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1.
a) x : \(\left(\dfrac{3}{4}\right)^3\) =\(\left(\dfrac{3}{4}\right)^3\)
x = \(\left(\dfrac{3}{4}\right)^3.\left(\dfrac{3}{4}\right)^3\)
x = \(\dfrac{3}{4}^{3+3}\)
x = \(\dfrac{3}{4}^6\)
x = \(\dfrac{729}{4096}\)
b) \(\left(\dfrac{2}{5}\right)^5.x=\left(\dfrac{2}{5}\right)^8\)
x = \(\left(\dfrac{2}{5}\right)^8:\left(\dfrac{2}{5}\right)^5\)
x = \(\dfrac{2}{5}^{8-5}\)
x = \(\dfrac{2}{5}^3\)
x = \(\dfrac{8}{5}\)
2.
(0,36)\(^8\) \([\left(0,6\right)^3]^8\) = (0,6)\(^{3.8}\) = ( 0,6)\(^{24}\)
( 0,216)\(^4\) = \([\left(0,6\right)^3]^4\) = (0.6)\(^{3.4}\) = ( 0,6)\(^{12}\)
\(x:\left(\dfrac{3}{4}\right)^3=\left(\dfrac{3}{4}\right)^2\)
\(x=\left(\dfrac{3}{4}\right)^2.\left(\dfrac{3}{4}\right)^3\) <=> \(x=\left(\dfrac{3}{4}\right)^{2+3}\)
=> \(x=\left(\dfrac{3}{4}\right)^5\)
b, \(\left(\dfrac{2}{5}\right)^5.x=\left(\dfrac{2}{5}\right)^8\)
\(x=\left(\dfrac{2}{5}\right)^8:\left(\dfrac{2}{5}\right)^5\Leftrightarrow x=\left(\dfrac{2}{5}\right)^{8-5}\)
=>\(x=\left(\dfrac{2}{5}\right)^3\)
bài 2 : Với bài này ta cần áp dụng quy tắc: \(\left(x^m\right)^n=x^{m.n}\)
\(0,36^8=\left[\left(0,6\right)^2\right]^8=\left(0,6\right)^{16}\)
\(0,216^4=\left[\left(0,6\right)^3\right]^4=\left(0,6\right)^{12}\)
làm bài 3 BĐT
theo bảng xét dấu
còn bài 1,2 ở trên là 1.1 và 1.2 đều trg bài 1.2
bài 1.2 (tức bài 2 ở trên )làm a,b,c,d
\còn bài 2( tức bài 2 ở trên) làm hết
1) a) \(\left|7x-5y\right|+\left|2z-3y\right|+\left|xy+yz+xz-2000\right|\ge0\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}7x=5y\\2z=3y\\xy+yz+xz=2000\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{7}y\\z=\dfrac{3}{2}y\\xy+yz+xz=2000\end{matrix}\right.\)
Ta có: \(xy+yz+xz=2000\)
\(\Rightarrow\dfrac{5}{7}y^2+\dfrac{3}{2}y^2+\dfrac{15}{14}y^2=2000\)
\(\Rightarrow y^2\left(\dfrac{5}{7}+\dfrac{3}{2}+\dfrac{15}{14}\right)=2000\Leftrightarrow\dfrac{23}{7}y^2=2000\)
Tìm \(y\) và suy ra \(x;z\) là được,Bài này nghiệm khá xấu
b) \(\left|3x-7\right|+\left|3x+2\right|+8=\left|7-3x\right|+\left|3x+2\right|+8\ge\left|7-3x+3x+2\right|+8\ge9+8=17\)Dấu "=" xảy ra khi: \(-\dfrac{3}{2}\le x\le\dfrac{7}{3}\)
2) a)Ta có: \(\left\{{}\begin{matrix}\left|x-5\right|+\left|1-x\right|\ge\left|x-5+1-x\right|=4\\\dfrac{12}{\left|y+1\right|+3}\le\dfrac{12}{3}=4\end{matrix}\right.\)
Mà theo đề bài: \(\left|x-5\right|+\left|1-x\right|=\dfrac{12}{\left|y+1\right|+3}\)
\(\Rightarrow\left|x-5\right|+\left|1-x\right|=\dfrac{12}{\left|y+1\right|+3}=4\)
\(\Rightarrow\left\{{}\begin{matrix}1\le x\le5\\y=-1\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}\left|y+3\right|+5\ge5\\\dfrac{10}{\left(2x-6\right)^2+2}\le\dfrac{10}{2}=5\end{matrix}\right.\)
Mà theo đề bài: \(\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}\)
\(\Rightarrow\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}=5\)
\(\Rightarrow\left\{{}\begin{matrix}y=-3\\x=3\end{matrix}\right.\)
c) Ta có: \(\left\{{}\begin{matrix}\left|x-1\right|+\left|3-x\right|\ge\left|x-1+3-x\right|=2\\\dfrac{6}{\left|y+3\right|+3}\le\dfrac{6}{3}=2\end{matrix}\right.\)
Mà theo đề bài: \(\left|x-1\right|+\left|3-x\right|=\dfrac{6}{\left|y+3\right|+3}\)
\(\Rightarrow\left|x-1\right|+\left|3-x\right|=\dfrac{6}{\left|y+3\right|+3}=2\)
\(\Rightarrow\left\{{}\begin{matrix}1\le x\le3\\y=-3\end{matrix}\right.\)
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
Câu 1:
a)\(\left(\frac{2}{3}\right)^2=\frac{4}{9}\) b)\(\left(-2\frac{3}{4}\right)^2=\left(-\frac{11}{4}\right)^2=\frac{121}{16}\)
c)\(\left(0,6\right)^4=\left(\frac{3}{5}\right)^4=\frac{81}{625}\) d)\(\left(-\frac{1}{2}\right)^4=\frac{1}{16}\)
e)\(\left(-\frac{1}{5}\right)^5=\frac{-1}{3125}\)
a.\(\frac{5}{4}x^2y.\left(\frac{-5}{6}xy\right)^0\left(\frac{-7}{3}xy\right)\)= \(\frac{5}{4}x^2y.1.\left(\frac{-7}{3}xy\right)\)= \(\frac{-35}{12}x^3.y^2\)
câu b, c,d làm tương tự như trên nha ^.^
1)
a) \(12^8\) và \(8^{12}\)
Ta có: \(12^8=\left(2^4\right)^8=2^{32}.\)
\(8^{12}=\left(2^3\right)^{12}=2^{36}.\)
Vì \(32< 36\) nên \(2^{32}< 2^{36}.\)
=> \(12^8< 8^{12}.\)
b) \(\left(-5\right)^{39}\) và \(\left(-2\right)^{91}\)
Ta có: \(\left(-5\right)^{39}=\left[\left(-5\right)^3\right]^{13}=\left(-125\right)^{13}.\)
\(\left(-2\right)^{91}=\left[\left(-2\right)^7\right]^{13}=\left(-128\right)^{13}.\)
Vì \(\left(-125\right)>\left(-128\right)\) nên \(\left(-125\right)^{13}>\left(-128\right)^{13}.\)
=> \(\left(-5\right)^{39}>\left(-2\right)^{91}.\)
2)
a) Tích của hai lũy thừa: \(x^{16}=x^{15}.x\)
b) Lũy thừa của \(x^4\): \(x^{16}=\left(x^4\right)^4.\)
c) Thương của hai lũy thừa: \(x^{16}=x^{18}:x^2.\)
Chúc bạn học tốt!
\(a,3^{16}:3=3^{16-1}=3^{15}\)
\(b,3^6.3^4.3^2.3=3^{6+4+2+1}=3^{13}\)
\(c,\left(-\frac{1}{4}\right).\left(6\frac{2}{11}\right)+\left(3\frac{9}{11}\right).\left(-\frac{1}{4}\right)=\left(-\frac{1}{4}\right).\frac{68}{11}+\frac{42}{11}.\left(-\frac{1}{4}\right)\)
\(=\left(-\frac{1}{4}\right)\left(\frac{68}{11}+\frac{42}{11}\right)\)
\(=\left(-\frac{1}{4}\right).10\)
\(=-\frac{10}{4}=-\frac{5}{2}\)
\(d,\left(-\frac{1}{2}\right)^3+\frac{1}{2}:5=\left(-\frac{1}{2}\right)\left(\left(\frac{1}{2}\right)^2-\frac{1}{5}\right)\)
\(=-\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{5}\right)\)
\(=-\frac{1}{2}.\frac{1}{20}\)
\(=-\frac{1}{40}\)
\(g,1\frac{1}{25}+\frac{2}{21}-\frac{1}{25}+\frac{19}{21}=\frac{26}{25}+\frac{2}{21}-\frac{1}{25}+\frac{19}{21}\)
\(=\left(\frac{26}{25}-\frac{1}{25}\right)+\left(\frac{2}{21}+\frac{19}{21}\right)\)
\(=1+1\)
\(=2\)
1.
a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)
b) x=0
d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)
e) \(x=\frac{2}{3}\)
\(a)18^{20}.45^5.5^{25}.8^{10}\)
\(=\left(2.3^2\right)^{20}.\left(3^2.5\right)^5.5^{25}.\left(2^3\right)^{10}\)
\(=2^{20}.3^{40}.3^{10}.5^5.5^{25}.2^{30}\)
\(=2^{50}.3^{50}.5^{30}\)
\(=6^{50}.5^{30}\)
\(=\left(6^5\right)^{10}.\left(5^3\right)^{10}\)
\(=7776^{10}.125^{10}\)
\(=972000^{10}\)
\(b)\left(x^2.y\right)^5.\left(x^2.y^2\right)^7.\left(xy\right)^6.x^3\)
\(=x^{10}.y^5.x^{14}.y^{14}.x^6.y^6.x^3\)
\(=x^{33}.y^{25}\)
\(=x^{25}.y^{25}.x^8\left(?\right)\)
\(c)2^7.3^8.4^9.9^8\)
\(=2^7.3^8.\left(2^2\right)^9.\left(3^2\right)^8\)
\(=2^7.3^8.2^{18}.3^{16}\)
\(=2^{25}.3^{24}\)\(\left(?\right)\)
Bạn trả lời đúng rồi a, Mình tick cho bạn nha