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 rằng \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
Ta có: 1+2y/18=1+4y/24
=> 24(1+2y)=18(1+4y)
=>24+48y=18+72y
=>24-18=72y-48y
=>6=24y
=> y=1/4
thay y=1/4 vào đề ta có:
(1+ 1/2)/18=1+1/24=(1+3/2)/6x
=>1/12=(5/2)/6x
=> 12(5/2)=6x
=>30=6x
=>x=5
Vậy x=5
y=1/4
Tìm x, y, z biết: \(\frac{1+4y}{18}=\frac{1+5y}{24}=\frac{1+6y}{6x}\)
\(\frac{1+4y}{18}=\frac{1+5y}{24}\Rightarrow24+96y=18+90y\)
\(\Rightarrow6+6y=0\Leftrightarrow6\left(1+y\right)=0\)Vậy y = -1
Thay y = -1 ta có :
\(\frac{1-5}{24}=\frac{1-6}{6x}\Leftrightarrow\frac{-5}{30}=-\frac{5}{6x}\left(\frac{-4}{24}=-\frac{5}{30}=\frac{1-5}{24}\right)\)
Vậy 6x = 30 hay x = 5
TÌM X BIẾT RẰNG:
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{x}\)
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{^{6x}}\)
Tìm x,y
1+2y/18 = 1+4y/24
=> 24.(1+2y) = 18.(1+4y)
=> 24+48y = 18+72y
=> 24-18 = 72y-48y
=> 6 = 24y
=> y = 1/4
Lại có : 1+6y/6x = 1+4y/24
=> (1+6.1/4) : 6x = (1+4.1/4) : 24
=> 5/12x = 1/12
=> 12x = 5 : 1/12 = 60
=> x = 60 : 5 = 12
Vậy .........
Tk mk nha
Áp dụng t/c dãy tỉ số = nhau, ta có:
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
=> 1 + 4y/9 + 3x = 1 + 4y/24 => 9 + 3x = 24
=> 3x = 15
=> x = 15 : 3
=> x = 5
Tìm x : \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{1+2y+1+6y}{18+6x}=\frac{2+8y}{18+6x}\)
\(\Rightarrow\frac{1+4y}{24}=\frac{2+8y}{18+6x}\left(1\right)\)
Từ ( 1 )
=> \(\frac{1+4y}{2\left(1+4y\right)}=\frac{24}{18+6x}\)
\(=\frac{1}{2}=\frac{24}{18+6x}\)
\(\Rightarrow18+6x=48\)
\(6x=48-18\)
\(6x=30\)
\(x=30:6\)
\(x=5\)
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
tìm X
Ta có :
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+2y+1+4y}{18+24}=\frac{2+6y}{42}=\frac{2\left(1+6y\right)}{2.21}=\frac{1+6y}{21}\)
Lại có :
\(\frac{1+6y}{6x}=\frac{1+6y}{21}\)
\(\Leftrightarrow\)\(6x=21\)
\(\Leftrightarrow\)\(x=\frac{7}{2}\)
Vậy \(x=\frac{7}{2}\)
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\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
\(\Rightarrow\frac{1}{18}+\frac{2y}{18}=\frac{1}{24}+\frac{4y}{24}=\frac{1}{6x}+\frac{6y}{6x}\)
\(\Rightarrow\left(\frac{1}{18}+\frac{2y}{18}\right)-\left(\frac{1}{6x}+\frac{6y}{6x}\right)=0\)
\(\Rightarrow\frac{1}{18}+\frac{2y}{18}-\frac{1}{6x}-\frac{6y}{6x}=0\)
\(\frac{y}{9}-\frac{1}{6x}-\frac{6y}{6x}=\frac{1}{18}\)
\(\frac{\frac{2}{3}xy}{6x}-\frac{1}{6x}-\frac{6y}{6x}=\frac{1}{18}\)
\(\frac{\frac{2}{3}xy-1-6y}{6x}=\frac{1}{18}\)
\(6x=\left(\frac{2}{3}xy-1-6y\right)\cdot\frac{1}{18}\)
\(x=\frac{\left[\left(\frac{2}{3}xy-1-6y\right)\cdot\frac{1}{18}\right]}{6}\)
\(x=\left(\frac{\frac{\frac{2}{3}xy-1-6y}{18}}{6}\right)\)
\(x=\frac{\frac{2}{3}xy-1-6y}{108}\)
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
\(\frac{1+2y}{18}=\frac{1+4y}{24}+\frac{1+6y}{6x}\)