Tìm x thỏa mãn
\(\left(\dfrac{1}{3}\right)^x+\left(\dfrac{1}{3}\right)^{x+2}=\dfrac{10}{243}\)
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5 tháng 3 2022
a, \(\dfrac{4\left(x-3\right)^2-\left(2x-1\right)^2-12x}{12}< 0\)
\(\Rightarrow4\left(x^2-6x+9\right)-4x^2+4x-1-12x< 0\)
\(\Leftrightarrow-32x+35< 0\Leftrightarrow x>\dfrac{35}{32}\)
b, \(\dfrac{24+12\left(x+1\right)-36+3\left(x-1\right)}{12}< 0\)
\(\Rightarrow-12x+15x+9< 0\Leftrightarrow3x< -9\Leftrightarrow x>-3\)
23 tháng 1
a: \(\left(\sqrt{3}\right)^x=243\)
=>\(3^{\dfrac{1}{2}\cdot x}=3^5\)
=>\(\dfrac{1}{2}\cdot x=5\)
=>x=10
b: \(0,1^x=1000\)
=>\(\left(\dfrac{1}{10}\right)^x=1000\)
=>\(10^{-x}=10^3\)
=>-x=3
=>x=-3
c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)
=>\(\left(0,2\right)^{x+3}< 0,2\)
=>x+3>1
=>x>-2
d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)
=>\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{3}{5}\right)^{-2}\)
=>2x+1<-2
=>2x<-3
=>\(x< -\dfrac{3}{2}\)
e: \(5^{x-1}+5^{x+2}=3\)
=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)
=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)
=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)
14 tháng 1 2021
Áp dụng bất đẳng thức AM - GM:
\(P\ge3\sqrt[3]{\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}\).
Áp dụng bất đẳng thức AM - GM ta có:
\(xy+1=xy+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\ge5\sqrt[5]{\dfrac{xy}{4^4}}\).
Tương tự: \(yz+1\ge5\sqrt[5]{\dfrac{yz}{4^4}};zx+1\ge5\sqrt[5]{\dfrac{zx}{4^4}}\).
Do đó \(\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)\ge125\sqrt[5]{\dfrac{\left(xyz\right)^2}{4^{12}}}\)
\(\Rightarrow\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{1}{4^{12}\left(xyz\right)^3}}\).
Mà \(xyz\le\dfrac{\left(x+y+z\right)^3}{27}=\dfrac{1}{8}\)
Nên \(\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{8^3}{4^{12}}}=125\sqrt[5]{\dfrac{1}{2^{15}}}=\dfrac{125}{8}\)
\(\Rightarrow P\ge\dfrac{15}{2}\).
Vậy...
HN
17 tháng 1 2021
Áp dụng bất đẳng thức AM - GM:
P≥33√(xy+1)(yz+1)(zx+1)xyz.
Áp dụng bất đẳng thức AM - GM ta có:
xy+1=xy+14+14+14+14≥55√xy44.
Tương tự: yz+1≥55√yz44;zx+1≥55√zx44.
Do đó (xy+1)(yz+1)(zx+1)≥1255√(xyz)2412
⇒(xy+1)(yz+1)(zx+1)xyz≥1255√1412(xyz)3.
Mà xyz≤(x+y+z)327=18
Nên (xy+1)(yz+1)(zx+1)xyz≥1255√83412=1255√1215=1258
⇒P≥152.
12 tháng 2 2022
\(\Leftrightarrow-\dfrac{3}{4}< =x< =\dfrac{1}{2}\)
hay x=0
AH
Akai Haruma
Giáo viên
4 tháng 11 2023
Lời giải:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}$
$\Rightarrow (\frac{1}{x}+\frac{1}{y})+(\frac{1}{z}-\frac{1}{x+y+z})=0$
$\Leftrightarrow \frac{x+y}{xy}+\frac{x+y}{z(x+y+z)}=0$
$\Leftrightarrow (x+y)(\frac{1}{xy}+\frac{1}{z(x+y+z)})=0$
$\Leftrightarrow (x+y).\frac{z(x+y+z)+xy}{xyz(x+y+z)}=0$
$\Leftrightarrow (x+y).\frac{(z+x)(z+y)}{xyz(x+y+z)}=0$
$\Leftrightarrow (x+y)(y+z)(x+z)=0$
$\Leftrightarrow x=-y$ hoặc $y=-z$ hoặc $z=-x$
Nếu $x=-y$ thì:
$P=\frac{3}{4}+[(-y)^8-y^8](y^9+z^9)(z^{10}-x^{10})=\frac{3}{4}+0.(y^9+z^9)(z^{10}-x^{10})=\frac{3}{4}$
Nếu $y=-z$ thì:
$P=\frac{3}{4}+(x^8-y^8)[(-z)^9+z^9](z^{10}-x^{10})=\frac{3}{4}+(x^8-y^8).0.(z^{10}-x^{10})=\frac{3}{4}$
Nếu $z=-x$ thì:
$P=\frac{3}{4}+(x^8-y^8)(y^9+z^9)[(-x)^{10}-x^{10}]=\frac{3}{4}+(x^8-y^8)(y^9+z^9).0=\frac{3}{4}$
24 tháng 6 2021
`đk:x ne +-3,x ne -2`
`B=(21/(x^2-9)-(x-4)/(3-x)-(x-1)/(3+x)):(1-1/(x+3))`
`=(21/(x^2-9)+(x-4)/(x-3)-(x-1)/(x+3)):((x+3-1)/(x+3))`
`=((21+x^2-x-12-x^2+4x-3)/((x-3)(x+3))):(x+2)/(x+3)`
`=(3x+6)/((x-3)(x+3))*(x+3)/(x+2)`
`=(3x+6)/((x-3)(x+2))`
`=3/(x-3)`
`b)|2x+1|=5`
`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2(tm)\\x=-3(l)\end{array} \right.\)
`=>B=3/(2-3)=-3`
`c)B=-3/5`
`<=>3/(x-3)=3/(-5)`
`<=>x-3=-5`
`<=>x=-2(l)`
`d)B<0`
`<=>3/(x-3)<0`
Mà `3>0`
`=>x-3<0<=>x<3`
24 tháng 6 2021
a) đk: \(x\ne\pm3\)
\(B=\left[\dfrac{21}{\left(x-3\right)\left(x+3\right)}+\dfrac{x-4}{x-3}-\dfrac{x-1}{x+3}\right]:\left(\dfrac{x+3-1}{x+3}\right)\)
= \(\left[\dfrac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right]:\dfrac{x+2}{x+3}\)
= \(\dfrac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}\)
= \(\dfrac{3x+6}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}=\dfrac{3}{x-3}\)
b) Để \(\left|2x+1\right|=5\)
<=> \(\left[{}\begin{matrix}2x+1=5< =>x=2\left(c\right)\\2x+1=-5< =>x=-3\left(l\right)\end{matrix}\right.\)
Thay x = 2, ta có;
B = \(\dfrac{3}{2-3}=-3\)
c) Để B = \(\dfrac{-3}{5}\)
<=> \(\dfrac{3}{x-3}=\dfrac{-3}{5}\)
<=> x - 3 = -5
<=> x = -2
d) Để B < 0
<=> \(\dfrac{3}{x-3}< 0\)
<=> x - 3 < 0
<=> x < 3
AH
Akai Haruma
Giáo viên
13 tháng 3 2021
Thay $x=\sqrt{\frac{1}{2,5}}; y=z=\sqrt{\frac{1}{0,25}}$ ta thấy đề sai bạn nhé!
\(\left(\dfrac{1}{3}\right)^x+\left(\dfrac{1}{3}\right)^{x+2}=\dfrac{10}{243}\)
\(\left(\dfrac{1}{3}\right)^x+\left(\dfrac{1}{3}\right)^x.\left(\dfrac{1}{3}\right)^2=\dfrac{10}{243}\)
\(\left(\dfrac{1}{3}\right)^x.\left[1+\left(\dfrac{1}{3}\right)^2\right]=\dfrac{10}{243}\)
\(\left(\dfrac{1}{3}\right)^x.\dfrac{10}{9}=\dfrac{10}{243}\)
\(\left(\dfrac{1}{3}\right)^x=\dfrac{10}{243}:\dfrac{10}{9}\)
\(\left(\dfrac{1}{3}\right)^x=\dfrac{1}{27}\)
\(\left(\dfrac{1}{3}\right)^x=\left(\dfrac{1}{3}\right)^3\)
\(\Rightarrow x=3\)