Bài 2. Tìm x biết
\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
Tìm x, biết:
a) \(\dfrac{2}{3}\)x - \(\dfrac{1}{2}\)x = \(\left(-\dfrac{7}{12}\right)\) . \(1\dfrac{2}{5}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2\) = \(\dfrac{9}{4}\)
c) (1,25 - \(\dfrac{4}{5}\)x)3 = -125
a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)
\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{49}{10}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)
+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)
\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{13}{15}\)
+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)
\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)
\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{125}{16}\)
a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)
\(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)
\(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)
\(x\) = - \(\dfrac{49}{60}\).6
\(x\) = -\(\dfrac{49}{10}\)
Bài 1 : Thực hiện phép tính:
a, \(\dfrac{6}{7}.[\left(\dfrac{-7}{5}-\dfrac{3}{2}:\dfrac{-5}{-4}\right)+\left(\dfrac{3}{2}\right)^2]\)
Bài 2 : Tìm x :
a , \(\dfrac{2}{3}:x=1,4-\dfrac{12}{5}\)
b , \(\dfrac{x}{5}=\dfrac{5}{6}+\dfrac{-19}{30}\)
Bài 3 : Lớp 6A có 40 h/s . Cuối năm xếp loại giỏi , khá , TB . Biết số h/s giỏi bằng \(\dfrac{1}{4}\)
số h/s cả lớp . Số h/s khá bằng \(\dfrac{3}{5}\)số h/s còn lại
Tính số h/s giỏi , khá , TB?
Bài 4 : Lớp 6B có h/s xếp loại giỏi , khá , TB , yếu . Số h/s giỏi chiếm \(\dfrac{1}{6}\) , số h/s khá chiếm \(\dfrac{1}{3}\)
số h/s cả lớp . Số h/s TB chiếm \(\dfrac{7}{8}\) số h/s giỏi . Còn lại là 17 h/s yếu . Tính số h/s lớp 6B
2:
a: =>2/3:x=1,4-2,4=-1
=>x=-2/3
b: =>x/5=25/30-19/30=6/30=1/5
=>x=1
3:
Số học sinh giỏi là 40*1/4=10 bạn
Số học sinh khá là 30*3/5=18 bạn
Số học sinh TB là 30-18=12 bạn
5. \(3-1\dfrac{1}{2}-x+\dfrac{5}{4}=2-\left|1\dfrac{1}{8}-\dfrac{5}{12}\right|\) 6. \(3\dfrac{1}{14}-5\dfrac{1}{3}-\dfrac{4}{7}+\dfrac{11}{21}=x-\dfrac{1}{2}\) 7. \(\dfrac{11}{-40}+\dfrac{4}{5}-\left|\dfrac{3}{4}-1\dfrac{5}{12}\right|=\dfrac{3}{20}-x\)
chắc h có mấy thành cay r nên ko làm bn lên mạng tải phẩn mêm có cánh iair đó :D
\(3-1\dfrac{1}{2}-x+\dfrac{5}{4}=2-\left|1\dfrac{1}{8}-\dfrac{5}{12}\right|\)
\(3-1\dfrac{1}{2}-x+\dfrac{5}{4}=2-\dfrac{17}{24}\)
\(x=3-1\dfrac{1}{2}+\dfrac{5}{4}-2+\dfrac{17}{24}\)
\(x=\dfrac{35}{24}\)
bn làm từng bước nhé :D
Tính:
a) \(\dfrac{11}{10}+\dfrac{3}{5}:\dfrac{2}{3}\)
b) \(\dfrac{4}{3}\) + 5 x \(\dfrac{5}{8}\)
c) \(\left(\dfrac{2}{5}+\dfrac{3}{7}\right)x\dfrac{25}{29}\)
d) \(\dfrac{1}{4}x\dfrac{5}{12}+\dfrac{5}{12}x\dfrac{4}{5}\)
a) \(\dfrac{11}{10}+\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{11}{10}+\dfrac{3}{5}\times\dfrac{3}{2}=\dfrac{11}{10}+\dfrac{9}{10}=\dfrac{20}{10}=2\)
b) \(\dfrac{4}{3}+5\times\dfrac{5}{8}=\dfrac{4}{3}+\dfrac{25}{8}=\dfrac{32}{24}+\dfrac{75}{24}=\dfrac{107}{24}\)
c) \(\left(\dfrac{2}{5}+\dfrac{3}{7}\right)\times\dfrac{25}{29}=\left(\dfrac{14}{35}+\dfrac{15}{35}\right)\times\dfrac{25}{39}=\dfrac{29}{35}\times\dfrac{25}{39}=\dfrac{145}{274}\)
d) \(\dfrac{1}{4}\times\dfrac{5}{12}+\dfrac{5}{12}\times\dfrac{4}{5}=\dfrac{5}{12}\times\left(\dfrac{1}{4}+\dfrac{4}{5}\right)=\dfrac{5}{12}\times\dfrac{21}{20}=\dfrac{105}{240}=\dfrac{7}{16}\)
a) \(\dfrac{11}{10}+\dfrac{3}{5}x\dfrac{3}{2}=\dfrac{11}{10}+\dfrac{9}{10}=\dfrac{20}{10}=2\)
b) \(\dfrac{4}{3}+\dfrac{25}{8}=\dfrac{32}{24}+\dfrac{75}{24}=\dfrac{107}{24}\)
c) \(\dfrac{29}{35}x\dfrac{25}{29}=\dfrac{5}{7}\)
\(=\dfrac{5}{12}x\left(\dfrac{1}{4}+\dfrac{4}{5}\right)=\dfrac{5}{12}x\dfrac{21}{20}=\dfrac{7}{16}\)
5. \(3-1\dfrac{1}{2}-x+\dfrac{5}{4}=2-\left|1\dfrac{1}{8}-\dfrac{5}{12}\right|\) 6. \(3\dfrac{1}{14}-5\dfrac{1}{3}-\dfrac{4}{7}+\dfrac{11}{21}=-\dfrac{1}{2}\) 7. \(\dfrac{11}{-40}+\dfrac{4}{5}-\left|\dfrac{3}{4}-1\dfrac{5}{12}\right|=\dfrac{3}{20}-X\)
Bài 2:Tìm x biết:
a)\(\dfrac{1}{7}+x=\dfrac{-2}{3}\)
b)\(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)
c)\(\left\{\dfrac{3}{5}-2x\right\}.\dfrac{5}{8}=1\)
d)\(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)
e)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
f)\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
g)\(\left|X+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)
h)\(\left(\dfrac{1}{32}\right)^x.8^{2x}=512\)
i)\(5,3x+\left(-3,3\right)x+1,7=-4,9\)
a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)
hay \(x=-\dfrac{17}{21}\)
Vậy: \(x=-\dfrac{17}{21}\)
b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)
\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)
Vậy: \(x=\dfrac{4}{5}\)
c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)
\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)
\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)
hay \(x=-\dfrac{1}{2}\)
Vậy: \(x=-\dfrac{1}{2}\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)
\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)
hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)
Vậy: \(x=-\dfrac{2}{3}\)
e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)
Vậy: \(x=-\dfrac{5}{7}\)
f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)
\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)
\(\Leftrightarrow-x-\dfrac{9}{60}=0\)
\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)
hay \(x=-\dfrac{3}{20}\)
Vậy: \(x=-\dfrac{3}{20}\)
g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)
tìm x
\(\dfrac{3-x}{5-x}=\dfrac{6}{11}\) \(\left(1\dfrac{1}{3}-25\%.x-\dfrac{5}{12}\right)-2x=1,6:\dfrac{3}{5}\)
\(\dfrac{1}{2}.\left(x-\dfrac{2}{3}\right)-\dfrac{1}{3}.\left(2x-3\right)=x\)
\(2.\left(\dfrac{1}{2}-x\right)-3\left(x-\dfrac{1}{3}\right)=\dfrac{7}{2}\)
a: =>11(x-3)=6(x-5)
=>11x-33=6x-30
=>5x=3
=>x=3/5
b: =>(4/3-1/4x-5/12)-2x=8/5*5/3=8/3
=>-9/4x+11/12=8/3
=>-9/4x=32/12-11/12=21/12=7/4
=>x=-7/9
c: =>1/2x-1/3-2/3x-1=x
=>-1/6x-4/3=x
=>-7/6x=4/3
=>x=-4/3:7/6=-4/3*6/7=-24/21=-8/7
d: =>1-2x-3x+1=7/2
=>-5x=3/2
=>x=-3/10
MỌI NGƯỜI GIÚP EM VỚI
Bài 1: tìm x
a)\(\left|3x-5\right|=4\)
b)\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
c)\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
Bài 2: Tính
a)\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
b)\(\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
c)\(\dfrac{\left(\dfrac{3}{10}-\dfrac{4}{15}-\dfrac{7}{20}\right).\dfrac{5}{19}}{\left(\dfrac{1}{14}+\dfrac{1}{7}-\dfrac{-3}{35}\right).\dfrac{-4}{3}}\)
Bài 1:
a) \(\left|3x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x=-2004\)( do \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\))
Bài 2:
a) \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}=1\)
b) \(=-\left(\dfrac{1}{99.100}+\dfrac{1}{98.99}+\dfrac{1}{97.98}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)
\(=-\left(\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{98}-\dfrac{1}{99}+...+1-\dfrac{1}{2}\right)\)
\(=-\left(1-\dfrac{1}{100}\right)=-\dfrac{99}{100}\)
Bài 1:
a) \(\left|3x-5\right|=4\) (1)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
b) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\) \(\left(do\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\) \(\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)
\(\Leftrightarrow x=-2004\)
tìm x biết:
a) \(5^x.\left(5^3\right)^2=625\)
b)\(\left(\dfrac{12}{15}\right)^x=\left(\dfrac{5}{3}\right)^{-5}-\left(-\dfrac{3}{5}\right)^4\)
c)\(\left(-\dfrac{3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
d)\(172x^2-7^9:98^3=2^{-3}\)