Tìm X:
\(\frac{5}{X+1}\)= \(\frac{15}{18}\).
Tìm các số x, y biết
a) \(x=-\frac{18}{24}+\frac{15}{21}\)
b) \(\frac{-1}{3}-x=\frac{1}{2}-\frac{1}{-4}\)
c) \(\frac{-5}{x}=\frac{-y}{8}=\frac{18}{72}\)
Help me, please!!!
\(\frac{-5}{x}=\frac{-y}{8}=\frac{18}{72}\)
\(\Leftrightarrow\frac{-5}{x}=\frac{-y}{8}=\frac{1}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}-\frac{5}{x}=\frac{1}{4}\\-\frac{y}{8}=\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5.4:1\\-y=8.1:4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-20\\-y=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-20\\y=-2\end{cases}}}\)
vậy x=-20 và y=-2
\(-\frac{1}{3}-x=\frac{1}{2}-\frac{1}{-4}\)
\(-\frac{1}{3}-x=\frac{1}{2}-\frac{-1}{4}\)
\(-\frac{1}{3}-x=\frac{2}{4}-\frac{-1}{4}\)
\(-\frac{1}{3}-x=\frac{3}{4}\)
\(x=-\frac{1}{3}-\frac{3}{4}\)
\(x=-\frac{4}{12}-\frac{9}{12}\)
\(x=-\frac{13}{12}\)
\(x=-\frac{18}{24}+\frac{15}{21}\)
\(x=-\frac{3}{4}+\frac{5}{7}\)
\(x=-\frac{21}{28}+\frac{20}{28}\)
\(x=-\frac{1}{28}\)
Tìm x
\(\left(2\frac{1}{3}:x+\frac{-2}{3}\right):1\frac{3}{4}=-3\frac{2}{5}-\frac{15}{18}:\frac{3}{4}\)
\(\Rightarrow\left(\frac{7}{3}:x-\frac{2}{3}\right):\frac{7}{4}=-\frac{17}{5}-\frac{10}{9}\Rightarrow\left(\frac{7}{3}:x-\frac{2}{3}\right):\frac{7}{4}=-\frac{203}{45}\Rightarrow\frac{7}{3}:x-\frac{2}{3}=-\frac{1421}{180}\Rightarrow\frac{7}{3}:x=-\frac{1301}{180}\Rightarrow x=-\frac{420}{1301}\)
Tìm x:
\(\frac{\left(13\frac{2}{9}-15\frac{2}{3}\right)\cdot\left(30^2-5^4\right)}{\left(18\frac{3}{7}-17\frac{1}{4}\right)\cdot\left(25-12\cdot5^2\right)}\cdot x=\frac{\frac{2}{11}+\frac{3}{13}+\frac{4}{15}+\frac{5}{17}}{4\frac{1}{11}+\frac{5}{13}+\frac{9}{15}+\frac{13}{17}}\)
\(1-\left[5\frac{4}{9}+x-7\frac{7}{18}\right]:15\frac{3}{4}=0\)
TÌM X NHA
Tìm ba số x, y, z biết \(\frac{x}{5}=\frac{y}{-7},\frac{y}{4}=\frac{z}{15}\) và x + 3y - 4z = 18
Ta có:
\(\begin{cases}\frac{x}{5}=\frac{y}{-7}\\\frac{y}{4}=\frac{z}{15}\end{cases}\)\(\Rightarrow\begin{cases}\frac{x}{-20}=\frac{y}{28}\\\frac{y}{28}=\frac{z}{105}\end{cases}\)\(\Rightarrow\frac{x}{-20}=\frac{y}{28}=\frac{z}{105}=\frac{3y}{84}=\frac{4z}{420}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{x}{-20}=\frac{y}{28}=\frac{z}{105}=\frac{3y}{84}=\frac{4z}{420}=\frac{x+3y-4z}{-20+84-420}=\frac{18}{-356}=\frac{-9}{178}\)
\(\Rightarrow\begin{cases}x=\frac{-9}{178}.\left(-20\right)=\frac{90}{89}\\y=\frac{-9}{178}.28=\frac{-126}{89}\\z=\frac{-9}{178}.105=\frac{-945}{178}\end{cases}\)
Vậy \(x=\frac{90}{89};y=\frac{-126}{89};z=\frac{-945}{178}\)
Tìm x,y biết:\(\frac{1+3y}{15+x}=\frac{1+6y}{18}=\frac{1+9y}{9x}\)
\(x+\frac{-1}{5}=\frac{-2}{15}\)
Em cứ lấy máy tính bấm bài 1 đi là đc
16, giải phương trình.
1, \(\frac{x+5}{65}+\frac{x+10}{60}=\frac{x+15}{55}+\frac{x+20}{50}\)
2, \(\frac{x+91}{81}+\frac{x+92}{82}+\frac{x+93}{83}=3\)
3, \(\frac{59-x}{19}+\frac{58-x}{18}=\frac{57-x}{17}+\frac{56-x}{16}\)
4, \(\frac{x}{15}+\frac{x+1}{16}+\frac{x+2}{17}+\frac{x+3}{18}+\frac{x+4}{19}=5\)
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!!!!