Bài 1 Tìm x biết :
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
Tìm x biết \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
Tìm x biết \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
ta có: \(\frac{1+2y}{18}=\frac{1+4y}{24}\)
\(\Rightarrow\left(1+2y\right).24=18.\left(1+4y\right)\)
\(24+48y=18+72y\)
\(48y-72y=18-24\)
\(-24y=-6\)
\(y=\frac{1}{4}\)
thay vào \(\frac{1+4.\frac{1}{4}}{24}=\frac{1+6.\frac{1}{4}}{6x}\)
\(\frac{1+1}{24}=\frac{1+\frac{3}{2}}{6x}\)
\(\frac{1}{12}=\frac{5}{2}:6x\)
\(6x=\frac{5}{2}:\frac{1}{12}\)
\(6x=30\)
\(x=30:6\)
\(x=5\)
KL: x =5; y = 1/4
Tìm x biết rằng ; \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{1+2y+1+6y}{18+6x}=\frac{2+8y}{6\left(3+x\right)}=\frac{1+4y}{3\left(3+x\right)}\)
\(\Rightarrow3\left(3+x\right)=24\)\(\Rightarrow3+x=8\)\(\Rightarrow x=5\)
Vậy \(x=5\)
Ta có: \(\frac{1+2y}{18}=\frac{1+4y}{24}\)
\(\Leftrightarrow24\left(1+2y\right)=18\left(1+4y\right)\)
\(\Leftrightarrow24+48y=18+72y\)
\(\Leftrightarrow24y-6=0\Leftrightarrow y=\frac{1}{4}\)
\(\Rightarrow\frac{1+2y}{18}=\frac{1+6y}{6x}\Leftrightarrow\frac{1+\frac{1}{2}}{18}=\frac{1+\frac{3}{2}}{6x}\)
\(\Leftrightarrow x=5\)
Vậy x = 5 và \(y=\frac{1}{4}\)
Tìm x biết:
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
1 + 2y/18=1 + 6y/6x=1 + 2y + 1 + 6y/18 + 6x=2 + 8y/18 + 6x=2.(1 + 4y)/2.(9 + 3x)=1 + 4y/9 + 3x
Suy ra:1 + 4y/9 + 3x=1 + 4y/24=>9 + 3x=24
3x=15
x=5
Tìm x biết rằng \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
\(\Rightarrow\)\(\frac{1+2y+1+4y+1+6y}{18+24+6x}\)=\(\frac{\left(1+1+1\right)+2y+4y+6y}{6\left(3+4+x\right)}=\frac{y\left(2+4+6\right)+3}{6\left(3+4+x\right)}=\frac{3+y.12}{6\left(7+x\right)}\)
=\(\frac{3\left(1+4y\right)}{3.2\left(7+x\right)}=\frac{1+4y}{14+2x}\)
\(\Rightarrow\)\(\frac{1}{14}=\frac{2y}{x}\Rightarrow x=14.2y=28y\)
\(\frac{x}{y}=28\)
Tìm x, biết rằng : \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
Ta có: \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{1+2y+1+6y}{18+6x}=\frac{2\left(1+4y\right)}{6\left(x+3\right)}=\frac{1+4y}{3x+9}\)
\(=>\frac{1+4y}{24}=\frac{1+4y}{3x+9}\)\(=>3x+9=24\)
<=>3x=15
<=>x=5
Vậy x có giá trị bằng 5
Chúc bạn học tốt!
Tìm x, y biết
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
Ta có: 1 + 2y/18 = 2.(1+2y)/2.18 = 2+4y/36
Sử dụng tc dãy tỉ số bằng nhau ta có:
2+4y/36 = 1+4y/24 = 2+4y-1-4y/36-24 = 1/12
Do 1+2y/18 = 1/12=> y = 1/4
1+6y/6x = 1/12=> x = 5
Vậy x = 5; y = 1/4
Tìm x biết rằng:
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{1+2y+1+6y}{18+6x}\)
\(=\frac{2\left(1+4y\right)}{2\left(3x+9\right)}=\frac{1+4y}{3x+9}\)
\(\Rightarrow\frac{1+4y}{24}=\frac{1+4y}{3x+9}\)
\(\Rightarrow3x+9=24\)
\(\Rightarrow x+3=24:3=8\)
\(\Rightarrow x=5\)
\(\frac{1+2y}{18}=\frac{2+4y}{36}=\frac{1+4y}{24}=\frac{2+4y-\left(1+4y\right)}{36-24}=\frac{1}{12}\)
\(\Rightarrow\frac{1+4y}{24}=\frac{1}{12}\Rightarrow4y+1=2\Rightarrow y=\frac{1}{4}\)
\(\Rightarrow\frac{1+6y}{6x}=\frac{1}{12}\Rightarrow x=2\cdot\left(1+6y\right)=2+12y=2+12\cdot\frac{1}{4}=5\)
Vậy x = 5.
tìm x,y biết
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{1+2y+1+6y}{18+6x}=\frac{8y+2}{18+6x}\)
suy ra
\(\frac{1+4y}{24}=\frac{8y+2}{18+6x}=\frac{2\left(1+4y\right)}{2\left(9+3x\right)}=\frac{1+4y}{9+3x}\)
=>9+3x=24
3x=24-9
3x=15
x=15:3
x=5
TK CHO MÌNH NHA