Cmr \(\forall x>0\) thì:
\(\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2}{\left(x+\dfrac{1}{x}\right)^3+x^3+\dfrac{1}{x^3}}\ge6\)
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MH
19 tháng 2 2022
1) \(\left(x-2\right)\left(3+2x\right)-2x\left(x+5\right)=6\)
\(3x+2x^2-6-4x-2x^2-10x-6=0\)
\(-11x=12\)
\(x=-\dfrac{12}{11}\)
2) \(x^2-4-\left(x-5\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(x+2-x+5\right)=0\)
\(7\left(x-2\right)=0\)
\(\Leftrightarrow x=2\)
19 tháng 2 2022
1, \(3x+2x^2-6-4x-2x^2-10x=0\Leftrightarrow-11x-6=0\Leftrightarrow x=-\dfrac{6}{11}\)
2, \(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-x+5\right)=0\Leftrightarrow x=2\)
3, bạn xem lại đề
5, đk x khác -4 ; 4
\(96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)-6\left(x^2-16\right)\)
\(\Leftrightarrow96=2x^2-9x+4+3x^2+11x-4-6x^2+96\)
\(\Leftrightarrow-x^2+2x=0\Leftrightarrow-x\left(x-2\right)=0\Leftrightarrow x=0;x=2\)(tm)
N
25 tháng 2 2019
Có: \(x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3x^3.\dfrac{1}{x^3}\left(x+\dfrac{1}{x}\right)\)\(=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)\)
Có: \(x^6+\dfrac{1}{x^6}=\left(x^2+\dfrac{1}{x^2}\right)^3-3\left(x^2+\dfrac{1}{x^2}\right)\)\(=\left[\left(x+\dfrac{1}{x}\right)^2-2\right]^3-3\left[\left(x+\dfrac{1}{x}\right)^2-2\right]\)
Đặt \(a=x+\dfrac{1}{x}\left(a\ge2\right)\)
\(P=\dfrac{a^6-\left[a^2-2\right]^3+3a^2+4}{a^3+a^3-3a}\)
\(P=\dfrac{-6a^4+15a^2+4}{2a^3-3a}\)
\(\Rightarrow6a^4+2Pa^3-15a^2-3Pa-4=0\)
\(\Rightarrow a^2\left(6a^2+2P+14\right)-\left(14a^2+3Pa+4\right)=0\)
Để pt \(\left\{{}\begin{matrix}6a^2+2P+14\\14a^2+3Pa+4\end{matrix}\right.\) có nghiệm thì
\(\left\{{}\begin{matrix}4P^2-336\ge0\\9P^2-224\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}P\le-2\sqrt{21}\\P\ge2\sqrt{21}\end{matrix}\right.\\\left[{}\begin{matrix}P\le-\dfrac{4\sqrt{14}}{3}\\P\ge\dfrac{4\sqrt{14}}{3}\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow P_{min}=\dfrac{4\sqrt{14}}{3}\)
28 tháng 2 2022
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
29 tháng 10 2021
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
14 tháng 3 2018
Lời giải:
a) Nếu không điều kiện gì của $x$ thì biểu thức không có GTNN
vì cho $x$ chạy từ \(-100\) đến âm vô cùng thì giá trị $A$ càng nhỏ (âm) vô cùng
b) Điều kiện: \(x>0\)
\(B=\frac{\left ( x+\frac{1}{x} \right )^6-\left ( x^6+\frac{1}{x^6} \right )-2}{\left ( x+\frac{1}{x} \right )^3+\left ( x^3+\frac{1}{x^3} \right )}=\frac{\left ( x+\frac{1}{x} \right )^6-\left [ (x^3+\frac{1}{x^3})^2-2 \right ]-2}{\left ( x+\frac{1}{x}\right )^3+\left ( x^3+\frac{1}{x^3} \right )}\)
\(=\frac{\left ( x+\frac{1}{x} \right )^6-\left ( x^3+\frac{1}{x^3} \right )^2}{\left ( x+\frac{1}{x} \right )^3+\left ( x^3+\frac{1}{x^3} \right )}=\frac{\left [ \left ( x+\frac{1}{x} \right )^3-\left ( x^3+\frac{1}{x^3} \right ) \right ]\left [ \left ( x+\frac{1}{x} \right )^3+\left ( x^3+\frac{1}{x^3} \right ) \right ]}{\left ( x+\frac{1}{x} \right )^3+\left ( x^3+\frac{1}{x^3} \right )}\)
\(=\left ( x+\frac{1}{x} \right )^3-\left ( x^3+\frac{1}{x^3} \right )=\left ( x+\frac{1}{x} \right )^3-\left [ \left ( x+\frac{1}{x} \right )^3-3.x.\frac{1}{x}\left ( x+\frac{1}{x} \right ) \right ]\) (sd hằng đẳng thức đáng nhớ \(x^3+y^3=(x+y)^3-3xy(x+y)\) )
\(=3\left(x+\frac{1}{x}\right)\geq 3.2\sqrt{x.\frac{1}{x}}=6\) (theo BĐT Cô-si cho hai số dương)
Vậy \(B_{\min}=6\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} x=\frac{1}{x}\\ x>0\end{matrix}\right.\Leftrightarrow x=1\)
T
1 tháng 9 2023
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
T
1 tháng 9 2023
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
\(\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2}{\left(x+\dfrac{1}{x}\right)+x^3+\dfrac{1}{x^3}}\)
\(=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+2+\dfrac{1}{x^6}\right)}{\left(x+\dfrac{1}{x}\right)+\left(x^3+\dfrac{1}{x^3}\right)}\)
\(=\dfrac{\left[\left(x+\dfrac{1}{x}\right)^3\right]^2-\left(x^3+\dfrac{1}{x^3}\right)^2}{\left(x+\dfrac{1}{x}\right)^3+\left(x^3+\dfrac{1}{x^3}\right)}\)
\(=\left(x+\dfrac{1}{x}\right)^3-\left(x^3+\dfrac{1}{x^3}\right)\)
\(=3x+\dfrac{3}{x}\)
\(=3\left(x+\dfrac{1}{x}\right)\)
Áp dụng bất đẳng thức \(x+\dfrac{1}{x}\ge2\forall x>0\)
\(\Rightarrow3\left(x+\dfrac{1}{x}\right)\ge6\)
\(\Rightarrowđpcm\)
Akai Haruma Ace Legona Unruly Kid
ai đi ngang qua cứu e vs :((