}\)
\(\Leftrightarrow\frac{\frac{3}{2}}{12}=\frac{\frac{5}{2}}{4x}\Leftrightarrow6x=30\Leftrightarrow x=5\)
Vậy x = 5; y = 1/6
Tìm x thỏa mãn:
\(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)
\(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)
\(\Leftrightarrow12\left(1+6y\right)=16\left(1+3y\right)\)
\(\Leftrightarrow3\left(1+6y\right)=4\left(1+3y\right)\)
\(\Leftrightarrow3+18y=4+12y\)
\(\Leftrightarrow3-4=12y-18y\)
\(\Leftrightarrow-1=-6y\)
\(\Rightarrow y=\frac{1}{6}\)
\(\Rightarrow\frac{1+3y}{12}=\frac{1+3.\frac{1}{6}}{12}=\frac{1+9.\frac{1}{6}}{4x}\)
\(\Leftrightarrow\frac{\frac{3}{2}}{12}=\frac{\frac{5}{2}}{4x}\)
\(\Leftrightarrow\frac{1}{8}=\frac{\frac{5}{2}}{4x}\)
\(\Rightarrow4x=8.\frac{5}{2}=20\)
\(\Rightarrow x=20:4=5\)
Vậy \(x=5\)
tìm x,y
\(\frac{1+3y}{12}\) = \(\frac{1+6y}{16}\) = \(\frac{1+9y}{4x}\)
ta có: \(\frac{1+3y}{12}=\frac{1+6y}{16}\)
\(\Rightarrow\frac{2.\left(1+3y\right)}{24}=\frac{1+6y}{16}\)
\(\Rightarrow\frac{2+6y}{24}=\frac{1+6y}{16}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có
\(\frac{2+6y}{24}=\frac{1+6y}{16}=\frac{2+6y-\left(1+6y\right)}{8}=\frac{2+6y-1+6y}{8}=\frac{1}{8}\)
\(\Rightarrow8.\left(1+6y\right)=16\)
\(\Rightarrow8+48y=16\)
\(\Rightarrow48y=8\)
=> y=\(\frac{1}{8}\)
Ta có
\(\frac{2+6y}{24}=\frac{1+6y}{16}=\frac{1}{8}\)
\(\Rightarrow\frac{1+9y}{4x}=\frac{1}{8}\)
Thế y=1/6 vào biểu thức, ta có
\(\frac{1+9y}{4x}=\frac{1}{8}\)
\(\Rightarrow\frac{1+9.\frac{1}{6}}{4x}=\frac{1}{8}\)
\(\Rightarrow\frac{\frac{5}{2}}{4x}=\frac{1}{8}\)
\(\Rightarrow20=4x\)
\(\Rightarrow x=5\)
Vậy x=5
\(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)
tim x
giai nhanh an 2 tick ( minh co toi 2 nick )
\(\frac{1+3y}{12}\)= \(\frac{1+6y}{16}\)=\(\frac{1+9y}{4x}\)
mình cần cách giải nha
Giải bình thường:
(1)\(\frac{1+3y}{12}=\frac{1+6y}{16}\Leftrightarrow16.\left(1+3y\right)=12.\left(1+6y\right)\) tích trung tỷ = tích ngoại tỷ
\(\Leftrightarrow4.\left(1+3y\right)=3.\left(1+6y\right)\)
\(\Leftrightarrow4+3.4y=3.+6.3y\Leftrightarrow\left(18-12\right)y=4-3=1\)
\(\Rightarrow y=\frac{1}{6}\)
(2)\(\frac{1+6y}{16}=\frac{1+9y}{4x}\Leftrightarrow\frac{2}{16}=\frac{1+\frac{3}{2}}{4x}=\frac{5}{8x}\Rightarrow x=5\)