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a/ \(\left(3x-\dfrac{2}{4}\right)\left(x+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{2}{4}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy ................
b/ \(\left(2x-5\right).\left(\dfrac{3}{2}x+9\right).\left(0,3x-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\\dfrac{3}{2}x+9=0\\0,3x-12=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\\dfrac{3}{2}x=-9\\0,3x=12\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-6\\x=40\end{matrix}\right.\)
Vậy ..
\(a)\left(3x-\dfrac{2}{4}\right).\left(x+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{2}{4}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
\(b)\left(2x-5\right).\left(\dfrac{3}{2}x+9\right).\left(0,3x-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\\dfrac{3}{2}x+9=0\\0,3x-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\\dfrac{3}{2}x=-9\\0,3x=12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-6\\x=40\end{matrix}\right.\)
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a,\(\dfrac{2}{3}x-\dfrac{4}{9}=-\dfrac{5}{27}\)
\(\dfrac{2}{3}x=\dfrac{17}{27}\)
\(x=\dfrac{17}{18}\)
b,\(\dfrac{2}{3}x+\dfrac{4}{9}=\dfrac{5}{27}\)
\(\dfrac{2}{3}x=-\dfrac{7}{27}\)
\(x=-\dfrac{7}{18}\)
c,\(x:1,2=5,4:6\)
\(x:1,2=0,9\)
\(x=1,08\)
d,\(1,68:1,2=5,4:x\)
\(1,4=5,4:x\)
\(x=\dfrac{27}{7}\)
e,\(\left(1-2x\right)^2+1=10\)
\(\left(1-2x\right)^2=9\)
\(1-2x=3\)
\(2x=-2\)
\(x=-1\)
f,\(\left(1-2x\right)^2-6=10\)
\(\left(1-2x\right)^2=16\)
\(1-2x=4\)
\(2x=-3\)
\(x=-\dfrac{3}{2}\)
\(\frac{5}{3}x-0,4=\frac{9}{40}\)
\(\Rightarrow\frac{5}{3}x=\frac{9}{40}+0,4\)
\(\Rightarrow\frac{5}{3}x=\frac{5}{8}\)
\(\Rightarrow x=\frac{5}{8}:\frac{5}{3}\)
\(\Rightarrow x=\frac{3}{8}\)
vậy x=\(\frac{3}{8}\)
d: =>6x^2+2x-3x-1+9x-6x^2+12-8x=5
=>13=5(loại)
e: =>0,6x^2-0,3x-0,6x^2-0,39x=0,38
=>-0,69x=0,38
=>x=-38/69
\(5-\left|3x-1\right|=3\)
\(\left|3x-1\right|=2\)
\(\Rightarrow\orbr{\begin{cases}3x-1=2\\3x-1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}3x=3\\3x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)
vậy \(\orbr{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)
\(\left|x+\frac{3}{4}\right|-5=-2\)
\(\left|x+\frac{3}{4}\right|=3\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=3\\x+\frac{3}{4}=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=-\frac{15}{4}\end{cases}}\)
\(\left(1-2x\right)^2=9\)
\(\left(1-2x\right)^2=3^2\)
\(\Rightarrow1-2x=3\)
\(\Rightarrow2x=-2\)
\(\Rightarrow x=-1\)
vậy \(x=-1\)
\(\left(x+5\right)^3=-64\)
\(\left(x+5\right)^3=\left(-4\right)^3\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=-9\)
vậy \(x=-9\)
\(\left(2x+1\right)^2=\frac{4}{9}\)
\(\left(2x+1\right)^2=\left(\frac{2}{3}\right)^2\)
\(\Rightarrow2x+1=\frac{2}{3}\)
\(\Rightarrow2x=\frac{-1}{3}\)
\(\Rightarrow x=\frac{-1}{6}\)
vậy \(x=-\frac{1}{6}\)
a) \(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(\Rightarrow\left(\frac{3}{5}-\frac{2}{3}-1\right).x=\frac{-5}{21}:\frac{1}{7}=\frac{-5}{3}\)
\(\Rightarrow\frac{-16}{15}.x=\frac{-5}{3}\Rightarrow x=\frac{-5}{3}:\frac{-16}{15}=\frac{25}{16}\)
b) \(\left(x-\frac{1}{4}\right)^2=\frac{1}{36}\)
\(\Rightarrow\left(x-\frac{1}{4}\right)^2=\left(±\frac{1}{6}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{4}=\frac{1}{6}\\x-\frac{1}{4}=\frac{-1}{6}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{12}\\x=\frac{1}{12}\end{cases}}\)
5/3 + (-2/7) - (-1,2)
=5/3 + (-2/7) - (-12/10)
=5/3 + (-2/7) - (-6/5)
= 35/21 + (-6/21) -(-6/5)
= 29/21 - (-6/5)
= 29/21 + 6/5
= 145/105 + 126/105
= 271/105
`@` `\text {Ans}`
`\downarrow`
`0,3x+1,2=2/3x+9/10`
`=> 0,3x + 1,2 - 2/3x - 9/10 = 0`
`=> (0,3-2/3)x + (1,2-9/10) = 0`
`=> (-11/30x) + 3/10 = 0`
`=> -11/30x = -3/10`
`=> x = -3/10 \div (-11/30)`
`=> x = 9/11`
Vậy, `x=9/11`
`@` `\text {Kaizuu lv uuu}`
Áp dụng phương pháp chuyển vế đổi dấu, em chuyển tất cả các hạng tử chứa ẩn \(x\) sang một bên, các hạng tử không chứa \(x\) sang một bên, đồng thời đổi dấu các hạng tử vừa chuyển.
0,3\(x+1,2=\dfrac{2}{3}x+\dfrac{9}{10}\)
\(\dfrac{2}{3}x-0,3x=1,2-\dfrac{9}{10}\)
\(\left(\dfrac{2}{3}-0,3\right)x\) = 0,3
\(\dfrac{11}{30}x\) = 0,3
\(x\) = 0,3 : \(\dfrac{11}{30}\)
\(x\) = \(\dfrac{9}{11}\)