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\(\left(x^2.y\right)^5.\left(x^2.y^2\right)^7.\left(x.y^2\right)^6.x^3\)
\(=x^{10}.y^5.x^{14}.y^{14}.x^6.y^{12}.x^3\)
\(=x^{33}.y^{31}\)
a) \(\left(\dfrac{1}{2}x-3\right)\left(-\dfrac{1}{3}+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=0+3\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{1}{2}\\x=0-\left(-\dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)
d) \(9x^2=1\)
\(\Leftrightarrow x^2=1:9\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
\(\Leftrightarrow x^2=\left(\dfrac{1}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
a ) \(\left(3\times x-15\right)^7=0.\)
\(3\times x-15=0\)
\(3\times x=15\)
\(x=5\)
b ) \(10-\left\{\left[\left(x\div3+17\right)\div10+3\times2^4\right]\div10\right\}=5\)
\(10-\left\{\left[\left(x\div3+17\right)\div10+3\times16\right]\div10\right\}=5\)
\(10-\left\{\left[\left(x\div3+17\right)\div10+48\right]\div10\right\}=5\)
\(\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)
\(\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)
\(\left(x\div3+17\right)\div10+48=50\)
\(\left(x\div3+17\right)\div10=2\)
\(x\div3+17=20\)
\(x\div3=3\)
\(x=9\)
\(a,\)\(-\frac{3}{5}\cdot x=\frac{1}{4}+0,75\)
\(-\frac{3}{5}\cdot x=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)
\(x=1\div\left(-\frac{3}{5}\right)\)
\(x=-\frac{5}{3}\)
\(b,\)\(\left(\frac{1}{7}-\frac{1}{3}\right)\cdot x=\frac{28}{5}\times\left(\frac{1}{4}-\frac{1}{7}\right)\)
\(\left(\frac{3}{21}-\frac{7}{21}\right)\cdot x=\frac{28}{5}\cdot\left(\frac{7}{28}-\frac{4}{28}\right)\)
\(-\frac{4}{21}\cdot x=\frac{28}{5}\cdot\frac{3}{28}\)
\(-\frac{4}{21}\cdot x=\frac{3}{5}\)
\(x=\frac{3}{5}\div\left(-\frac{4}{21}\right)\)
\(x=-\frac{63}{20}\)
\(c,\)\(\frac{5}{7}\cdot x=\frac{9}{8}-0,125\)
\(\frac{5}{7}\cdot x=\frac{9}{8}-\frac{1}{8}\)
\(\frac{5}{7}\cdot x=1\)
\(x=1\div\frac{5}{7}\)
\(x=\frac{7}{5}\)
\(d,\)\(\left(\frac{2}{11}+\frac{1}{3}\right)\cdot x=\left(\frac{1}{7}-\frac{1}{8}\right)\cdot36\)
\(\left(\frac{6}{33}+\frac{11}{33}\right)\cdot x=\left(\frac{8}{56}-\frac{7}{56}\right)\cdot36\)
\(\frac{17}{33}\cdot x=\frac{1}{56}\cdot36\)
\(\frac{17}{33}\cdot x=\frac{9}{14}\)
\(x=\frac{9}{14}\div\frac{17}{33}\)
\(x=\frac{9}{14}\cdot\frac{33}{17}=\frac{297}{238}\)
Ta thấy : \(x^3+5\) < \(x^3+10\) < \(x^3+15\) < \(x^3+30\)
Nếu có 1 thừa số âm : \(x^3+5
Để (x3 + 5) . (x3 + 10) . (x3 + 15) x (x3 + 30) < 0
Mà x3 + 5 < x3 + 10 < x3 + 15 < x3 + 30 nên
<=> x3 + 5 < 0 => x3 < -5 => x \(\le\) -2
hoặc x3 + 5 < 0 và x3 + 10 < 0 và x3 + 15 < 0
=> x3 + 15 < 0 => x3 < -15 => x \(\le-3\)
Vậy \(x\le2\) với \(x\in Z\)
\(a,x \times 23-6 \times 23+x \times 69=230\)
\(x \times 23+x \times 69=230+6 \times 23\)
\(x \times (23+69)=368\)
\(x \times 92=368\)
\(x =368:92=4\)
~~~~~~~~~~~~~~~~~~~~~~~~~
`b,(x+1).(x+2)=72`
`x^2+2x+x+2=72`
`x^2+3x-70=0`
`x^2+10x-7x-70=0`
`x(x+10)-7(x+10)=0`
`(x+10)(x-7)=0`
`@TH1:x+10=0=>x=-10`
`@TH2:x-7=0=>x=7`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`c,(x+2).16x=160x`
`16x^2+32x-160x=0`
`16x^2-128x=0`
`16x(x-8)=0`
`@TH1:16x=0=>x=0`
`@TH2:x-8=0=>x=8`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`d,(x-5)(3x-18)=0`
`3x^2-18x-15x+90=0`
`3x^2-33x+90=0`
`x^2-11x+30=0`
`x^2-5x-6x+30=0`
`x(x-5)-6(x-5)=0`
`(x-5)(x-6)=0`
`@TH1:x-5=0=>x=5`
`@TH2:x-6=0=>x=6`
a, 23X - 6 x 23 + X x 69 = 230
23(X - 6 + 3X) = 230
4X - 6 = 230 : 23
4X - 6 = 10
4X = 10 + 6
4X = 16
X = 16 : 4
X = 4
b, (X+1)(X+2) = 72
(X +1)(X+2) = 8x9
X + 1 = 8
X = 8 - 1
X = 7
c, (X+2) x 16 x X = 160 x X
( X + 2 ) x 16 x X - 160 x X = 0
16X ( X + 2 - 10) = 0
\(\left[{}\begin{matrix}x=0\\x+2-10=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
x ϵ {0; 8}
d, (X - 5 )(3x -18) = 0
⇔ \(\left[{}\begin{matrix}x-5=0\\3x-18=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=5\\3x=18\end{matrix}\right.\)\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
x ϵ {5; 6}