Tính:

2) \(\left(\frac{2}{3}\right)^3-\left(\frac{3}{4}\right)^2.\left(-1\right)^5\)

\(=\frac{8}{27}-\frac{9}{16}.\left(-1\right)\)

\(=\frac{8}{27}-\left(-\frac{9}{16}\right)\)

\(=\frac{371}{432}.\)

Xin lỗi, anh chỉ làm câu này thôi em.

Chúc em học tốt!

b: \(\left(-0,25\right)^3:\left(-0,25\right)=\left(-0,25\right)^{3-1}=\left(-0,25\right)^2=\left(-\frac14\right)^2=\frac{1}{16}\)

c: \(\left\lbrack\left(-0,5\right)^2\right\rbrack^2=\left\lbrack\left(\frac12\right)^2\right\rbrack^2=\left(\frac12\right)^4=\frac{1}{16}\)

d: \(a^5\cdot a^2=a^{5+2}=a^7\)

f: \(\frac{3^2}{\left(0,375\right)^2}=\left(\frac{3}{0,375}\right)^2=8^2=64\)

e: \(\left(\frac15\right)^5\cdot5^5=\left(\frac15\cdot5\right)^5=1^5=1\)

g: \(\frac{120^3}{40^3}=\left(\frac{120}{40}\right)^3=3^3=27\)

h: \(\left(0,125\right)^3\cdot512=\left(\frac18\right)^3\cdot8^3=1\)

mọi người trả lời nhanh đi ạ, e tối đi học rùi, thanks trc

a: =>||12x-1/2|-2|=-2/3x3/4=-6/12=-1/2(loại)

b: =>2/3-1/3x-1/2+2/3x=2x+2/3

=>-5/3x=1/2

=>x=-1/2:5/3=-1/2x3/5=-3/10

c: =>|3/2x+1/4|=2+3/4=11/4

=>3/2x+1/4=11/4 hoặc 3/2x+1/4=-11/4

=>3/2x=5/2 hoặc 3/2x=-3

=>x=3/5 hoặc x=-3:3/2=-2

a) \(\left(\frac{3}{5}\right)^{15}:\left(\frac{9}{25}\right)^5=\left(\frac{3}{5}\right)^{15}:\left(\left(\frac{3}{5}\right)^2\right)^5=\left(\frac{3}{5}\right)^{15}:\left(\frac{3}{5}\right)^{10}=\left(\frac{3}{5}\right)^5\)

b) \(5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{9}:3=4+\frac{1}{27}=4\frac{1}{27}\)

c) \(2^3+3.\left(\frac{1}{2}\right)^0+\left(-2\right)^2:\frac{1}{2}.8=8+3.1+4:\frac{1}{2}.8=8+3+64=75\)

d) \(\left(-1\right)^{-1}-\left(-\frac{3}{5}\right)^0+\left(\frac{1}{2}\right)^{2:2}=-1-1+\left(\frac{1}{2}\right)^1=-2+\frac{1}{2}=-\frac{3}{2}\)

bạn giải thích cho mình phần a nhé

What???

Nà ní!!!!!!!!!

1.

a) \(3^3.9^{-1}\)

\(=27.\frac{1}{9}\)

\(=3.\)

b) \(25.5^{-1}.5^0\)

\(=25.\frac{1}{5}.1\)

\(=5.1\)

\(=5.\)

c) \(3^2.\frac{1}{243}.81^2.3^{-3}\)

\(=9.\frac{1}{243}.6561.\frac{1}{27}\)

\(=\frac{1}{27}.6561.\frac{1}{27}\)

\(=243.\frac{1}{27}\)

\(=9.\)

Chúc bạn học tốt!

a, \(3^3.9^{-1}\)

\(=27.\frac{1}{9}\)

\(=\frac{27}{9}=3\)

b, \(25.5^{-1}.5^0\)

\(=25.\frac{1}{5}.1\)

\(=\frac{25}{5}.1\)

\(=5.1\)

\(=5\)

c, \(3^2.\frac{1}{143}.81^2.3^{-3}\)

\(=9.\frac{1}{143}.6561.\frac{1}{3^3}\)

\(=9.\frac{1}{143}.6561.\frac{1}{27}\)

\(=9.\frac{1}{143}\left(6561.\frac{1}{27}\right)\)

\(=9.\frac{1}{143}.243\)

\(=\frac{9}{143}.243\)

\(=\frac{2187}{143}\)

Câu d tương tự các câu trên

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.3^9.2^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(=\dfrac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1+5\right)}=\dfrac{3^8-3^9}{3^8.6}=\dfrac{3^8.\left(1-3\right)}{3^8.6}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

~ Học tốt ~

Bài 1:

1) \(3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^3}\)

\(=3^2.\left(\dfrac{1}{3}\right)^5.\left(3^4\right)^2.\dfrac{1}{3^3}\)

\(=3^2.\dfrac{1}{3^5}.3^8.\dfrac{1}{3^3}\)

\(=3^2=9\)

2) \(\left(4.2^5\right):\left(2^3.\dfrac{1}{16}\right)\)

\(=\left(2^2.2^5\right):[2^3.\left(\dfrac{1}{2}\right)^4]\)

\(=2^7:2^3:\dfrac{1}{2^4}\)

\(=2^4.2^4=256\)

3)\(\left(2^{-1}+3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)

\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}.1:2^3\)

\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2^4}\)

\(=\dfrac{43}{48}\)

4)\(\left(-\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

\(=-3-1+\dfrac{1}{4}.\dfrac{1}{2}\)

\(=-3-1+\dfrac{1}{8}\)

\(=-4+\dfrac{1}{8}\\ \)

\(=-\dfrac{31}{8}\)

5)\([\left(0,1\right)^2]^0+[\left(\dfrac{1}{7}\right)^{-1}]^2.\dfrac{1}{49}.[\left(2^2\right)^3:2^5]\\ =1+7^2.\dfrac{1}{7^2}.2^6:2^5\\ =1+1.2\\ =3\)

Chúc bạn học tốt

Ta có :

\(A=\frac{1}{3}+\frac{2}{3^2}+......+\frac{100}{3^{100}}\) \(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+.....+\frac{100}{3^{99}}\)

\(\Rightarrow3A-A=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)= 2A

Đặt \(B=1+\frac{1}{3}+...+\frac{1}{3^{99}}\) \(\Rightarrow3B=3+1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}\)

\(\Rightarrow3B-B=3-\frac{1}{3^{99}}=2B\) \(\Rightarrow B=\frac{3}{2}-\frac{1}{3^{99}.2}\)

\(\Rightarrow2A=\frac{3}{2}-\frac{1}{3^{99}.2}-\frac{100}{3^{100}}\)\(\Rightarrow A=\frac{3}{4}-\frac{1}{3^{99}.4}-\frac{100}{3^{100}}< \frac{3}{4}\Rightarrow\left(đpcm\right)\)

Ta có :

\(C=1+3+3^2+....+3^{100}\) \(\Rightarrow C-1=3+3^2+....+3^{100}\)

\(\Rightarrow3\left(C-1\right)=3^2+3^3+.....+3^{101}\)\(\Rightarrow3C-3-\left(C-1\right)=3^{101}-3\)

\(\Rightarrow2C-2=3^{101}-3\Rightarrow2C=3^{101}-1\)\(\Rightarrow C=\frac{3^{101}-1}{2}\)

Ta có :

\(D=2^{100}-2^{99}+2^{98}-.....-2\) \(\Rightarrow2D=2^{101}-2^{100}+2^{99}-.....-2^2\)

\(\Rightarrow2D+D=2^{101}-2=3D\) \(\Rightarrow D=\frac{2^{101}-2}{3}\)

\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)

\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

\(2A=1+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)-\frac{100}{3^{100}}\)

Ta thấy biểu thức trong dấu ngoặc nhỏ hơn 1/2 ( tự chứng minh ) nên 2A < 1 + 1/2 

\(\Rightarrow A< \frac{3}{4}\)

\(C=1+3+3^2+3^3+...+3^{100}\)

\(3C=3+3^2+3^3+3^4+...+3^{101}\)

\(3C-C=\left(3+3^2+3^3+3^4+...+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{100}\right)\)

\(2C=3^{101}-1\)

\(C=\frac{3^{101}-1}{2}\)