\(14:\left(0,4+\frac{0,16}{x}\right)=7\)
TÌM X
14:(0,4+0,16:x)=7
0,4+0,16:x=14:7
0,4+0,16:x=2
0,16:x=2-0,4
0,16:x=1,6
x=0,1
14:(0,4+0,16:x)=7
=> (0,4+0,16:x)=14:7
=>(0,4+0,16:x)=2
=>0,16:x=2-0,4
=> 0,16:x =1,6
=>x= 0,16:1,6
=>x=0,1
14 : ( 0,4 + 0,16 : x) = 7
0,4 + 0,16 : x = 14 : 7
0,4 + 0,16 : x = 2
0,16 : x = 2 - 0,4
0,16 : x = 1,6
x =0,16 : 1,6
x =0,1
tìm x
14: ( 0,4+0,16 : x ) =7
14: ( 0,4+0,16 : x ) =7
0,4 + 0,16 : x = 14 : 7 = 2
0,16 : x = 2 - 0,4 = 1,6
x = 0,16 : 1,6 = 0,1
Vậy x = 0,1
14: ( 0,4+0,16 : x ) =7
=>\(0,4+0,16:x\)= 14:7
=>\(\left(0,4+0,16:x\right)=2\)
=>\(0,16:x=2-0,4\)
=>\(0,16:x=1,6\)
=> x=0,1
14/(0,4+0,16:x)=7
0,4+0,16:x=2
0,16:x=1,6
x=0,16:1,6
x=0,1
Chuc ban vao nam hoc moi dc nhieu thanh h cao!!
Tìm x :
a ) 14 : ( 0,4 + \(\frac{0,16}{x}\) ) = 7
Mik sẽ like cho bạn nào trả lời trước ^^
\(a.\)
\(14:\left(0,4+\frac{0,16}{x}\right)=7\)\(=>0,4+\frac{0,16}{x}=14:7\)
\(=>0,4+\frac{0,16}{x}=2=>\frac{0,16}{x}=2-0,4\)
\(=>\frac{0,16}{x}=1,6=>0,16:x=1,6\)
\(=>x=0,16:1,6=>x=0,1\)
Vậy \(x=0,1\)
Tìm x biết
a, 14 : ( 0,4 + 0,16/x ) = 7
b, 520 +7,5 × 4 = x + 174 /50 + 50
a, 14 : (0,4 + \(\dfrac{0,16}{x}\)) = 7
0,4 + \(\dfrac{0,16}{x}\) = 14 : 7
0,4 + \(\dfrac{0,16}{x}\) = 2
\(\dfrac{0,16}{x}\) = 2 - 0,4
\(\dfrac{0,16}{x}\) = 1,6
\(x\) = 0,16 : 1,6
\(x\) = 0,1
b, 520 + 7,5 x 4 = \(x\) + \(\dfrac{174}{50}\) + 50
520 + 30 = \(x\) + 3,48 + 50
\(x\) + 53,48 = 550
\(x\) = 550 - 53,48
\(x\) = 496,52
Bài 3: Giải các phương trình sau bằng cách đưa về dạng ax+b =0 :
a) \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\)
b) \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{5}\)
c) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x-1\right)}{7}-5\)
d) \(14\frac{1}{2}-\frac{2\left(x+3\right)}{5}=\frac{3x}{2}-\frac{2\left(x-7\right)}{3}\)
a) Ta có: \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\)
\(\Leftrightarrow\frac{63\left(3x-11\right)}{693}-\frac{231x}{693}-\frac{99\left(3x-5\right)}{693}+\frac{77\left(5x-3\right)}{693}=0\)
\(\Leftrightarrow189x-693-231x-297x+495+385x-231=0\)
\(\Leftrightarrow46x-429=0\)
\(\Leftrightarrow46x=429\)
hay \(x=\frac{429}{46}\)
Vậy: \(x=\frac{429}{46}\)
b) Ta có: \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{5}\)
\(\Leftrightarrow\frac{9x-0,7}{4}-\frac{5x-1,5}{7}-\frac{7x-1,1}{6}+\frac{5\left(0,4-2x\right)}{5}=0\)
\(\Leftrightarrow105\left(9x-0,7\right)-60\left(5x-1,5\right)-70\left(7x-1,1\right)+420\left(0,4-2x\right)=0\)
\(\Leftrightarrow945x-\frac{147}{2}-300x+90-490x+77+168-840x=0\)
\(\Leftrightarrow-685x+261.5=0\)
\(\Leftrightarrow-685x=-261.5\)
hay \(x=\frac{523}{1370}\)
Vậy: \(x=\frac{523}{1370}\)
c) Ta có: \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x-1\right)}{7}-5\)
\(\Leftrightarrow\frac{14\left(5x-3\right)}{84}-\frac{21\left(7x-1\right)}{84}-\frac{24\left(2x-1\right)}{84}+\frac{420}{84}=0\)
\(\Leftrightarrow70x-42-147x+21-48x+24+420=0\)
\(\Leftrightarrow-125x+423=0\)
\(\Leftrightarrow-125x=-423\)
hay \(x=\frac{423}{125}\)
Vậy: \(x=\frac{423}{125}\)
d) Ta có: \(14\frac{1}{2}-\frac{2\left(x+3\right)}{5}=\frac{3x}{2}-\frac{2\left(x-7\right)}{3}\)
\(\Leftrightarrow\frac{435}{30}-\frac{12\left(x+3\right)}{30}-\frac{45x}{30}+\frac{20\left(x-7\right)}{30}=0\)
\(\Leftrightarrow435-12x-36-45x+20x-140=0\)
\(\Leftrightarrow-37x+259=0\)
\(\Leftrightarrow-37x=-259\)
hay \(x=7\)
Vậy: x=7
Tính :
a) \(1\frac{7}{20}:2,7+2,7:1,35+\left(0,4:2\frac{1}{2}\right).\left(4,8-1\frac{3}{40}\right)\)
b) \(\left[\left(6\frac{3}{5}-3\frac{3}{14}\right):5\frac{5}{6}\right]:\left[\left(21-1,25\right):2,5\right]\)
Bài 1:
\(a,\frac{-5}{14}-\frac{37}{14}\le x
14:(0,4*x+0,16)*x+5=17
ai giúp mình được ko, đúng mình cho.
Rút gọn các biểu thức sau:
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)\) b) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
c) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}\) d) \(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)