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Ta có:
(-1/5)300 = (-1)300/5300 = 1/(53)100 = 1/125100
(-1/3)500 = (-1)500/3500 = 1/(35)100 = 1/243100
Vì 125100 < 243100
=> 1/125100 > 1/243100
=> (-1/5)300 > (-1/3)500
Ta có : \(\left(-\frac{1}{5}\right)^{300}=\left(-\frac{1}{5}\right)^{3.100}=\left(-\frac{1}{125}\right)^{100}=\left(\frac{1}{125}\right)^{100}\)
\(\left(-\frac{1}{3}\right)^{500}=\left(-\frac{1}{3}\right)^{5.100}=\left(-\frac{1}{243}\right)^{100}=\left(\frac{1}{243}\right)^{100}\)
Mà \(125< 243\Rightarrow\frac{1}{125}>\frac{1}{243}\Rightarrow\left(\frac{1}{125}\right)^{100}>\left(\frac{1}{243}\right)^{100}\)
\(=>\left(-\frac{1}{5}\right)^{300}>\left(-\frac{1}{3}\right)^{500}\)
1/ Ta có: \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{2^2}{4}=\frac{4}{4}=1\)
Dấu "=" xảy ra khi x=y=1
Máy mình bị lỗi nên ko nhìn được các bài tiếp theo
Chúc bạn học tốt :)
Ta có : x+y=2 => x=2-y. Thay vào bt ta đc : xy= (2-y).y = 2y -y^2
Vì y^2 >= 0 =>2y-y^2 nhỏ hơn hoặc bằng 0
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(\Leftrightarrow-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
=> x \(\in\) {-1;0;1;2;3;4;5;6}
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(\Leftrightarrow\)\(\frac{9-10}{12}\le\frac{x}{12}< 1-\left(\frac{8-3}{12}\right)\)
\(\Leftrightarrow\)\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
\(\Leftrightarrow-1\le x< 7\)
Mà x nguyên
=>x={-1;0;1;2;3;4;5;6}
\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5+\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
\(=3-\frac{1}{4}+\frac{2}{3}-5-\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
\(=\left(3-5-6\right)+\left(-\frac{1}{4}+\frac{7}{4}-\frac{3}{2}\right)+\left(\frac{2}{3}-\frac{1}{3}\right)+\frac{6}{5}\)
\(=-8+\frac{1}{3}+\frac{6}{5}\)
\(=-\frac{97}{15}\)
= 3 - 1/4 +2/3 - 5 - 1/3 + 6/5 - 6 + 7/4 - 3/2
= 2/3 . -3/2 . ( 3 + 5 + 6 ) . ( 2/3 + 1/3 ) . ( -1/4 - 7/4)
= -1 . 14 . 1 . 6/4
= -14 . 1 . 6/4
= -14 . 6/4
= -84/4 = -21
a) Ta có :\(\left(\frac{-1}{5}\right)^{300}=\frac{-1^{300}}{5^{300}}=\frac{1}{125^{100}}\)
\(\left(\frac{-1}{3}\right)^{500}=\frac{-1^{500}}{3^{500}}=\frac{1}{243^{100}}\)
Mà \(\frac{1}{125^{100}}>\frac{1}{243^{100}}\)
\(\Rightarrow\left(\frac{-1}{5}\right)^{300}>\left(\frac{-1}{3}\right)^{500}\)
b)Ta có :\(2^{90}=\left(2^{15}\right)^6=32768^6\)
\(5^{36}=\left(5^6\right)^6=15625^6\)
Vì \(32768^6>15625^6\Rightarrow2^{90}>5^{36}\)
a.Ta có: \(\left(\frac{-1}{5}\right)^{300}=\left(\frac{-1}{5}^3\right)^{100}=\left(\frac{-1}{125}\right)^{100}=\left(\frac{1}{125}\right)^{100}\)
\(\left(\frac{-1}{3}\right)^{500}=\left(\frac{-1}{3}^5\right)^{100}=\left(\frac{-1}{243}\right)^{100}=\left(\frac{1}{234}\right)^{100}\)
Mà: \(\frac{1}{125}>\frac{1}{234}\Rightarrow\left(\frac{1}{125}\right)^{100}>\left(\frac{1}{234}\right)^{100}\)
Vậy \(\left(\frac{-1}{5}\right)^{300}>\left(\frac{-1}{3}\right)^{500}\)
b.Ta có: \(2^{90}=\left(2^{10}\right)^9=1024^9\)
\(5^{36}=\left(5^4\right)^9=625^9\)
Mặt khác: \(1024>625\Rightarrow1024^9>625^9\)
Vậy \(2^{90}>5^{36}\)