12:\(\frac{x}{4}\)+4=15:(-5)
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\(B=1\frac{6}{41}\cdot\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}}\div\frac{4+\frac{4}{15}+\frac{4}{4}+\frac{4}{2013}}{5+\frac{5}{15}+\frac{5}{4}+\frac{5}{2013}}\right)\cdot\frac{124242423}{237373735}\)
\(B=\frac{47}{41}\cdot\left[\frac{12\left(1+\frac{1}{19}-\frac{1}{37}-\frac{1}{53}\right)}{3\left(1+\frac{1}{19}-\frac{1}{37}-\frac{1}{53}\right)}\div\frac{4\left(1+\frac{1}{15}+\frac{1}{4}+\frac{1}{2013}\right)}{5\left(1+\frac{1}{15}+\frac{1}{4}+\frac{1}{2013}\right)}\right]\cdot\frac{123}{235}\)
\(B=\frac{47}{41}\cdot\left[\frac{12}{3}\div\frac{4}{5}\right]\cdot\frac{123}{235}\)
\(B=\frac{3}{5}\cdot3\cdot\frac{5}{4}\)
\(B=\frac{9}{4}\)
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
a) \(\frac{6x-5}{-7}=\frac{5x-3}{-5}\)
=> -5(6x - 5) = -7(5x - 3)
=> -30x + 25 = -35x + 21
=> -30x + 25 + 35x - 21 = 0
=> (-30x + 35x) + (25 - 21) = 0
=> 5x + 4 = 0
=> 5x = -4
=> x = -4/5
b) \(\frac{12-7x}{-13}=\frac{4-3x}{-5}\)
=> -5(12 - 7x) = -13(4 - 3x)
=> -60 + 35x = -52 + 39x
=> -60 + 35x + 52 - 39x = 0
=> (-60 + 52) + (35x - 39x) = 0
=> -8 - 4x = 0
=> -8 = 4x
=> x = -2
c) \(\frac{2x+4}{7}=\frac{4x-2}{15}\)
=> 15(2x + 4) = 7(4x - 2)
=> 30x + 60 = 28x - 14
=> 30x + 60 - 28x + 14 = 0
=> 2x + 74 = 0
=> 2x = -74
=> x = -37
Áp dụng BĐT AM-GM ta có:
\(\left(\frac{12}{5}\right)^x+\left(\frac{15}{4}\right)^x\ge2\sqrt{9^x}=2\cdot3^x\)
\(\left(\frac{15}{4}\right)^x+\left(\frac{20}{3}\right)^x\ge2\sqrt{25^x}=2\cdot5^x\)
\(\left(\frac{20}{3}\right)^x+\left(\frac{12}{5}\right)^x\ge2\sqrt{16^x}=2\cdot4^x\)
Cộng theo vế 3 BĐT trên ta có:
\(2\left[\left(\frac{12}{5}\right)^x+\left(\frac{15}{4}\right)^x+\left(\frac{20}{3}\right)^x\right]\ge2\left(3^x+4^x+5^x\right)\)
\(\Rightarrow\left(\frac{12}{5}\right)^x+\left(\frac{15}{4}\right)^x+\left(\frac{20}{3}\right)^x\ge3^x+4^x+5^x\)
\(\left(\frac{12}{5}\right)^x+\left(\frac{15}{4}\right)^x\ge2\sqrt{\left(\frac{12}{5}\right)^x.\left(\frac{15}{4}\right)^x}=2.3^x;\left(\frac{20}{3}\right)^x+\left(\frac{12}{5}\right)^x\ge2.4^x\)
Cộng các vế tương ứng => đpcm
\(\dfrac{3}{7}+\dfrac{1}{2}=\dfrac{-15}{12}x+\dfrac{6}{5}x\)
\(\Leftrightarrow\)\(\dfrac{-15.5+12.6}{60}x=\dfrac{3.2+1.7}{14}\)
\(\Leftrightarrow\)\(\dfrac{-3}{60}x=\dfrac{13}{14}\)
\(\Leftrightarrow\)\(\dfrac{-1}{20}x=\dfrac{13}{14}\)
\(\Leftrightarrow\)x=-\(\dfrac{13}{14}:\dfrac{1}{20}=-\dfrac{13.20}{14}=-\dfrac{130}{7}\)
3(12+x)=7(x-4)
\(\Leftrightarrow\)36+3x=7x-28
\(\Leftrightarrow\)7x-3x=36+28
\(\Leftrightarrow\)4x=64
\(\Leftrightarrow\)x=16
a)\(-\frac{15}{12}x+\frac{3}{7}=\frac{6}{4}x-\frac{1}{2}\)
\(\Leftrightarrow\frac{15}{12}x+\frac{6}{4}x=\frac{1}{2}+\frac{3}{7}\)
\(\Leftrightarrow\frac{11}{4}x=\frac{9}{14}\Leftrightarrow x=\frac{36}{154}\)
b) \(\frac{2}{5}\left(x+1\right)-\frac{4}{5}x=0\)
\(\Leftrightarrow\frac{2}{5}x+\frac{2}{5}-\frac{4}{5}x=0\)
\(\Leftrightarrow-\frac{2}{5}x=-\frac{2}{5}\Leftrightarrow x=1\)
12:x/4+4=15:(-5)
12:x/4=-3-4
x/4=-12/7
<=>x*7=-12*4
<=>7x= -48
<=>x= -48/7
Vậy x=-48/7