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a: \(\dfrac{x}{0.9}=\dfrac{5}{6}\)
\(\Leftrightarrow x=\dfrac{3}{4}\)
b: \(\dfrac{-6}{x}=\dfrac{9}{-15}\)
\(\Leftrightarrow x=10\)
c: \(\dfrac{\dfrac{14}{15}}{\dfrac{9}{10}}=\dfrac{x}{\dfrac{3}{7}}\)
\(\Leftrightarrow x=\dfrac{3}{7}\cdot\dfrac{14}{15}:\dfrac{9}{10}=\dfrac{2}{5}\cdot\dfrac{10}{9}=\dfrac{20}{45}=\dfrac{4}{9}\)
a) \(\frac{x}{-4}=\frac{-3}{5}\)
\(\Rightarrow5x=\left(-3\right).\left(-4\right)\)
\(\Rightarrow5x=12\)
\(\Rightarrow x=\frac{12}{5}\)
những câu khác làm tương tự. Chỉ cần tích chéo là được
\(1\frac{1}{3}:0,8=\frac{2}{3}:x\)
\(\frac{4}{3}:\frac{4}{5}=\frac{2}{3}:x\)
\(\frac{2}{3}:x=\frac{4}{3}.\frac{5}{4}\)
\(\frac{2}{3}:x=\frac{5}{3}\)
\(x=\frac{2}{3}:\frac{5}{3}\)
\(x=\frac{2}{3}.\frac{3}{5}\)
\(x=\frac{2}{5}\)
vậy \(x=\frac{2}{5}\)
a, \(2,5:4x=0,5:0,2\)
\(\Rightarrow4x=\dfrac{2,5.0,2}{0,5}\)
\(\Rightarrow4x=1\Rightarrow x=\dfrac{1}{4}\)
b, \(\dfrac{1}{5}x:3=\dfrac{2}{3}:0,5\)
\(\Rightarrow\dfrac{1}{5}x=\dfrac{3.\dfrac{2}{3}}{0,5}\)
\(\Rightarrow\dfrac{1}{5}x=4\Rightarrow x=20\)
c, \(1,25:0,8=\dfrac{3}{8}:0,2x\)
\(\Rightarrow0,2x=\dfrac{0,8.\dfrac{3}{8}}{1,25}\)
\(\Rightarrow0,2x=0,24\Rightarrow x=1,2\)
Chúc bạn học tốt!!
a, \(2,5:4x=0,5:0,2\)
\(2,5:4x=2,5\)
\(4x=2,5:2,5\)
\(4x=1\)
\(x=1:4\)
\(x=\dfrac{1}{4}\)
Vậy .............
b, \(\dfrac{1}{5}x:3=\dfrac{2}{3}\)
\(\dfrac{1}{5}x=\dfrac{2}{3}.3\)
\(\dfrac{1}{5}x=2\)
\(x=2:\dfrac{1}{5}\)
\(x=10\)
Vậy .....
c, \(1,25:0,8=\dfrac{3}{8}:0,2x\)
\(1,5625=\dfrac{3}{8}:0,2x\)
\(0,2x=\dfrac{3}{8}:1,5625\)
\(0,2x=0,24\)
\(x=0,24:0,2\)
\(x=1,2\)
Vậy ...
a. 2(x-1) + (x+2) - (x+3) = 15 - (x+1)
=>2x-2+x+2-x-3=15-x-1
=>(2x+x-x)-2+2-3=15-1-x
=>2x-3=14-x
=>3x=17
=>x=17/3
b. x+1/15 + x+2/14 = x+4/12 + x+5/11
\(\Rightarrow\frac{x+1}{15}+1+\frac{x+2}{14}+1=\frac{x+4}{12}+1+\frac{x+5}{11}+1\)
\(\Rightarrow\frac{x+16}{15}+\frac{x+16}{14}=\frac{x+16}{12}+\frac{x+16}{11}\)
\(\Rightarrow\frac{x+16}{15}+\frac{x+16}{14}-\frac{x+16}{12}-\frac{x+16}{11}=0\)
\(\Rightarrow\left(x+16\right)\left(\frac{1}{15}+\frac{1}{14}-\frac{1}{12}-\frac{1}{11}\right)=0\)
\(\Rightarrow x+16=0\).Do \(\frac{1}{15}+\frac{1}{14}-\frac{1}{12}-\frac{1}{11}\ne0\)
=>x=-16
Sorry mink mới lớp 5 nên ko thể giúp bn lm bài toán này thành thật xin lỗi
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}+\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Dễ thấy \(\frac{1}{10}>\frac{1}{11}>\frac{1}{12}>\frac{1}{13}>\frac{1}{14}\)nên biểu thức trong ngoặc thứ hai \(\ne\)0
Do đó \(x+1=0\)\(\Rightarrow x=0-1=-1\)
b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+4}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right).\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)
Vì \(\frac{1}{2000}>\frac{1}{2001}>\frac{1}{2002}>\frac{1}{2003}\)nên biểu thức trong ngoặc thứ hai phải \(\ne\)0
Do đó \(x+2004=0\)\(\Rightarrow x=0-2004=-2004\)
\(\frac{x+2}{0,8}=\frac{0,2}{x+2}\)
=> \(\left(x+2\right)^2=0,2\cdot0,8\)
=> \(\left(x+2\right)^2=0,16\)
=> \(x+2=0,4\) và \(x+2=-0,4\)
=> \(x=0,4-2=-1,6\) và \(x=-0,4-2=-2,4\)
\(\frac{x+2}{0,8}=\frac{0,2}{x+2}\Leftrightarrow\left(x+2\right)\left(x+2\right)=0,2.0,8\Leftrightarrow\left(x+2\right).\left(x+2\right)=0,16\)
\(\Leftrightarrow\left(x+2\right)^2=0,16\)
(+) x + 2 =\(\sqrt{0,16}\) => x + 2 = 0,4 => x = 0,4 - 2 = - 1 ,6
(+) x + 2 = - căn 0,16 => x + 2 = - 0,4 => x = -0,4 - 2 = -2,4
a) \(\frac{0,5}{0,2}=\frac{1,25}{0,1x}\Leftrightarrow0,1x.0,5=0,2.1,25\)
\(\Leftrightarrow0,1x.0,5=0,25\Leftrightarrow0,1x=0,5\Leftrightarrow x=5\)
b) \(x-\frac{3}{2}=2x-\frac{4}{3}\Leftrightarrow x-2x=\frac{-4}{3}+\frac{3}{2}\)
\(\Leftrightarrow x-2x=\frac{1}{6}\Leftrightarrow-x=\frac{1}{6}\Leftrightarrow x=\frac{-1}{6}\)
c) \(x+\frac{13}{14}=\frac{4}{7}\Rightarrow x=\frac{4}{7}-\frac{13}{14}\Rightarrow x=\frac{-5}{14}\)
d)\(-3\left(x-2\right)=2x+1\)
\(\Leftrightarrow-3x+6=2x+1\Leftrightarrow-3x-2x=1-6\)
\(\Leftrightarrow-5x=-5\Leftrightarrow x=1\)
e) \(\left(x-1\right)^2-4=0\Leftrightarrow\left(x-1\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=\left(-2\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
cậu có thể tham khảo bài trên ạ, nếu thấy đúng thì cho mk 1 t.i.c.k ạ, thank nhiều
\(d,-3\left(x-2\right)=2x+1\)
\(< =>-3x+6=2x+1\)
\(< =>-3x-2x+6-1=0\)
\(< =>5-5x=0\)
\(< =>5\left(1-x\right)=0< =>x=1\)
\(e,\left(x-1\right)^2-4=0\)
\(< =>\left(x-1+2\right)\left(x-1-2\right)=\left(x+1\right)\left(x-3\right)=0\)
\(< =>\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}< =>\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)
a,\(\Leftrightarrow\frac{x}{0,2}=\frac{x}{0,8}\Leftrightarrow0,8x=0,2x\Leftrightarrow0,6x=0\Leftrightarrow x=0\)
b,\(\Leftrightarrow\frac{x+11}{14-x}=\frac{2}{3}\)
\(\Leftrightarrow3\left(x+11\right)=2\left(14-x\right)\)
\(\Leftrightarrow3x+33=28-2x\)
\(\Leftrightarrow3x+2x=28-33\)
\(\Leftrightarrow5x=-5\Leftrightarrow x=-1\)