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a) ( x + 3 )3 : 3 - 1 = -10
( x + 3 )3 : 3 = -10 + 1
( x + 3 )3 = -9 * 3
x + 3 = \(\sqrt[3]{-27}\)
x = -3 - 3
x = -6
b) 3 | x - 1 | + 5 = 17
3 | x - 1 | = 17 - 5
| x - 1 | = 12 : 3
| x - 1 | = 4
( 1 ) x - 1 > 0 => x - 1 = 4 => x = 5
( 2 ) x - 1 < 0 => x - 1 = -4 => x = -3
Vậy S = { -3 ; 5 }
\(a)\dfrac{1}{3}x+\dfrac{2}{5}\left(x+1\right)=0\)
\(\Leftrightarrow\dfrac{1}{3}x+\dfrac{2}{5}x+\dfrac{2}{5}=0\)
\(\Leftrightarrow x\left(\dfrac{5}{15}+\dfrac{6}{15}\right)=\dfrac{-2}{5}\)
\(\Leftrightarrow x.\dfrac{11}{15}=\dfrac{-2}{5}\)
\(\Leftrightarrow x=\dfrac{-2}{5}.\dfrac{15}{11}\)
\(\Leftrightarrow x=\dfrac{-6}{11}\)
a ) \(5\left(x^2\right)+7x+2\)
\(\Leftrightarrow5x^2+7x+2=0\)
\(\Leftrightarrow5x^2+5x+2x+2=0\)
\(\Leftrightarrow\left(5x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=-1\end{matrix}\right.\)
Vậy .............
b ) \(\dfrac{x+1}{17}+\dfrac{x+2}{16}=\dfrac{x+3}{15}+\dfrac{x+4}{14}\)
\(\Leftrightarrow\dfrac{x+1}{17}+1+\dfrac{x+2}{16}+1=\dfrac{x+3}{15}+1+\dfrac{x+4}{14}+1\)
\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}=\dfrac{x+18}{15}+\dfrac{x+18}{14}\)
\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}-\dfrac{x+18}{15}-\dfrac{x+18}{14}=0\)
\(\Leftrightarrow\left(x+18\right)\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\)
Vì \(\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)\ne0\)
Ta có : \(x+18=0\Leftrightarrow x=-18\)
Vậy ......
c ) \(\dfrac{x-1}{x-3}=\dfrac{x-4}{x-7}\)
\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=\left(x-3\right)\left(x-4\right)\)
\(\Leftrightarrow x^2-7x-x+7=x^2-4x-3x+12\)
\(\Leftrightarrow-x=5\)
\(\Leftrightarrow x=-5\)
Vậy ..
a, \(\frac{1}{6}x+\frac{1}{10}-\frac{4}{15}x+1=0\)
\(\Leftrightarrow-\frac{1}{10}x=-\frac{11}{10}\)
\(\Leftrightarrow x=11\)
b,\(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Leftrightarrow\frac{1}{7}x-\frac{2}{7}=0\)hoặc \(-\frac{1}{5}x+\frac{3}{5}=0\)hoặc \(\frac{1}{3}x+\frac{4}{3}=0\)
+) \(\frac{1}{7}x-\frac{2}{7}=0\Leftrightarrow\frac{1}{7}x=\frac{2}{7}\Leftrightarrow x=2\)
+)\(-\frac{1}{5}x+\frac{3}{5}=0\Leftrightarrow-\frac{1}{5}x=-\frac{3}{5}\Leftrightarrow x=3\)
+)\(\frac{1}{3}x+\frac{4}{3}=0\Leftrightarrow\frac{1}{3}x=-\frac{4}{3}\Leftrightarrow x=-4\)
c, \(\frac{1}{2}x-\frac{11}{15}:\frac{33}{35}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{9}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}x=\frac{4}{9}\)
\(\Leftrightarrow x=\frac{8}{9}\)
a/ \(\frac{1}{6}x+\frac{1}{10}-\frac{4}{15}x+1=0\)
\(\Rightarrow-\frac{1}{10}x=-\frac{11}{10}\)
\(\Rightarrow x=11\)
b/ \(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Rightarrow\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{1}{7}x=\frac{2}{7}\Rightarrow x=2\)
hoặc \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow-\frac{1}{5}x=-\frac{3}{5}\Rightarrow x=3\)
hoặc \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{1}{3}x=-\frac{4}{3}\Rightarrow x=-4\)
Vậy x = 2, x = 3, x = -4
c/ \(\frac{1}{2}x-\frac{11}{15}:\frac{33}{35}=-\frac{1}{3}\)
\(\Rightarrow\frac{1}{2}x-\frac{7}{9}=-\frac{1}{3}\)
\(\Rightarrow\frac{1}{2}x=\frac{4}{9}\Rightarrow x=\frac{8}{9}\)
Vậy x = 8/9
a) -5(x - 3) - 2(5 - 3x) = -(x - 1)
=> -5x + 15 - 10 + 6x = -x + 1
=> x + 5 = -x + 1
=> x + x = 1 - 5
=> 2x = -4
=> x = -4 : 2
=> x = -2
\(-5\left(x-3\right)-2\left(5-3x\right)=-\left(x-1\right)\)
\(\Leftrightarrow-5x+15-10+6x=-x+1\)
\(\Leftrightarrow x+5=1-x\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-4\div2\)
\(\Leftrightarrow x=-2\)
a) x + 15 = 36 - 2x
x + 15 = 36 - (x + x )
15 =36 - ( x + x) - x
15 = 36 - x - x - x
15 = 36 - 3x
3x = 36 - 15
3x = 21
x = 21 : 3
=> x = 7
b) (x - 7) - (2x +5) = -14
x - 7 -( 2x + 5) = -14
x - (2x + 5) = -14 + 7 = -7
x - 2x - 5 = -7
x - 2x = -7 + 5 = -2
x - x + x = 2
x = 2 (-x + x cũng bằng chính nó)
=> x = 2
c) (x - 12) - 15 = (-7 + 20) - (18+x)
(x - 12) - 15 = 13 - (18 + x)
(x - 12) - 15 = 13 - 18 - x
(x - 12) - 15 = -5 - x
15 = (x - 12 ) - (-5 - x)
15 = x - 12 + 5 + x
15 = x + (-12) + 5 + x
15 = 2x + [(-12) + 5]
15 = 2x + -7
2x = -7 + 15
2x = 8
x = 8 : 2
=> x = 4
..................
a/ \(13x-1=10x+5\)
\(\Rightarrow13x-10x=5+1\)
\(\Rightarrow3x=6\Rightarrow x=\dfrac{6}{3}=2\)
Vậy x = 2
b/ \(3\cdot\left|x-1\right|-1=5\cdot\left|x-1\right|-7\)
\(\Rightarrow3\cdot\left|x-1\right|-5\cdot\left|x-1\right|=-7+1\)
\(\Rightarrow\left(3-5\right)\cdot\left|x-1\right|=-6\)
\(\Rightarrow\left|x-1\right|=-6:\left(-2\right)=3\)
\(\Rightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
Vậy...................
c/ \(x^2+8x+15=0\)
\(\Leftrightarrow x^2+3x+5x+15=0\)
\(\Leftrightarrow x\left(x+3\right)+5\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=-3\end{matrix}\right.\)
Vậy............
a) 13x - 1 = 10x + 5
13x - 10x = 5 + 1
3x = 6
x = 2
Vậy x = 2 là giá trị cần tìm
b) 3|x - 1| - 1 = 5|x - 1| - 7
-1 + 7 = 5|x - 1| - 3|x - 1|
6 = 2|x - 1|
3 = |x - 1|
=> \(\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\)
TH1 : x - 1 = 3
x = 3 + 1
x = 4
TH1 : x - 1 = -3
x = -3 + 1
x = -2
Vậy \(\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\) là giá trị cần tìm
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