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\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+4}{96}+\frac{x+1}{99}=-4\)
\(\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+1}{99}+1\right)=-4+4\)
\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{96}+\frac{x+100}{99}=0\)
\(\left(x+100\right).\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}\right)=0\)
=> \(\orbr{\begin{cases}x+100=0\\\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}=0\end{cases}}\)
Mà \(\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}\ne0\)
=> x + 100 = 0
=> x = -100
Vậy x = -100
Câu b trừ mỗi số đi 1 tức là trừ cả cụm đó cho 3 rùi lm tương tự câu a
\(\frac{x+2}{x}=\frac{1}{2}\)
\(\Rightarrow2.\left(x+2\right)=x\)
\(\Rightarrow2x+4=x\)
\(\Rightarrow2x-x=-4\)
\(\Rightarrow x=-4\)
\(b,\frac{x+3}{x+4}=\frac{3}{5}\)
\(\Rightarrow5.\left(x+3\right)=3.\left(x+4\right)\)
\(\Rightarrow5x+15=3x+12\)
\(\Rightarrow5x-3x=12-15\)
\(\Rightarrow2x=-3\)
\(\Rightarrow x=-\frac{3}{2}\)
\(\frac{x+5}{6}=\frac{6}{x+5}\)
\(\Rightarrow\left(x+5\right).\left(x+5\right)=6.6\)
\(\Rightarrow\left(x+5\right)^2=6^2\)
\(\Rightarrow\orbr{\begin{cases}x+5=6\\x+5=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-11\end{cases}}\)
Vậy x = 1 hoặc x= - 11
\(\frac{x+1}{3}=\frac{12}{x+1}\)
\(\Rightarrow\left(x+1\right).\left(x+1\right)=3.12\)
\(\Rightarrow\left(x+1\right)^2=36\)
\(\Rightarrow\left(x+1\right)^2=6^2\)
\(\Rightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}\)
\(\frac{-3}{7}.\frac{5}{12}+\frac{3}{7}.\frac{7}{12}+\frac{-3}{12}\)
\(=\frac{3}{7}.\frac{-5}{12}+\frac{3}{7}.\frac{7}{12}+\frac{-1}{4}\)
\(=\frac{3}{7}.\left(\frac{-5}{12}+\frac{7}{12}\right)+\frac{-1}{4}\)
\(=\frac{3}{7}.\frac{2}{12}+\frac{-1}{4}\)
\(=\frac{3}{7}.\frac{1}{6}+\frac{-1}{4}\)
\(=\frac{3}{42}+\frac{-1}{4}\)
\(=\frac{1}{14}-\frac{1}{4}\)
\(=\frac{4}{56}-\frac{14}{56}\)
\(=\frac{-10}{56}=\frac{-5}{28}\)
a) x - 2/3 = -5/12
=> x = -5/12 + 2/3
=> x = 1/4
b) \(\frac{x+5}{3}=\frac{5}{9}\)
=> \(\frac{3\left(x+5\right)}{9}=\frac{5}{9}\)
=> 3(x + 5) = 5
=> x + 5 = 5 : 3
=> x + 5 = 5/3
=> x = 5/3 - 5
=> x = 20/3
c) \(\frac{x-2}{27}=\frac{3}{x-2}\)
=> (x - 2)(x - 2) = 27 . 3
=> (x - 2)2 = 81
=> (x - 2)2 = 92
=> \(\orbr{\begin{cases}x-2=9\\x-2=-9\end{cases}}\)
=> \(\orbr{\begin{cases}x=11\\x=-7\end{cases}}\)
Vậy ...
a,x/2=y/5
<=> 2x/4=y/5=2x+y/4+5=18/9=2
+,x/2=2 => x=4
+, y/5=2 => y=10
g, x/2=y/5
đặt x/2=y/5=k
=> x=2k ; y=5k
ta có 2k.5k=90
k2.10=90
k2=9
=> k=3 k=-3
+, x/2=2=> x=4 x/2=-2 => x=-4
+, y/5=2 => y=10 y/5=-2 => y=-10
CÁC Ý SAU BN LÀM NỐT NHÉ DỄ MÀ
a) Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\frac{x}{2}=\frac{y}{5}=\frac{2x+y}{4+5}=\frac{18}{9}=2\)
\(\Rightarrow x=4;y=10\)
mấy bài còn lại tương tự
a, \(\frac{x+1}{5}+\frac{x+1}{7}=\frac{x+1}{9}\)
\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{7}-\frac{x+1}{9}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
b, \(\frac{x+4}{96}+\frac{x+3}{97}=\frac{x+2}{98}+\frac{x+1}{99}\)
\(\Leftrightarrow\left(\frac{x+4}{96}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+1}{99}+1\right)\)
\(\Leftrightarrow\frac{x+100}{96}+\frac{x+100}{97}=\frac{x+100}{98}+\frac{x+100}{99}\)
\(\Leftrightarrow\frac{x+100}{96}+\frac{x+100}{97}-\frac{x+100}{98}-\frac{x+100}{99}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{96}+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
a) x + 1/5 + x + 1/7 = x + 1/9
<=> 1/5x + 1/5 + 1/7x + 1/7 = 1/9x + 1/9
<=> (1/5x + 1/7x) + (1/5 + 1/7) = 1/9x + 1/9
<=> 12/35x + 12/35 = 1/9x + 1/9
<=> 12/35x + 12/35 - 1/9x = 1/9
<=> 73/315x + 12/35 = 1/9
<=> 73/315x = 1/9 - 12/35
<=> 73/315x = -73/315
<=> x = 73/315 : -73/315 = -1
=> x = -1
b) làm tương tự