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a: \(\Leftrightarrow\dfrac{x+1}{2x+1}=\dfrac{x+4}{2x+6}\)
=>(x+1)(2x+6)=(2x+1)(x+4)
\(\Leftrightarrow2x^2+6x+2x+6=2x^2+8x+x+4\)
=>9x+4=8x+6
=>x=2
b: \(x^2+5x=0\)
=>x(x+5)=0
=>x=0 hoặc x=-5
a) \(-5\cdot\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\cdot\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}\cdot x-\dfrac{5}{6}\\ -5\cdot x+1-\dfrac{1}{2}\cdot x-\dfrac{1}{3}=\dfrac{3}{2}\cdot x-\dfrac{5}{6}\\ x\cdot\left(-5-\dfrac{1}{2}\right)+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{3}{2}\cdot x\\ x\cdot\dfrac{-11}{2}+\dfrac{7}{6}=\dfrac{3}{2}\cdot x\\ \dfrac{3}{2}\cdot x-\dfrac{-11}{2}\cdot x=\dfrac{7}{6}\\ x\cdot\left(\dfrac{3}{2}-\dfrac{-11}{2}\right)=\dfrac{7}{6}\\ x\cdot7=\dfrac{7}{6}\\ x=\dfrac{7}{6}:7\\ x=\dfrac{1}{6}\)
Vậy x = \(\dfrac{1}{6}\)
b, \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2^x\\ \dfrac{1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot30\cdot31}{2^{30}\cdot\left(1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot30\cdot31\right)\cdot64}=2^x\\ \dfrac{1}{2^{30}\cdot2^6}=2^x\\ \dfrac{1}{2^{36}}=2^x\\ 2^{-36}=2^x\\ \Rightarrow x=-36\)
a.
| x | = 5,6
=>\(\left[{}\begin{matrix}x=5,6\\x=-5,6\end{matrix}\right.\)
Vậy \(x\in\left\{-5,6;5,6\right\}\)
b, \(\left|x-3,5\right|=5\)
=>\(\left[{}\begin{matrix}x-3,5=5\\x-3,5=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8,5\\x=-1,5\end{matrix}\right.\)
Vậy \(x\in\left\{-1,5;8,5\right\}\)
c,\(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=> \(\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{4};\dfrac{5}{4}\right\}\)
d,\(\left|4x\right|-\left(\left|-13,5\right|\right)=\left|\dfrac{1}{4}\right|\)
=> \(\left|4x\right|-13,5=\dfrac{1}{4}\)
=> \(\left|4x\right|=13,75\)
=>\(\left[{}\begin{matrix}4x=13,75\\4x=-13,75\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3,4375\\x=-3,4375\end{matrix}\right.\)
Vậy \(x\in\left\{-3,4375;3,4375\right\}\)
e, ( x - 1 ) 3 = 27
=> x - 1 = 3
=> x = 4
Vậy x = 4
f, ( 2x - 3)2 = 36
=> \(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=4,5\\x=-1,5\end{matrix}\right.\)
Vậy x\(\in\left\{-1,5;4,5\right\}\)
g, \(5^{x+2}=625\)
=> \(5^{x+2}=5^4\)
=> x + 2 = 4
=> x = 2
Vậy x = 2
h, ( 2x - 1)3 = -8
=> 2x - 1 = -2
=> x = \(\dfrac{-1}{2}\)
Vậy x = \(\dfrac{-1}{2}\)
i, \(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}...\dfrac{30}{62}.\dfrac{31}{64}=2^x\)
=> \(\dfrac{1.2.3.4.5...30.31}{4.6.8.10.12...62.64}=2^x\)
=>\(\dfrac{1.2.3.4.5...30.31}{\left(2.3.4.5...30.31.32\right)\left(2.2.2.2...2.2_{ }\right)}=2^x\)(có 31 số 2)
=> \(\dfrac{1}{32.2^{31}}=2^x\)
=> \(\dfrac{1}{2^{36}}=2^x\)
=> x = -36
Vậy x = -36
a, \(\dfrac{3}{4}+x=\dfrac{8}{13}\)
\(x=\dfrac{8}{13}-\dfrac{3}{4}\)
\(x=-\dfrac{7}{52}\)
b,\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=-\dfrac{3}{20}\)
c, \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(2x-\dfrac{1}{7}=0\)
\(x-\dfrac{1}{7}=0:2\)
\(x-\dfrac{1}{7}=0\)
\(x=0-\dfrac{1}{7}\)
\(x=\dfrac{1}{7}\)
d, \(\dfrac{3}{4}+\dfrac{1}{4}\div x=\dfrac{2}{5}\)
\(\left(\dfrac{3}{4}+\dfrac{1}{4}\right):x=\dfrac{2}{5}\)
\(1:x=\dfrac{2}{5}\)
\(x=1:\dfrac{2}{5}\)
\(x=\dfrac{5}{2}\)
a) \(\dfrac{3}{4}+x=\dfrac{8}{13}\)\(\Leftrightarrow\) \(x=\dfrac{8}{13}-\dfrac{3}{4}=\dfrac{-7}{52}\) vậy \(x=\dfrac{-7}{52}\)
b) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\) \(\Leftrightarrow\) \(\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\) \(\Leftrightarrow\) \(x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=\dfrac{-3}{20}\) vậy \(x=\dfrac{-3}{20}\)
c) \(2x\left(x-\dfrac{1}{7}\right)=0\) \(\Leftrightarrow\) \(2x^2-\dfrac{2}{7}x=0\)
\(\Delta\) = \(\left(\dfrac{-2}{7}\right)^2-4.2.0=\dfrac{4}{49}>0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(x_1=\dfrac{\dfrac{2}{7}+\sqrt{\dfrac{4}{49}}}{4}=\dfrac{1}{7}\)
\(x_2=\dfrac{\dfrac{2}{7}-\sqrt{\dfrac{4}{49}}}{4}=0\)
vậy \(x=0;x=\dfrac{1}{7}\)
Các câu dễ tự làm :v
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
Bài 7:
x/1=z/2 nên x/6=z/12
=>x/6=y/9=z/12
=>x/2=y/3=z/4
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{27}{9}=3\)
=>x=6; y=9; z=12
\(a)\dfrac{-1}{4}.13\dfrac{9}{11}-0,25.6\dfrac{2}{11}\)
\(=\dfrac{-1}{4}.\dfrac{152}{11}-\dfrac{1}{4}.\dfrac{68}{11}\)
\(=-38-\dfrac{17}{11}\)
\(=\dfrac{-418}{11}+\dfrac{-17}{11}\)
\(=\dfrac{-435}{11}\)
\(b)\dfrac{31}{9}.\left|x\right|-\dfrac{5}{2}=\dfrac{8}{3}\)
\(\Leftrightarrow\dfrac{31}{9}.\left|x\right|=\dfrac{16}{6}+\dfrac{15}{6}\)
\(\Leftrightarrow\dfrac{31}{9}.\left|x\right|=\dfrac{31}{6}\)
\(\Leftrightarrow\left|x\right|=\dfrac{31}{6}.\dfrac{9}{31}\)
\(\Leftrightarrow\left|x\right|=1,5\)
\(\Leftrightarrow x\in\left\{1,5;-1,5\right\}\)
Vậy \(x\in\left\{1,5;-1,5\right\}\)
a) \(2\cdot31\cdot12+4\cdot6\cdot41+8\cdot28\cdot3\)
\(=24\cdot31+24\cdot41+24\cdot28\)
\(=24\cdot\left(31+41+28\right)\)
\(=24\cdot\left(72+28\right)\)
\(=24\cdot100\)
\(=2400\)
b) \(3,9\cdot\dfrac{13}{10}+\dfrac{0,87}{0,01}\cdot0,39\)
\(=\dfrac{3,9}{10}\cdot13+87\cdot0,39\)
\(=0,39\cdot13+87\cdot0,39\)
\(=0,39\cdot\left(13+87\right)\)
\(=0,39\cdot100\)
\(=39\)