a) \(2^{x-1}=16\)

\(\Rightarrow2^{x-1}=2^4\)

\(\Rightarrow x-1=4\)

\(\Rightarrow x=4+1\)

\(\Rightarrow x=5\)

Vậy \(x=5.\)

c) \(\left(x+20\right)^{100}+\left|y+4\right|=0\)

Ta có:

\(\left\{{}\begin{matrix}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{matrix}\right.\forall x,y.\)

\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\) \(\forall x,y\)

\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+20=0\\y+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0-20\\y=0-4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-20\\y=-4\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{-20;-4\right\}.\)

Chúc bạn học tốt!

=> 

=> 

=>  hoặc 

+)  => x = 1

+)  =>  hoặc 

=> x = 2 hoặc x = 0

Vậy x = 1 hoặc x = 2 hoặc x = 0

(x-1)^(x+2)=(x-1)^(x+6)

(x-1)^(x+2)-(x-1)^(x+6)=0

(x-1)^(x+2) . [1-(x-1)^4]=0

=> (x-1)^(x+2)=0 hoặc 1-(x-1)^4=0

    x-1=0                        (x-1)^4=1

    x=1                           x-1=1 hoặc x-1=-1

                                        x=2 hoặc x=0

vậy x  {0;1;2} 

a, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)

\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)

\(\Rightarrow\left(x-1\right)^{x+2}\left(1-\left(x-1\right)^4\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-1=0\\1-\left(x-1\right)^4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\x=2\end{cases}}\)

b, \(\left(x+20\right)^{100}+|y+4|=0\)

\(\Rightarrow\hept{\begin{cases}x+20=0\\y+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-20\\y=-4\end{cases}}\)

1/ \(\left(x-1\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{25}\\x+1=-\sqrt{25}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=5\\x+1=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-6\end{matrix}\right.\)

Vậy...

2/ \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)

\(\Leftrightarrow\left(x-1\right)^{x+6}-\left(x-1\right)^{x+2}=0\)

\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^{x+4}-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^{x+4}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\end{matrix}\right.\)

Vậy...

3/ Với mọi x, y ta có :

\(\left\{{}\begin{matrix}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\)

Mà \(\left(x+20\right)^{100}+\left|y+4\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+20=0\\y+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-20\\y=-4\end{matrix}\right.\)

Vậy..

1) (x - 1)² = 25
(x - 1)² = 5²
=> x - 1 = 5
x = 5 + 1
x = 6

a, \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)\(\Rightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)\(\Rightarrow x=\frac{5}{6}\)

b, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)

\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)

\(\Rightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^4\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^4=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^4=1\end{cases}}\)

Giải: \(\left(x-1\right)^4=1\)\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

c, Vì \(\left(x+20\right)^{100}\ge0\)\(\forall x\inℝ\); \(\left|y+4\right|\ge0\)\(\forall y\inℝ\)

\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\)\(\forall x,y\inℝ\)

Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+20=0\\y+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-20\\y=-4\end{cases}}\)

d, \(2^{x-1}=16\)\(\Rightarrow2^{x-1}=2^4\)=> x - 1 = 4 => x = 5 

a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\)( do \(x^2\ge0,\left(y-\dfrac{1}{10}\right)^4\ge0\))

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)

b) \(\left(\dfrac{1}{2}.x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\)( do \(\left(\dfrac{1}{2}x-5\right)^{20}\ge0,\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\))

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)

\(a,\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\\ b,\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\end{matrix}\right.\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)

Mà \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)

\(\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)

a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)

Mà \(x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\forall x;y\)

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

Vậy \(\left(x;y\right)=\left(0;\dfrac{1}{10}\right)\)

b) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)

Mà \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\forall x;y\)

\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}=0\\\left(y^2-\dfrac{1}{4}\right)^{10}=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=10\\\left[{}\begin{matrix}y=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(10;\dfrac{1}{2}\right);\left(10;-\dfrac{1}{2}\right)\right\}\)

Bài 1:

\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)

Ta thấy:

\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)

\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)

\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)

\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)

\(\Rightarrow10x+\frac{10}{11}=0\)

\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)

Bài 2:

Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)

\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)

Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)

Bài 3:

a)\(2009-\left|x-2009\right|=x\)

\(\Rightarrow\left|x-2009\right|=2009-x\)

\(\Rightarrow\left|x-2009\right|=-\left(x-2009\right)\)

Vì GTTĐ của số âm bằng số đối của nó

\(\Rightarrow x-2009\le0\)

\(\Rightarrow x\le2009\)

Vậy với mọi \(x\le2009\) đều thỏa mãn

b)\(\left|3x+2\right|=\left|5x-3\right|\)

\(\Rightarrow3x+2=5x-3\) hoặc \(3x+2=3-5x\)

\(\Rightarrow2x=5\) hoặc \(8x=1\)

\(\Rightarrow x=\frac{5}{2}\) hoặc \(x=\frac{1}{8}\)

mk chưa lên lớp 7

Áp dụng tính chất: \(a^{2n}+b^{2m}=0\Leftrightarrow\hept{\begin{cases}a=0\\b=0\end{cases}}\)(2n và 2m là các số chẵn)