Bài 5: Lũy thừa của một số hữu tỉ

\(\left(x-5\right)^2=\left(1-3x\right)^2\)

\(\Leftrightarrow x^2-10x+25=1-6x+9x^2\)

\(\Leftrightarrow8x^2+4x-24=0\)

\(\Leftrightarrow4\left(2x^2+x-6\right)=0\)

\(\Leftrightarrow2x^2+x-6=0\)

\(\Leftrightarrow2x^2+4x-3x-6=0\)

\(\Leftrightarrow2x\left(x+2\right)-3\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy.....

(x-5)²=(1-3x)²

=>x-5=1-3x

x+3x=1+5

4x=6

x=6:4

x=1.5

Ta thấy \(\left|a\right|\ge0\) \(\Leftrightarrow\left|a\right|\ge0^5\) Vì 5 là số mũ lẻ.

Hay \(b^5\ge0^5\)

=>\(b\ge0\) Vậy b dương

Mà a và b trái dấu => a âm

A=

\(\dfrac{10^{101}-1}{10^{102}-1}< \dfrac{10^{101}-1+11}{10^{102}-1+11}=\dfrac{10^{101}+10}{10^{102}+10}=\dfrac{10\left(10^{100}+1\right)}{10\left(10^{101}+1\right)}=B\)

Vậy A<B

CHÚC BẠN HỌC TỐT

Lời giải:

\((3y-1)^{20}=(3y-1)^{10}\)

\(\Leftrightarrow (3y-1)^{20}-(3y-1)^{10}=0\)

\(\Leftrightarrow (3y-1)^{10}.(3y-1)^{10}-(3y-1)^{10}=0\)

\(\Leftrightarrow (3y-1)^{10}[(3y-1)^{10}-1]=0\)

\(\Rightarrow \left[\begin{matrix} (3y-1)^2=0\\ (3y-1)^{10}=1\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} 3y-1=0\\ 3y-1=1\\ 3y-1=-1\end{matrix}\right.\Rightarrow \left[\begin{matrix} y=\frac{1}{3}\\ y=\frac{2}{3}\\ y=0\end{matrix}\right.\)

(3y - 1)² = 3y - 1

(3y - 1) (3y - 1) = (3y - 1).1

=> 3y - 1= 1

=> 3y = 2

=> y = \(\)^{\(\dfrac{2}{3}\)}

Có: (2x-4)^x+1=(2x-4)^x+5

<=> (2x-4)^x+1 - (2x-4)^x+5=0

<=> (2x-4)^x+1\([1-\left(2x-4\right)^4]=0\)

<=> \(\left[{}\begin{matrix}\left(2x-4\right)^{x+1}=0\\1-\left(2x-4\right)^4=0\end{matrix}\right.< =>\left[{}\begin{matrix}2x-4=0\\\left(2x-4\right)^4=1\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}2x=4\\2x-4=1hoặc2x-4=-1\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=2\\2x=5hoặc2x=3\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=\dfrac{5}{2};x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy x\(\in\left\{2;\dfrac{5}{2};\dfrac{3}{2}\right\}\)

ý b làm tương tự

Có: A=\(\dfrac{3}{1.5}+\dfrac{3}{5.10}+...+\dfrac{3}{100.105}\)

=> A=\(3.\dfrac{5}{5}\left(\dfrac{1}{1.5}+\dfrac{1}{5.10}+...+\dfrac{1}{100.105}\right)\)

=> A= \(3.\dfrac{1}{5}\left(\dfrac{5}{1.5}+\dfrac{5}{5.10}+...+\dfrac{5}{100.105}\right)\)

=> A=\(\dfrac{3}{5}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{105}\right)\)

=> A= \(\dfrac{3}{5}\left(1-\dfrac{1}{105}\right)\)=\(\dfrac{3}{5}.\dfrac{104}{105}=\dfrac{312}{525}\)

Có:\(\dfrac{25^5.4^2}{5^5.\left(-2\right)^5}=\dfrac{\left(5^2\right)^5.2^4}{5^5.\left(-2\right)^5}=\dfrac{5^{10}.2^4}{5^5.\left(-2\right)^5}=\dfrac{5^5}{-2}=\dfrac{3125}{2}\)

\(\dfrac{6^2.6^3}{3^5}=\dfrac{3^2.2^2.3^3.2^3}{3^5}=\dfrac{2^5.3^5}{3^5}=2^5=32.\)

\(\dfrac{25^2.4^2}{5^5.\left(-2\right)^5}=\dfrac{5^4.2^4}{\left(-10\right)^5}=\dfrac{10^4}{10^4.\left(-10\right)}=\dfrac{-1}{10}.\)

\(\dfrac{6^2.6^3}{3^5}=\dfrac{6^{2+3}}{3^5}=\dfrac{6^5}{3^5}=2^5=32\)

\(\dfrac{25^2.4^2}{5^5.\left(-2\right)^5}=\dfrac{\left(5^2\right)^2.\left(2^2\right)^2}{\left[5.\left(-2\right)\right]^5}=\dfrac{5^4.2^4}{-10^5}=\dfrac{10^4}{-10^5}=\dfrac{-1}{10}\)

\(\dfrac{0,125^5.2,4^5}{-0,3^5.0,001^3}=\dfrac{\left(0,125.2,4\right)^5}{-0,3^5.0,001^3}=\dfrac{0,3^5}{-0,3^5.0,001^3}=-\dfrac{1}{0,001^3}=-1000000000\)

\(\left(2\dfrac{3}{4}+\dfrac{1}{2}\right)=\dfrac{11}{4}+\dfrac{1}{2}=\dfrac{11}{4}+\dfrac{2}{4}=\dfrac{13}{4}\)

\(\dfrac{6^2.6^3}{3^5}=\dfrac{2^2.3^2.2^3.3^3}{3^5}=2^5=32\)

\(\dfrac{25^2.4^2}{5^5.-2^5}=\dfrac{5^4.2^4}{5^5.-2^5}=\dfrac{10^4}{-10^5}=\dfrac{1}{-10}=\dfrac{-1}{10}\)

\(\dfrac{0,125^5.2,4^5}{-0,3^5.0,001^3}=\dfrac{\left(\dfrac{1}{8}\right)^5.\left(\dfrac{12}{5}\right)^5}{-0,3^5.0,001^3}=\dfrac{0,3^5}{-0,3.0,001^5}\)

= \(\dfrac{1}{-0,001^5}\)

\(\left(2\dfrac{3}{4}+\dfrac{1}{2}\right)=\dfrac{11}{4}+\dfrac{1}{2}=\dfrac{11+2}{4}=\dfrac{13}{4}\)

Đặt B=6x³-3x²+2|x|+4

Thay x=\(\dfrac{-2}{3}\) vào biểu thức B ta có

B=6(\(\dfrac{-2}{3}\))³ - 3(\(\dfrac{-2}{3}\))²+2|\(\dfrac{-2}{3}\)|+4

=6.\(\dfrac{-8}{27}\)- 3.\(\dfrac{4}{9}\)+2.\(\dfrac{2}{3}\)+4

=\(\dfrac{-16}{9}\)-\(\dfrac{4}{3}\)+\(\dfrac{4}{3}\)+4

=4-\(\dfrac{16}{9}\)

=\(\dfrac{20}{9}\)