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Lm câu 2 trc nhé:
\(x-3+x-3=\left(x-3\right)+\left(x-3\right)=2\left(x-3\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3\)
Chỉ lm tắt thôi ạ, hiểu rồi tự trình bày nha~
\(\frac{x}{3}-\frac{4}{y}=\frac{1}{5}\)
\(\Leftrightarrow\frac{4}{y}=\frac{x}{3}-\frac{1}{5}\)
\(\Leftrightarrow\frac{4}{y}=\frac{x5}{15}-\frac{3}{15}\)
\(\Leftrightarrow\frac{4}{y}=\frac{x5-3}{15}\)
\(\Leftrightarrow4.15=x5-3y\)
\(\Leftrightarrow60=x5-3y\)
\(\Leftrightarrow x5-3y=60\)
tìm x,y như bt nhé
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+y+z}{2+3+4}=\frac{81}{9}=9\)
\(\Rightarrow\begin{cases}\frac{x}{2}=9\\\frac{y}{3}=9\\\frac{z}{4}=9\end{cases}\)\(\Rightarrow\begin{cases}x=18\\y=27\\z=36\end{cases}\)
Vậy x=18;y=27;z=36
Áp dụng tc dãy tỉ
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+y+z}{2+3+4}=\frac{81}{9}=9\)
Với \(\frac{x}{2}=9\Rightarrow x=18\)
Với \(\frac{y}{3}=9\Rightarrow y=27\)
Với \(\frac{z}{4}=9\Rightarrow z=36\)
a) \(\frac{3}{4}x+1=2\)
\(\Leftrightarrow\)\(\frac{3}{4}x=1\)
\(\Leftrightarrow\)\(x=\frac{4}{3}\)
b) \(\frac{2}{3}+\frac{1}{3}:x=-1\)
\(\Leftrightarrow\)\(\frac{1}{3}:x=-\frac{5}{3}\)
\(\Leftrightarrow\)\(x=-\frac{1}{5}\)
c) \(\left|2x+1\right|=5\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-1=5\\2x-1=-5\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x=6\\2x=-4\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=-2\end{array}\right.\)
A/ \(\frac{3}{4}\)x+1=2 \(\Rightarrow\)\(\frac{3}{4}x=2-1\Rightarrow\frac{3}{4}x=1\Rightarrow x=1:\frac{3}{4}\Rightarrow x=\frac{4}{3}\)
Vậy x=...
B/ \(\frac{2}{3}\)+\(\frac{1}{3}\):x=−1 \(\Rightarrow\frac{1}{3}:x=-1-\frac{2}{3}\Rightarrow\frac{1}{3}:x=\frac{-5}{3}\Rightarrow x=\frac{1}{3}:\frac{-5}{3}\Rightarrow x=\frac{-1}{5}\)
x = 8
y = 3
ai tick cho mình may mắn cả năm
không tick cho mình vận xui sẽ đến
X-4/y-3=4/3
=>3.(x-4)=4.(y-3)
=>3x-12=4y-12
Hay 3x=4y
Mà x-y=5
=>x=y+5
=>3x=4y<=>3.(y+5)=4y
<=>3y+15=4y
<=>4y-3y=15 hay y=15
Thay vào x=y+5=>x=20
vậy x=20;y=15
a)\(\frac{-2}{3}.x+\frac{1}{5}=\frac{3}{10}\)
\(\frac{-2}{3}x=\frac{3}{10}-\frac{1}{5}=\frac{1}{10}\)
\(\frac{-2}{3}x=\frac{1}{10}\)
\(x=\frac{1}{10}\div\frac{-2}{3}=\frac{-3}{20}\)
b)\(\left(50\%.x+2\frac{1}{4}\right).\frac{-2}{3}=\frac{17}{6}\)
\(50\%.x+2\frac{1}{4}=\frac{17}{6}\div\frac{-2}{3}\)\(=\frac{-17}{4}\)
\(50\%.x+2\frac{1}{4}=\frac{-17}{4}\)
\(50\%.x=\frac{-17}{4}-2\frac{1}{4}=\frac{-17}{4}-\frac{9}{4}=\frac{-26}{4}=\frac{-13}{2}\)
\(50\%.x=\frac{-13}{2}\)
\(x=\frac{-13}{2}\div50\%=-13\)
\(\frac{-2}{3}\)\(\hept{\begin{cases}.\\.\\.\end{cases}ads}\)
Bài 1:
\(\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
= \(\left[\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
= \(\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
=\(\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
=\(\frac{1}{26}+\frac{1}{27}+....+\frac{1}{26}\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
......????