Tính :
a) 2(1-x)+3(x+2)+x=10+x
b) 8-2|x-3|=|-2|
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1, Ta có :
\(x+\frac{3}{5}=\frac{4}{7}\div\frac{8}{21}\)
\(x+\frac{3}{5}=\frac{4}{7}\times\frac{21}{8}\)
\(x+\frac{3}{5}=\frac{3}{2}\)
\(x=\frac{3}{2}-\frac{3}{5}\)
\(x=\frac{15}{10}-\frac{6}{10}\)
\(x=\frac{9}{10}\)
Vậy x = \(\frac{9}{10}\)
2, Ta có :
\(\frac{2}{3}+\frac{3}{4}\div x=-\frac{1}{6}\)
\(\frac{3}{4}\div x=-\frac{1}{6}-\frac{2}{3}\)
\(\frac{3}{4}\div x=-\frac{1}{6}-\frac{4}{6}\)
\(\frac{3}{4}\div x=-\frac{5}{6}\)
\(x=\frac{3}{4}\div\left(-\frac{5}{6}\right)\)
\(x=\frac{3}{4}\times\left(-\frac{6}{5}\right)\)
\(x=-\frac{9}{10}\)
Vậy x = \(-\frac{9}{10}\)
a)=1/2 . 8/15 - 3/4.47/9
=4/15 - 47/12
=-73/20
b)=2-1/3 . -21/20
=2+7/20
=47/20
\(a,=\dfrac{x^3-\left(x-1\right)\left(x^2+x+1\right)}{1-x}=\dfrac{x^3-x^3+1}{1-x}=\dfrac{1}{1-x}\\ b,=\dfrac{2x+x^2+3x+2+2-x}{\left(x+2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x+2\right)^2}=1\)
`(x+3)(x^2-5x+8)=(x+3).x^2`
`<=>(x+3)(x^2-5x+8-x^2)=0`
`<=>(x+3)(8-5x)=0`
`<=>` \(\left[ \begin{array}{l}x+3=0\\8-5x=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac85\\x=-3\end{array} \right.\)
Vậy `S={-3,8/5}`
`(x+3)(x^2-5x+8)=(x+3).x^2`
`<=>(x+3)(x^2-5x+8-x^2)=0`
`<=>(x+3)(-5x+8)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\-5x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{8}{5}\end{matrix}\right.\)
Vậy `S={-3;8/5}`.
\(2x^4-x^3+2x^2+1=2x^4-2x^3+2x^2+x^3-x^2+x+x^2-x+1\\ \)
\(=2x^2\left(x^2-x+1\right)+x\left(x^2-x+1\right)+\left(x^2-x+1\right)=\left(x^2-x+1\right)\left(2x^2+x+1\right)\)
Vậy a = 2; b = 1; c = 1.
Lời giải:
Vì $x=9$ nên $x-9=0$
Ta có:
$F=(x^{2017}-9x^{2016})-(x^{2016}-9x^{2015})+(x^{2015}-9x^{2014})-....-(x^2-9x)+x-10$
$=x^{2016}(x-9)-x^{2015}(x-9)+x^{2014}(x-9)-....-x(x-9)+x-10$
$=x^{2016}.0-x^{2015}.0+x^{2014}.0-...-x.0+x-10$
$=x-10=9-10=-1$
a) 2(1-x)+3(x+2)+x=10+x
→2-2x+3x+6+x=10+x
→2x+8=10+x
→2x-x=10-8 →x=2
giúp mình đi nha ~~~