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\(\left(2x-3\right)^2=25\)
\(\Rightarrow\left(2x-3\right)^2=5^2\)
\(\Rightarrow2x-3=5\)
\(\Rightarrow2x=5+3\)
\(\Rightarrow2x=8\)
\(\Rightarrow x=4\)
Bài 1
A = \(\frac{3}{7}.\left(\frac{3}{7}\right)^{19}\)= \(\left(\frac{3}{7}\right)^{20}\)
B = \(\left[\left(-\frac{3}{7}\right)^5\right]^4\)= \(\left(-\frac{3}{7}\right)^{20}\)
Bài 2
a. (2x - 3)2 = 25
<=> \(\orbr{\begin{cases}2x-3=5\\2x-3=-5\end{cases}}\)
<=> \(\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)
Vậy ...
b. \(\frac{27}{3^x}\)= 3
<=> 27 = 31+x
<=> 33 = 31+x
<=> 3 = 1 + x
<=> x = 2
a) Ta co :
\(2^{30}=\left(2^3\right)^{10}=8^{10}\)
\(3^{20}=\left(3^2\right)^{10}=9^{10}\)
Vì \(8^{10}< 9^{10}\)
\(\Rightarrow2^{30}< 3^{20}\)
b)
Ta có :
\(7^6+7^5-7^4\)
\(=7^4\left(7^2+7-1\right)\)
\(=7^4.55\)
=> đpcm
a)Ta có:
230 = (23)10 = 810
320 = ( 32 )10 = 910
Vì 810 < 910 => 230 < 320
b) 76 + 75 - 74
= 74 (72 + 7 - 1 )
= 74 *55 chia hết 55
Đpcm
Ta có: \(\frac{-11}{3^7.7^3}=\frac{-11}{\frac{3^7.7^4.1}{7}}=-\frac{77}{3^7.7^4}=\frac{-78+1}{3^7.7^4}=-\frac{78}{3^7.7^4}+\frac{1}{3^7.7^4}\)
Do \(3^7.7^4>3^4.7^4\) => \(\frac{78}{3^7.7^4}< \frac{78}{3^4.7^4}\) => \(-\frac{78}{3^7.7^4}>-\frac{78}{3^4.7^4}\)=> \(-\frac{78}{3^7.7^4}+\frac{1}{3^7.7^4}>-\frac{78}{3^4.7^4}\)
=> \(-\frac{11}{3^7.7^4}>-\frac{78}{3^4.7^4}\)
Ta có: \(\frac{-1987}{-1986}=\frac{1986+1}{1986}=1+\frac{1}{1986}>1\)
\(\frac{-1984}{-1985}=\frac{1985-1}{1985}=1-\frac{1}{1985}< 1\)
=> \(\frac{-1987}{-1986}>\frac{-1984}{-1985}\)
Ta có: \(\frac{x}{5}< \frac{5}{4}< \frac{x+2}{5}\)
<=> \(x< \frac{25}{4}< x+2\)
Xét: \(x+2>\frac{25}{4}\) => \(x>\frac{17}{4}\)
=> \(\frac{17}{4}< x< \frac{25}{4}\)
Do x thuộc Z => x \(\in\){5; 6}
a. 3111 < 3211 = (25)11 = 255
1714 > 1614 = (24)14 = 256
Mà 255 < 256
=> 3111 < 255 < 256 < 1714
Vậy 3111 < 1714.
b. 3500 = (35)100 = 243100
7200 = (72)100 = 49100
Mà 243100 > 49100
Vậy 3500 > 7200
c. 85 = (23)5 = 215 = 2.214
3.47 = 3.(22)7 = 3.214
Mà 2 < 3 => 2.214 < 3.214
Vậy 85 < 3.47.
a) Ta có: \(31^{11}< 32^{11}=\left(2^5\right)^{11}=2^{55}\)
\(17^{14}>16^{14}=\left(2^4\right)^{14}=2^{56}\)
Vì 255<256 => \(31^{11}< 2^{55}< 2^{56}< 17^{14}\)nên 3111<1714
b) Ta có: \(3^{500}=\left(3^5\right)^{100}=243^{100}\)
\(7^{200}=\left(7^2\right)^{100}=49^{100}\)
Vì \(243^{100}>49^{100}\)nên 3500>7200
c) Ta có: \(8^5=\left(2^3\right)^5=2^{15}=2.2^{14}\)
\(3.4^7=3.\left(2^2\right)^7=3.2^{14}\)
Vì 2<3 => 2.214<3.214 =>85<3.47
Bài 1: a) (2x+1)2 = 25
(2x+1)2 = 52
=> 2x + 1 = 5 hoặc 2x+1 = -5
=> x=2 hoặc x=-3
b) 2x+2 - 2x = 96
<=> 2x . 22 - 2x = 96
<=> 2x(4-1) =96
<=>2x = 96 :3 = 32 = 25
<=> x = 5
c) (x-1)3 = 125
<=> (x-1)3 = 53
<=> x-1=5
<=>x= 5 +1 = 6
Câu 1:
a) 2225 và 3150
Ta có:2225=(29)25=51225
3150=(36)25=72925
Vì 51225<72925
Suy ra: 2225<3150
Câu 2:
a)\(25^3:5^2=\left(5^2\right)^3:5^2=5^6:5^2=5^4\)
b)\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
c)\(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3+\frac{1}{4}:2=3+\frac{1}{8}=\frac{25}{8}\)
Câu 3:
a)\(9.3^3.\frac{1}{81}.3^2=3^2.3^3.3^2.\left(\frac{1}{3^4}\right)=3^7:3^4=3^3\)
b)\(4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:\left(2^3.\frac{1}{2^4}\right)=2^7:\frac{1}{2}=2^8\)
c)\(3^2.2^5.\left(\frac{2}{3}\right)^2=288.\frac{4}{9}=2^7\)
d)\(\left(\frac{1}{3}\right)^3.\frac{1}{3}.9^2=\left(\frac{1}{3}\right)^4.\left(3^2\right)^2=3^4.\left(\frac{1}{3}\right)^4=3^4:3^4=1\)
Giải :
\(B=\left[\left(-\frac{3}{7}\right)^5\right]^4\)
\(B=\left(-\frac{3}{7}\right)^{20}\)
\(A=\frac{3}{7}\cdot\left(\frac{3}{7}\right)^{19}\)
\(A=\left(\frac{3}{7}\right)^{20}\)
\(\Rightarrow A>B\)
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