Tìm y biết:
\(y-\frac{1}{4}.y=\frac{1}{2}\)
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Ta có:
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{1+6y+1+2y}{18+6x}=\frac{2+8y}{18+6x}\)\(=\frac{2\left(1+4y\right)}{2\left(9+3x\right)}=\frac{1+4y}{9+3x}\Rightarrow9+3x=24\Rightarrow x=5\)
\(\Rightarrow\left(1+6y\right)18=\left(1+2y\right)30\Rightarrow18+108y=30+60y\)
\(\Rightarrow48y=12\Rightarrow y=\frac{12}{48}=\frac{1}{4}\)
Vậy x = 5 và y = \(\frac{1}{4}\)
\(\frac{3}{2}+\frac{13}{6}+\frac{37}{12}+\frac{81}{20}-y=\frac{4}{5}\)
\(\frac{54}{5}-y=\frac{4}{5}\)
y=10
a) \(\frac{1}{2}-|\frac{5}{4}-2x|=\frac{1}{3}\Leftrightarrow|\frac{5}{4}-2x|=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{6}\\\frac{5}{4}-2x=-\frac{1}{6}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{5}{4}-\frac{1}{6}=\frac{13}{12}\\2x=\frac{5}{4}+\frac{1}{6}=\frac{17}{12}\end{cases}}}\)
Tự làm nốt và kết luận
b) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)=0\)
Vì \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)\ne0\forall x\Rightarrow x+1=0\Leftrightarrow x=-1\)
Vậy ....
Câu 1,
x+y=-1/3 ; y+z=5/4 ; x+z= 4/3
=> 2(x+y+z)=9/4
=> x+y+z=9/8
Ta lại có: x+y=-1/3
=> z=9/8 -(-1/3)=35/24
Ta lại có: z+y=5/4
=> y=-5/24
=> x=.....
Câu 2:
\(-4\le x\le-\frac{11}{18}\)
Ta có : x+y=1/2=>x=1/2-y
y+z=1/3=>z=1/3-y
=>x-z=1/2-y-1/3+y=1/2-1/3-(y-y)=1/6
Vậy x = (1/6+1/4):2=5/24
z = (1/4-1/6):2=1/24
=> y = 1/3-1/24=7/24
1, ta co \(\frac{x}{5}=\frac{y}{6}=\frac{x}{20}=\frac{y}{24}\)
\(\frac{y}{8}=\frac{z}{7}=\frac{y}{24}=\frac{z}{21}\)
=>\(\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y-z}{20+24-21}=\frac{69}{23}=3\)
=>\(x=3\cdot20=60\)
\(y=3\cdot24=72\)
\(z=3\cdot21=63\)
3. ta co \(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}=\frac{x+y-z+t}{15-7+3-1}=\frac{10}{10}=1\)
=> \(x=1\cdot15=15\)
\(y=1\cdot7=7\)
\(z=1\cdot3=3\)
\(t=1\cdot1=1\)
\(y-\frac{1}{4}.y=\frac{1}{2}\)
\(\left(1-\frac{1}{4}\right).y=\frac{1}{2}\)
\(\frac{3}{4}.y=\frac{1}{2}\)
\(y=\frac{1}{2}:\frac{3}{4}\)
\(y=\frac{1}{2}.\frac{4}{3}\)
\(y=\frac{2}{3}\) Vậy: \(y=\frac{2}{3}\)
y=2/3