Ta có : \(P=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}-2\right)}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\)
\(=3\left(x+\frac{1}{x}\right)\ge6\) \(\left(x>0\right)\).
Vậy \(P_{Min}=6\) khi \(x=1.\)
Happy New year :)
a,\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|=4x\)
b,\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
Tìm x: \(3\left(x-2\right)^2-1\frac{1}{4}\left(1+3x\right)^2=1\frac{1}{3}\left(x-1\right)\left(3-x\right)-1\frac{1}{6}x\left(x-3\right)\)
Tìm x: \(3\left(x-2\right)^2-1\frac{1}{4}\left(1+3x\right)^2=1\frac{1}{3}\left(x-1\right)\left(3+x\right)-1\frac{1}{6}x\left(x-3\right)\)
Tìm x \(\in\) Z
a,\(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)< x< \frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\)
b, \(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{3}\right)\)
c, \(x:\left(\frac{1}{2}\right)^2=\frac{-1}{2}\)
d,\(\left(x^4\right)^2=\frac{x^{12}}{x^5}\left(x\ne0\right)\)
e,\(\frac{3^2.3^8}{27^3}=3^x\)
f,\(2^{x-1}=\left(16\right)^5\)
1. Tìm x ϵ Q sao cho:
a) (2x-3). (x+1) < 0.
b) \(\left(x-\frac{1}{2}\right).\left(x+3\right)\)> 0.
2. Tính:
S=\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{999.1001}\)
3. Tìm x: Biết x không thuộc{-2; -5; -10; -17}
\(\frac{3}{\left(x+2\right).\left(x+5\right)}+\frac{5}{\left(x+5\right).\left(x+10\right)}+\frac{7}{\left(x+10\right).\left(x+17\right)}=\frac{x}{\left(x+2\right).\left(x+17\right)}\)
Giúp mk vs đg cần gấp nha cảm ơn mấy bạn nhiều lắm !!!
\(\left(\frac{1}{2}\right)^5.x=\left(\frac{1}{2}\right)^7\)
\(\left(\frac{3}{7}\right)^2.x=\left(\frac{9}{21}\right)^2\)
Bài tìm x biết :
2x=2
3x=34
7x=77
(-3)x=(-3)5
(-5)x=(-5)4
2x=4
2x=8
2x=16
3x+1=32
5x-1=5
6x+4=610
52x-7=511
(-2)4x+2=64
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\)
\(\left(-\frac{1}{10}\right)^x=\frac{1}{100}\)
\(\left(-\frac{3}{2}\right)^x=\frac{9}{4}\)
\(\left(-\frac{3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(-\frac{2}{3}\right)^x=-\frac{8}{27}\)
\(a,\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(b,\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(c,\left(-3\right)^{x+5}=\frac{1}{81}\)
\(d,\left(\frac{1}{9}^x\right)=\left(\frac{1}{27}\right)^6\)
\(e,\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(f,5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(r,4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)
\(h,\left(\frac{1}{2}-\frac{1}{3}\right).6x+6^{x+2}=6^{10}+6^7\)
nhờ mấy bn giúp mk tối mình nộp rồi
Tìm x :
\(-5\left(x+\frac{1}{5}\right)+\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}_{ }-\frac{5}{6}\)
tìm x:
a,\(\left(x-\frac{7}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
b,\(\left(\frac{x}{2}-5\right)^5=-32\)
c,\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
