Đi 2, 3 thôi là đc.

Bài 3:

Bạn tham khảo

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a, x : 1/3 = ( 1/3 ) ³

x=(1/3)^4

HT

Trả lời:

a, \(x\div\frac{1}{3}=\left(\frac{1}{3}\right)^3\)

\(\Leftrightarrow x=\left(\frac{1}{3}\right)^3\times\left(\frac{1}{3}\right)\)

\(\Leftrightarrow x=\left(\frac{1}{3}\right)^4\)

\(\Leftrightarrow x=\frac{1}{81}\)

b, \(x\div\left(\frac{1}{2}\right)^2=\frac{1}{16}\)

\(\Leftrightarrow x\div\left(\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^4\)

\(\Leftrightarrow x=\left(\frac{1}{2}\right)^4\times\left(\frac{1}{2}\right)^2\)

\(\Leftrightarrow x=\left(\frac{1}{2}\right)^6\)

\(\Leftrightarrow x=\frac{1}{64}\)

1) Tính :

a) \(\left(0,5\right)^3=0,125=\frac{1}{8}\)

b) \(\left(-\frac{1}{4}\right)^0=1\)

c) ...........

d) \(2^3+3.\left(\frac{1}{2}\right)^0-8=3\)

Trả lời:

a, \(\left(0,5\right)^3=0,125\)

b, \(\left(-\frac{1}{4}\right)^0=1\)

c, \(\left(\frac{3}{5}\right)^{15}:\left(\frac{9}{25}\right)^{15}=\frac{3^{15}}{5^{15}}:\frac{9^{15}}{25^{15}}=\frac{3^{15}}{5^{15}}.\frac{25^{15}}{9^{15}}=\frac{3^{15}}{5^{15}}.\frac{\left(5^2\right)^{15}}{\left(3^2\right)^{15}}=\frac{3^{15}}{5^{15}}.\frac{5^{30}}{3^{30}}=\frac{5^{15}}{3^{15}}\)

d, \(2^3+3.\left(\frac{1}{2}\right)^0-8=8+3.1-8=3\)

1) Viết các số sau dưới dạng lũy thừa

a) \(625=5^4\)

b) \(\frac{4}{9}=\left(\frac{2}{3}\right)^2\)

c) \(0,81=0,9^2\)

d) \(\frac{9}{64}=\left(\frac{3}{8}\right)^2\)

\(625=5^4\)

\(\frac{4}{9}=\left(\frac{2}{3}\right)^2\)

\(0,81=\left(0,9\right)^2\)

\(\frac{9}{64}=\left(\frac{3}{8}\right)^2\)

8) Ta có  \(2.16\ge2^n>4\Rightarrow2^5\ge2^n>2^2\)

=> \(5\ge n>2\)

\(\Rightarrow n\in\left\{3;4;5\right\}\)

b) Ta có \(9.27\le3^n\le243\)

\(\Rightarrow3^5\le3^n\le3^5\)

=> n = 5

9) 81ⁿ : 3ⁿ = 27

=> (81 : 3)ⁿ = 27

=> 27ⁿ = 27

=> n = 1

b) 25 : 5ⁿ = 5

=> 5ⁿ = 5

=> n = 1

c) \(\frac{81}{\left(-3\right)^n}=-27\)

=> (-3)ⁿ = (-3) 

=> n = 1

d) \(\frac{1}{2}.2^n+4.2^n=9.2^5\)

=> \(\frac{9}{2}.2^n=9.2^5\)

=> 2ⁿ = 2⁶

=> n = 6

Trả lời :

7) \(\left(-1\frac{1}{4}x^2y\right)^2\left(-\frac{5}{6}xy\right)^0\left(-2\frac{1}{2}xy\right)^5\)

Đáp án là : \(-\frac{78125x^9y^7}{512}\)

8) \(\left(\frac{-8}{3}x^2z\right)^3\left(\frac{2}{3}xy^2z^2\right)^3\left(\frac{4}{5}x^3y\right)\)

Đáp án là : .....

Câu 8 :

Tham khảo!

# R

Khi x = 9 

=> x + 1 = 10

Khi đó P(9) = x⁵⁰ - (x + 1)x⁴⁹ + (x + 1)x⁴⁸ - (x + 1)x⁴⁷ + ... + (x + 1)x - 1

= x⁵⁰ - x⁵⁰ - x⁴⁹ + x⁴⁹ + x⁴⁸ - x⁴⁸ - x⁴⁷ + ... + x² + x - 1

= x - 1 

= 9 - 1 = 8 

=8

tốt

hok

nha

\(\left[\left(\frac{-7}{2}\right)^0\right]\)

\(=\left[1\right]^8\)

\(=1\)

=1 nha

       Bài giải

'THAM KHẢO

a,

Điều kiện: x+2≥0⇔x≥−2x+2≥0⇔x≥-2

|2x+3|=x+2|2x+3|=x+2

⇔[2x+3=x+22x+3=−x−2⇔[2x+3=x+22x+3=−x−2

⇔[x=−13x=−5⇔[x=−13x=−5

⇔⎡⎣x=−1(t/m)x=−53(t/m)⇔[x=−1(t/m)x=−53(t/m)

Vậy x∈{−1;−53}x∈{-1;-53}

b,

A=|x−2006|+|2007−x|≥|x−2006+2007−x|=|1|=1A=|x−2006|+|2007−x|≥|x−2006+2007−x|=|1|=1

Đẳng thức xảy ra ⇔(x−2006)(2007−x)≥0⇔(x−2006)(2007−x)≥0

⇔(x−2006)(x−2007)≤0⇔(x−2006)(x−2007)≤0

Vì x−2006>x−2007x−2006>x−2007

⇒{x−2006≥0x−2007≤0⇒{x−2006≥0x−2007≤0

⇔{x≥2006x≤2007⇔{x≥2006x≤2007

⇔2006≤x≤2007⇔2006≤x≤2007

Vậy Amin=1⇔2006≤x≤2007

a) |x - 1,7| = 2,3

=> x - 1,7 = 2,3 hoặc x - 1,7 = -2,3

=> x = 4 hoặc x = -0,6

b) |x + 3/4| - 1/3 = 0

=> |x + 3/4| = 1/3

=> x + 3/4 = 1/3 hoặc x + 3/4 = -1/3

=> x = -5/12 hoặc x = -13/12

Chúc e học tốt !

\(a,\left|x-1,7\right|=2,3\)

\(\Rightarrow x-1,7=2,3\text{ hoặc }x-1,7=\left(-2,3\right)\)

      \(x=2,3+1,7\)       \(x=\left(-2,3\right)+1,7\)

       \(x=4\)                        \(x=\left(-0,6\right)\)

Vậy ...