\(\dfrac{9+x}{13-x}=\dfrac{5}{6}\)
(-\(\dfrac{5}{9}\) x \(\dfrac{-5}{13}\))+( - \(\dfrac{5}{9}\) x \(\dfrac{6}{13}\))
\(=\dfrac{-5}{9}\left(\dfrac{-5}{13}+\dfrac{6}{13}\right)=\dfrac{-5}{9}\cdot\dfrac{1}{13}=\dfrac{-5}{117}\)
\(\left(-\dfrac{5}{9}\times\dfrac{-5}{13}\right)+\left(-\dfrac{5}{9}\times\dfrac{6}{13}\right)\)
\(=-\dfrac{5}{9}\times\dfrac{-5}{13}+-\dfrac{5}{9}\times\dfrac{6}{13}\)
\(=-\dfrac{5}{9}\times\left(\dfrac{-5}{13}+\dfrac{6}{13}\right)\)
\(=-\dfrac{5}{9}\times\dfrac{1}{13}\)
\(=\dfrac{-5}{117}\)
1.Thực hiện phép tính
a) \(\dfrac{4}{5}\) + \(\dfrac{6}{5}\) : \(\dfrac{7}{15}\) - \(\dfrac{3}{7}\)
b) \(\dfrac{9}{13}\) . \(\dfrac{5}{3}\) + \(\dfrac{-6}{13}\) .\(\dfrac{9}{13}\) - \(\dfrac{9}{13}\) . \(\dfrac{20}{13}\)
2.Tìm x,biết:
a) \(\dfrac{2}{5}\) . x - \(\dfrac{1}{3}\) = 1\(\dfrac{2}{3}\)
b) \(\dfrac{4}{9}\) - \(\dfrac{5}{9}\) : x = \(\dfrac{7}{3}\)
c) \(\dfrac{4}{5}\) . x + \(\dfrac{-7}{3}\) . x = \(\dfrac{\left(-2\right)^3}{5}\)
a)\(\dfrac{7}{8}\)x\(\dfrac{3}{13}\)+\(\dfrac{4}{9}\)x\(\dfrac{4}{13}\)
b)\(\dfrac{6}{5}\)+\(\dfrac{7}{3}\)+\(\dfrac{8}{9}\)
c)23: \(\dfrac{5}{14}\)+\(\dfrac{6}{7}\)+\(\dfrac{4}{9}\)
d)\(4\dfrac{1}{4}\)+\(7\dfrac{3}{7}\)-\(2\dfrac{4}{17}\)
e)8-(9\(\dfrac{2}{11}\)+\(\dfrac{8}{33}\))
a, \(\dfrac{7}{8}\) \(\times\) \(\dfrac{3}{13}\) + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{4}{13}\)
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{21}{8}\) + \(\dfrac{16}{9}\))
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{189}{72}\) + \(\dfrac{128}{72}\))
= \(\dfrac{1}{13}\) \(\times\) \(\dfrac{317}{73}\)
= \(\dfrac{317}{949}\)
b, \(\dfrac{6}{5}\) + \(\dfrac{7}{3}\) + \(\dfrac{8}{9}\)
= \(\dfrac{54}{45}\) + \(\dfrac{105}{45}\) + \(\dfrac{40}{45}\)
= \(\dfrac{199}{45}\)
c, 23 : \(\dfrac{5}{14}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{322}{5}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{20286}{315}\) + \(\dfrac{270}{315}\) + \(\dfrac{140}{315}\)
= \(\dfrac{20696}{315}\)
d, 4\(\dfrac{1}{4}\) + 7\(\dfrac{3}{7}\) - 2\(\dfrac{4}{17}\)
= 4 + \(\dfrac{1}{4}\) + 7 + \(\dfrac{3}{7}\) - 2 - \(\dfrac{4}{17}\)
= (4+7-2) + (\(\dfrac{1}{4}\) + \(\dfrac{3}{7}\) - \(\dfrac{4}{17}\))
= 9 + \(\dfrac{119}{476}\) + \(\dfrac{204}{476}\) - \(\dfrac{112}{476}\)
= 9\(\dfrac{211}{476}\) = \(\dfrac{4495}{476}\)
e, 8 - (9\(\dfrac{2}{11}\) + \(\dfrac{8}{33}\))
= 8 - 9 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= -1 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= \(\dfrac{-33}{33}\) - \(\dfrac{-6}{33}\) - \(\dfrac{8}{33}\)
= - \(\dfrac{47}{33}\)
\(\dfrac{21}{24}\)x\(\dfrac{2}{11}\):\(\dfrac{9}{8}\)=.......
\(\dfrac{17}{9}\)x\(\dfrac{5}{6}\):\(\dfrac{12}{13}\)=.......
\(\dfrac{21}{24}\cdot\dfrac{2}{11}:\dfrac{9}{8}=\dfrac{21}{24}\cdot\dfrac{2}{11}\cdot\dfrac{8}{9}=\dfrac{16}{99}\)
\(\dfrac{17}{9}\cdot\dfrac{5}{6}:\dfrac{12}{13}=\dfrac{17}{9}\cdot\dfrac{5}{6}\cdot\dfrac{13}{12}=\dfrac{1105}{648}\)
a)\(\dfrac{14}{99}\)
b)\(\dfrac{1105}{648}\)
Tìm x
\(\dfrac{9+x}{13-x}\) = \(\dfrac{5}{6}\)
\(\dfrac{9+x}{13-x}=\dfrac{5}{6}\\ \Rightarrow5\left(13-x\right)=6\left(9+x\right)\\ \Rightarrow65-5x=54+6x\\ \Rightarrow6x+5x=65-54\\ \Rightarrow11x=11\\ \Rightarrow x=1\)
Vậy x = 1
ĐKXĐ: \(x\ne13\)
\(\dfrac{9+x}{13-x}=\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{6\left(9+x\right)}{6\left(13-x\right)}=\dfrac{5\left(13-x\right)}{6\left(13-x\right)}\)
\(\Rightarrow6\left(9+x\right)=5\left(13-x\right)\)
\(\Leftrightarrow54+6x=65-5x\)
\(\Leftrightarrow6x+5x=65-54\)
\(\Leftrightarrow11x=11\)
\(\Leftrightarrow x=1\) (thỏa mãn ĐKXĐ)
\(\dfrac{9+x}{13-x}=\dfrac{5}{6}\\ \Rightarrow6\left(9+x\right)=5\left(13-x\right)\\ \Leftrightarrow54+6x=65-5x\\ \Leftrightarrow11x=11\\ \Leftrightarrow x=1\)
Vậy \(x=1\)
Giải các phương trình sau:
\(a.\dfrac{5x-2}{3}=\dfrac{5-3x}{2}\)
\(b.\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(c.2\left(x+\dfrac{3}{5}\right)=5-\left(\dfrac{13}{5}+x\right)\)
\(d.\dfrac{7}{8}x-5\left(x-9\right)=\dfrac{20x+1,5}{6}\)
\(e.\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\)
\(f.\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
a: =>10x-4=15-9x
=>19x=19
hay x=1
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x-32x=60-9
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)
=>3x=6/5
hay x=2/5
d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)
\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)
=>21x-120x+1080=80x+60
=>-179x=-1020
hay x=1020/179
e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>95x+6x=96+5
=>x=1
f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)
=>6x+24-30x+120=10x-15x+30
=>-24x+96=-5x+30
=>-19x=-66
hay x=66/19
\(\dfrac{6}{13}\) x \(\dfrac{8}{7}\) + \(\dfrac{6}{13}\) x \(\dfrac{9}{7}\) - \(\dfrac{3}{13}\) x \(\dfrac{6}{7}\)
Help me please
\(\dfrac{6}{13}.\dfrac{8}{7}+\dfrac{6}{13}.\dfrac{9}{7}-\dfrac{3}{13}.\dfrac{6}{7}=\dfrac{6}{13}.\left(\dfrac{8}{7}+\dfrac{9}{7}\right)-\dfrac{3}{13}.\dfrac{6}{7}=\dfrac{6}{13}.\dfrac{17}{7}-\dfrac{3}{13}.\dfrac{6}{7}=\dfrac{102}{91}-\dfrac{18}{91}=\dfrac{12}{13}\)
9 - 3 x ( X - 9 ) = 6
4 + 6 x ( X + 1 ) 70
\(\dfrac{X}{13}+\dfrac{15}{26}=\dfrac{46}{52}\)
\(\dfrac{11}{14}-\dfrac{3}{X}=\dfrac{5}{14}\)
5 x ( 3 + 7 x X ) = 40
X x 6 + 12 : 3 = 120
X x 3,7 + X x 6,3 = 120
( 15 x 24 - X ) : 0,25 = 100 : \(\dfrac{1}{4}\)
71 + 65 x 4 = \(\dfrac{X+140}{X}\)+ 260
( X +1 ) + ( X + 4 ) + ( x + 7 ) + ...... + (X + 28 ) = 155
đây là bài tìm X
Giải:
\(9-3\times\left(x-9\right)=6\)
\(3\times\left(x-9\right)=9-6\)
\(3\times\left(x-9\right)=3\)
\(x-9=3:3\)
\(x-9=1\)
\(x=1+9\)
\(x=10\)
\(4+6\times\left(x+1\right)=70\)
\(6\times\left(x+1\right)=70-4\)
\(6\times\left(x+1\right)=66\)
\(x+1=66:6\)
\(x+1=11\)
\(x=11-1\)
\(x=10\)
\(\dfrac{x}{13}+\dfrac{15}{26}=\dfrac{46}{52}\)
\(\dfrac{x}{13}=\dfrac{23}{26}-\dfrac{15}{26}\)
\(\dfrac{x}{13}=\dfrac{4}{13}\)
\(\Rightarrow x=4\)
\(\dfrac{11}{14}-\dfrac{3}{x}=\dfrac{5}{14}\)
\(\dfrac{3}{x}=\dfrac{11}{14}-\dfrac{5}{14}\)
\(\dfrac{3}{x}=\dfrac{3}{7}\)
\(\Rightarrow x=7\)
\(5\times\left(3+7\times x\right)=40\)
\(3+7\times x=40:5\)
\(3+7\times x=8\)
\(7\times x=8-3\)
\(7\times x=5\)
\(x=5:7\)
\(x=\dfrac{5}{7}\)
\(x\times6+12:3=120\)
\(x\times6+4=120\)
\(x\times6=120-4\)
\(x\times6=116\)
\(x=116:6\)
\(x=\dfrac{58}{3}\)
\(x\times3,7+x\times6,3=120\)
\(x\times\left(3,7+6,3\right)=120\)
\(x\times10=120\)
\(x=120:10\)
\(x=12\)
\(\left(15\times24-x\right):0,25=100:\dfrac{1}{4}\)
\(\left(360-x\right):0,25=400\)
\(360-x=400.0,25\)
\(360-x=100\)
\(x=360-100\)
\(x=260\)
\(71+65\times4=\dfrac{x+140}{x}+260\)
\(\left(x+140\right):x+260=71+260\)
\(x:x+140:x+260=331\)
\(1+140:x+260=331\)
\(140:x=331-1-260\)
\(140:x=70\)
\(x=140:70\)
\(x=2\)
\(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+...+\left(x+28\right)=155\)
\(10\times x+\left(1+4+7+...+28\right)=155\)
Số số hạng \(\left(1+4+7+...+28\right)\) :
\(\left(28-1\right):3+1=10\)
Tổng dãy \(\left(1+4+7+...+28\right)\) :
\(\left(1+28\right).10:2=145\)
\(\Rightarrow10\times x+145=155\)
\(10\times x=155-145\)
\(10\times x=10\)
\(x=10:10\)
\(x=1\)
Đều theo cách lớp 5 nha em!
a.\(\dfrac{x}{y}\) x \(\dfrac{3}{4}\) = \(\dfrac{5}{6}\) + \(\dfrac{1}{3}\)
b.\(\dfrac{7}{9}:\dfrac{x}{y}=\dfrac{10}{7}-\dfrac{13}{14}\)
a) \(\dfrac{x}{y}\times\dfrac{3}{4}=\dfrac{5}{6}+\dfrac{1}{3}\)
\(\dfrac{x}{y}\times\dfrac{3}{4}=\dfrac{7}{6}\)
\(\dfrac{x}{y}=\dfrac{7}{6}:\dfrac{3}{4}\)
\(\dfrac{x}{y}=\dfrac{14}{9}\)
b) \(\dfrac{7}{9}:\dfrac{x}{y}=\dfrac{10}{7}-\dfrac{13}{14}\)
\(\dfrac{7}{9}:\dfrac{x}{y}=\dfrac{1}{2}\)
\(\dfrac{x}{y}=\dfrac{7}{9}:\dfrac{1}{2}\)
\(\dfrac{x}{y}=\dfrac{14}{9}\)
Câu 1: Tìm x, biết:
a)\(x^2-\dfrac{16}{25}=0\) b)\(\dfrac{2}{5}-\left|\dfrac{1}{2}-x\right|=6\)
C2.Tính giá của biểu thức:
a)\(A=1\dfrac{5}{13}-0,25-\left(2\dfrac{5}{9}+\dfrac{18}{13}-\dfrac{1}{4}\right)\)
b)\(\dfrac{\dfrac{3}{5}.7^2-3.5^6+\dfrac{3}{5}.3^9}{\dfrac{3}{4}.7^2-\dfrac{3}{4}.5^7+\dfrac{3}{4}.3^9}\)
a)
x^2-16/25=0
x^2-4^2/5^2=0
=>x-4/5=0
x=0+4/5
x=0/5