Tìm x,y biết:\(\frac{1+3y}{15+x}=\frac{1+6y}{18}=\frac{1+9y}{9x}\)
Tìm x,y biết rằng\(\frac{1+3y}{15+x}=\frac{1+6y}{18}=\frac{1+9y}{9x}\)
đề có đúng như z ko bn:
ta có: \(\frac{1+3y}{15}=\frac{1+6y}{18}\)
\(\Rightarrow\left(1+3y\right).18=\left(1+6y\right).15\)
\(18+54y=15+90y\)
\(54y-90y=15-18\)
\(-36y=-3\)
\(y=-3:-36\)
\(y=\frac{1}{12}\)
ta có: \(\frac{1+6y}{18}=\frac{1+9y}{9x}\)
\(\Leftrightarrow\left(1+6y\right).9x=\left(1+9y\right).18\)
\(9x+54xy=18+162y\)
thay số: \(9x+54.\frac{1}{12}x=18+162.\frac{1}{12}\)
\(9x+\frac{9}{2}x=18+\frac{27}{2}\)
\(x.\left(\frac{9}{2}+9\right)=31\frac{1}{2}\)
\(x.13\frac{1}{2}=31\frac{1}{2}\)
\(x=31\frac{1}{2}:13\frac{1}{2}\)
\(x=45\)
KL: x =45 ; y= 1/12
CHÚC BN HỌC TỐT!!!!
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,y biết
\(\frac{1+3y}{12}=\frac{1+6y}{2x}=\frac{1+9y}{5x}\)
ta có: \(\frac{1+3y}{12}=\frac{1+6y}{2x}=\frac{1+9y}{5x}\)
\(\Rightarrow\frac{1+3y}{12}=\frac{1+6y}{2x}=\frac{1+6y-1-3y}{2x-12}=\frac{3y}{2x-12}\)
\(\Rightarrow\frac{3y}{2x-12}=\frac{1+9y}{5x}=\frac{9y+1-3y}{5x-2x+12}=\frac{1+6y}{3x+12}\)
\(\Rightarrow\frac{1+6y}{3x+12}=\frac{1+6y}{2x}\)
=> 3x + 12 = 2x
=> 3x - 2x = - 12
x = -12
xog r bn chỉ cần thay x = -12 vào 2 trong 3 p/s bất kì trên là tính đk y
\(\frac{1+3y}{12}=\frac{1+6y}{2x}=\frac{1+9y}{5x}=\frac{1+3y+1+9y}{12+5x}=\frac{2+12y}{12+5x}\)
\(\Rightarrow\frac{1+6y}{2x}=\frac{2+12y}{12+5x}\)
\(\Rightarrow\frac{12+5x}{2}=2x\)
\(\Rightarrow12+5x=4x\)
\(\Rightarrow12=-x\Leftrightarrow x=-12\)
Thay x vô mà tìm y
1+6y/2x=1+9y/5x=1+3y/12
=>1+6y/2=1+9y/5=1+3y/12 (nhân cả 2 vế với x)
=>12*(1+6y)=2*(1+3y)
=>12+72y=2+6y
=>72y-6y=2-12
=>66y=-10
=>y=-10/66
=>y=-5/33
RỒI BẠN TỰ THAY Y VÀO ĐỂ TIM X
HOK TOT
Tìm x,y biết :
\(\frac{1+2y}{18}=\frac{1+7y}{9x}=\frac{1+6y}{6x}\)
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,y
\(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)
tìm x , y , z biết
a,
\(\frac{x+y}{x}=\frac{y}{x+z}=\frac{z}{x+y}=x+y+z\)
b,
\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}\)
c,
\(\frac{1+3y}{12}=\frac{1+6y}{2x}=\frac{1+9y}{5x}\)
d,
\(\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)