0,5x + 1/3 - 2/3(x+1)=2/9
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6 tháng 7 2018
\(\hept{\begin{cases}\text{|}0,5x\text{|}=0,5x\\\sqrt{\left(0,5x\right)^2}=0,5x\\\left(0,5x\right)^2=\left(0,5x\right)^2\end{cases}}\)
2, tương tự
\(\hept{\begin{cases}\text{|}-\frac{2}{3}x\text{|}=\frac{2}{3}x\\\sqrt{\left(-\frac{2}{3}x\right)^2}=\frac{2}{3}x\\\left(-\frac{2}{3}x\right)^2=\left(\frac{2}{3}x\right)^2\end{cases}}\)
4, tương tự
PH
2
13 tháng 12 2016
a,
\(2\frac{2}{3}:x=1\frac{7}{9}:2\frac{2}{3}\)
\(\frac{8}{3}:x=\frac{16}{9}:\frac{8}{3}\)
\(\frac{8}{3}:x=\frac{2}{3}\)
\(\frac{8}{3}:\frac{2}{3}=x\)
\(x=4\)
Vậy x = 4
b,
\(-2^3+0,5x=1,5\)
\(-8+0,5x=1,5\)
\(0,5x=1,5+8\)
\(0,5x=9,5\)
\(x=9,5:0,5\)
\(x=19\)
Vậy x = 19
Q
20 tháng 6 2017
a) \(\left(2x^3-y^2\right)^3\)
\(=\left(2x^3\right)^3-3\cdot\left(2x^3\right)^2\cdot y^2+3\cdot2x^3\cdot\left(y^2\right)^{^2}-\left(y^2\right)^3\)
\(=8x^9-3\cdot4x^6y^2+3\cdot2x^3y^4-y^6\)
\(=8x^9-12x^6y^2+6x^3y^4-y^6\)
b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
c) \(\left(x+2y+z\right)\left(x+2y-z\right)\)
\(=\left(x+2y\right)^2-z^2\)
\(=x^2+4xy+4y^2-z^2\)
d) \(\left(2x^3y-0,5x^2\right)^3\)
\(=\left(2x^3y-\dfrac{1}{2}x^2\right)^3\)
\(=8x^9y^3-6x^8y^2+\dfrac{3}{2}x^7y-\dfrac{1}{8}x^6\)
e) \(\left(x^2-3\right)\left(x^4+3x^2+9\right)\)
\(=\left(x^2-3\right)\left(4x^2+9\right)\)
\(=4x^4+9x^2-12x^2-27\)
\(=4x^4-3x^2-27\)
f) \(\left(2x-1\right)\left(4x^2+2x+1\right)\)
\(=\left(2x\right)^3-1^3\)
\(=8x^3-1\)
TN
20 tháng 6 2017
\(a,\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)\(b,\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)
\(c,\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)\(d,\left(2x^3y-0,5x^2\right)^3=8x^9y^3-6x^4y^2x^2+3x^3yx^4-0,125x^6=8x^9y^3-6x^6y^2+3x^7y-0,125x^6\)
\(0,5x+\dfrac{1}{3}-\dfrac{2}{3}\left(x+1\right)=\dfrac{2}{9}\)
\(\dfrac{1}{2}x+\dfrac{1}{3}-\dfrac{2}{3}x-\dfrac{2}{3}=\dfrac{2}{9}\)
\(-\dfrac{1}{6}x-\dfrac{1}{3}=\dfrac{2}{9}\)
\(-\dfrac{1}{6}x=\dfrac{2}{9}+\dfrac{1}{3}\)
\(-\dfrac{1}{6}x=\dfrac{5}{9}\)
\(x=-\dfrac{10}{3}\)