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\(-\frac{9}{5}=\frac{-54}{30},\frac{11}{-6}=-\frac{55}{30}\)
\(-\frac{54}{30}>-\frac{55}{30}\Rightarrow-\frac{9}{5}>-\frac{11}{6}\)
\(-\frac{6}{11}=-\frac{30}{55}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có 290>289
<=> \(\sqrt{290}\) > \(\sqrt{289}\)
<=> \(\sqrt{290}\) > 17
Vậy ..........
\(a,290>289\)
\(\Rightarrow\sqrt{290}>\sqrt{289}\)
\(\Rightarrow\sqrt{290}>17\)
\(b,\sqrt{7}+\sqrt{15}< \sqrt{9}+\sqrt{16}\)
\(\Rightarrow\sqrt{7}+\sqrt{15}< 3+4\)
\(\Rightarrow\sqrt{7}+\sqrt{15}< 7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: 266 . 734 = 232 . 234 . 734 < (2.2.7)34 = 2834
Vậy 2834 > 266 . 734
Tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Ta\)\(có\)\(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
\(Mà\)\(125^{12}>121^{12}\)
\(=>5^{36}>11^{24}\)
Ta có:
\(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
\(Do125>121\)
\(\Rightarrow125^{12}>121^{12}\)
\(\Rightarrow5^{36}>11^{24}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(3.24^{100}=3.3^{100}.8^{100}=3^{101}.8^{100}\)
Xét : \(4^{300}\)và \(3^{101}.8^{100}\)ta có :
\(4^{300}=2^{300}.2^{300}=\left(2^2\right)^{150}.\left(2^3\right)^{100}=\)\(4^{150}.8^{100}\)
Vì \(8^{100}=8^{100}\)và \(4^{150}>3^{101}\Rightarrow4^{300}>3^{101}.8^{100}\)
\(\Rightarrow4^{300}+3^{400}>3.24^{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) So sánh
Ta có : 224 = 23.8 = (23)8 = 88
316 = 32.8 = (32)8 = 98
Vì 88 < 98
=> 224 < 316
2) Tính
\(\left(0,25\right)^4.1024=\left(\frac{1}{4}\right)^4.1024=\frac{1}{4^4}.2^{10}=\frac{1}{\left(2^2\right)^4}.2^{10}=\frac{1}{2^8}.2^{10}=\frac{2^{10}}{2^8}=2^2=4\)
3) Tìm x nguyên
(x - 1)x + 2 = (x - 1)x + 6
=> (x - 1)x + 6 - (x - 1)x + 2 = 0
=> (x - 1)x + 2.[(x - 1)4 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^4-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^4=1^4\end{cases}\Rightarrow}\orbr{\begin{cases}x-1=0\\x-1=\pm1\end{cases}}}\)
Nếu x - 1 = 0 => x = 1(tm)
Nếu x - 1 = - 1 => x = 0(tm)
Nếu x - 1 = 1 => x = 2(tm)
Vậy \(x\in\left\{1;0;2\right\}\)
Bài 1:Ta có:
2^24=2^(6.4)=64^4
3^16=3^(4.4)=81^4
Bài 2.Ta có:
(0.25)^4=1/4.1/4.1/4.1/4=1/256
=>1/256.1024=4
Bài 3:
Ta có:(x-1)^(x+2)=(x-1)^(x+6)
Chia hai vế cho (x-1)^(x+2),do đó:
1=(x-1)^(x+4)
<=>x-1=1
<=>x=2
Hoặc chia hai vế cho (x-1)^(x+6)
(x-1)^(x-4)=1
<=>x-1=1
<=>x=2
11^1979=11^1978x11=22^969x11<22^1320
mà 22^1320<37^1320
suy ra 11^1979<37^1320