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Tìm \(x\) câu a:
\(\frac13.x\) + \(\frac25.\left(x+1\right)\) = 0
\(\frac{5}{15}x\) + \(\frac{6}{15}x\) + \(\frac25\) = 0
\(\frac{11}{15}x\) = - \(\frac25\)
\(x=-\frac25:\frac{11}{15}\)
\(x\) = - \(\frac25\times\frac{15}{11}\)
\(x\) = - \(\frac{6}{11}\)
Vậy \(x=-\frac{6}{11}\)
Tìm \(x\) câu b:
\(x\) x 25% = 0,5
\(x\times0,25\) = 0,5
\(x=0,5:0,25\)
\(x=2\)
Vậy \(x=2\)
Câu 1a:
1/3x + 2/5(x + 1) = 0
1/3x + 2/5x + 2/5 = 0
1/3x + 2/5x = - 2/5
x(1/3 + 2/5) = -2/5
x.(5/15 + 6/15) = -2/5
x.11/15 = - 2/5
x = - 2/5 : 11/15
x = - 6/11
Vậy x = -6/11
Câu b:
x . 25%. x = 0,5
x.x = 0,5 : 25%
x^2 = 2
x = - \(\sqrt2\); x = \(\sqrt2\)
Vậy x ∈ {- \(\sqrt2\); \(\sqrt2\) )
Câu 1a:
1/3x + 2/5(x + 1) = 0
1/3x + 2/5x + 2/5 = 0
1/3x + 2/5x = - 2/5
x(1/3 + 2/5) = -2/5
x.(5/15 + 6/15) = -2/5
x.11/15 = - 2/5
x = - 2/5 : 11/15
x = - 6/11
Vậy x = -6/11
Câu b:
x . 25%. x = 0,5
x.x = 0,5 : 25%
x^2 = 2
x = - \(\sqrt2\); x = \(\sqrt2\)
Vậy x ∈ {- \(\sqrt2\); \(\sqrt2\) )
các bạn ơi mình đang cần gấp . Mình chỉ còn 20 phút thui . HUHU
Bài làm :
1) \(\frac{x+11}{4}=\frac{2x+4}{5}\)
\(\Leftrightarrow\left(x+11\right).5=4.\left(2x+4\right)\)
\(\Leftrightarrow5x+55=8x+16\)
\(\Leftrightarrow5x-8x=16-55\)
\(\Leftrightarrow-3x=-39\)
\(\Leftrightarrow x=\frac{-39}{-3}=\frac{39}{3}=13\)
2)\(\frac{x+4}{x+10}=\frac{3}{5}\)
\(\Leftrightarrow\left(x+4\right).5=\left(x+10\right).3\)
\(\Leftrightarrow5x+20=3x+30\)
\(\Leftrightarrow5x-3x=30-20\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=\frac{10}{2}=5\)
3)\(\frac{x+8}{x+14}=\frac{2}{3}\)
\(\Leftrightarrow\left(x+8\right).3=\left(x+14\right).2\)
\(\Leftrightarrow3x+24=2x+28\)
\(\Leftrightarrow3x-2x=28-24\)
\(\Leftrightarrow x=4\)
\(\frac{x}{-7}=\frac{5}{-35}\)
\(\frac{x.5}{-35}=\frac{5}{-35}\)
=> x . 5 = 5
x = 5 : 5
x = 1
\(a/\frac{7}{9}-\frac{x}{3}=\frac{1}{9}\)
\(\Rightarrow\frac{x}{3}=\frac{7}{9}-\frac{1}{9}\)
\(\Rightarrow\frac{x}{3}=\frac{2}{3}\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(b/\frac{1}{x}-\frac{-2}{15}=\frac{7}{15}\)
\(\Rightarrow\frac{1}{x}=\frac{7}{15}+\frac{-2}{15}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{3}\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
\(c/\frac{-11}{14}-\frac{-4}{x}=\frac{-3}{14}\)
\(\Rightarrow\frac{-4}{x}=\frac{-11}{14}-\frac{-3}{14}\)
\(\Rightarrow\frac{-4}{x}=\frac{-4}{7}\)
\(\Rightarrow x=7\)
Vậy \(x=7\)
\(d/\frac{x}{21}-\frac{2}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{x}{21}=\frac{5}{21}+\frac{2}{3}\)
\(\Rightarrow\frac{x}{21}=\frac{19}{21}\)
\(\Rightarrow x=19\)
Vậy \(x=19\)
#Mạt Mạt#
\(x^2-\frac{3}{5}x=0\)
=> \(xx-\frac{3}{5}x=0\)
=>\(x\left(x-\frac{3}{5}\right)=0\)
=> \(\orbr{\begin{cases}x-\frac{3}{5}=0\\x=0\end{cases}}\)=> \(\orbr{\begin{cases}x=0+\frac{3}{5}\\x=0\end{cases}}\)=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=0\end{cases}}\)
Vậy x = 0 hoặc x = \(\frac{3}{5}\)
\(x^2-\frac{3}{5}x=0\)
\(\Leftrightarrow x\left(x-\frac{3}{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-\frac{3}{5}=0\Leftrightarrow x=\frac{3}{5}\end{cases}}\)
Vậy x = 0 hoặc x = \(\frac{3}{5}\)
\(x^2-\frac{3}{5}x=0\)
\(\Leftrightarrow x.x-\frac{3}{5}x=0\)
\(\Leftrightarrow x.\left(x-\frac{3}{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-\frac{3}{5}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=0+\frac{3}{5}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=\frac{3}{5}\end{cases}}}\)
Vậy ......................
~ Hok tốt ~