Ta có: \(\left(x+20\right)^{100}\ge0;\left|y+4\right|\ge0\)

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

\(\Rightarrow x+20=0\text{ và }y+4=0\)

\(\Rightarrow x=-20\text{ và }y=-4\)

Vậy...

ĐÂY LÀ TOÁN 6:

a) Ta có : \(\orbr{\begin{cases}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{cases}}\)

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

Do đó \(\left(x+20\right)^{100}=0\)=> \(x=-20\)

\(y+4=0\Rightarrow y=-4\)

Vậy x = -20 và y = -4

b) \(\left(x-\frac{2}{5}\right)\left(x+\frac{3}{7}\right)=0\)

=> \(\orbr{\begin{cases}x-\frac{2}{5}=0\\x+\frac{3}{7}=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{3}{7}\end{cases}}\)

Ta có: (x+20)¹⁰⁰ >= 0 với mọi x thuộc Z

           Iy+4I >=0 với mọi x thuộc Z

Mà (x+20)¹⁰⁰+Iy+4I=0

=> (x+20)¹⁰⁰=0 và Iy+4I=0

<=> x+20=0 và y+4=0

<=> x=-20 và y=-4

do \(\hept{\begin{cases}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{cases}}\)

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

=>\(\hept{\begin{cases}x=-20\\y=-4\end{cases}}\)

Vì \(\hept{\begin{cases}\left(x+20\right)^{100}\ge0\forall x\\\left|y+4\right|\ge0\forall y\end{cases}}\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\)

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

Vậy x =  - 20 ; y = - 4

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

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

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)

=> 

=> 

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

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

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!

a) Ta có : (2x - 1)¹⁰⁰ + (x - y)¹⁰² = 0

<=> \(\hept{\begin{cases}2x-1=0\\x-y=0\end{cases}}\)

<=> \(\hept{\begin{cases}2x=1\\x=y\end{cases}}\)

<=> \(x=y=\frac{1}{2}\)

b) Ta có: |x - 3| + (x + y)²⁰²⁰ = 0

<=> \(\hept{\begin{cases}x-3=0\\x+y=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=3\\y=-x\end{cases}}\)

<=> \(\hept{\begin{cases}x=3\\y=-3\end{cases}}\)

Với x = 3 và y = -3 thay vào biểu thức A :

A = \(3^2.\left[3+\left(-3\right)\right]^{100}=9.0^{100}=0\)

a) Ta có (2x - 1)¹⁰⁰ \(\ge\)0 với mọi x

              (x - y)¹⁰²  \(\ge\)0 với mọi x,y

Do đó : (2x - 1)¹⁰⁰ + (x - y)¹⁰² \(\ge\)0 với mọi x,y

Và (2x-1)¹⁰⁰ + (x-y)¹⁰² = 0

<=> 2x - 1 = 0          <=> x = 1/2

và   x - y   = 0             và y = 1/2

b) Ta có : |x - 3| \(\ge\)0 với mọi x

           (x + y)²⁰²⁰\(\ge\)0 với mọi x,y

Do đó : |x - 3| + (x + y)²⁰²⁰ \(\ge\)0 với mọi x,y

Và |x - 3| + (x + y)²⁰²⁰ = 0

<=> x - 3 = 0                      <=> x = 3

   và x + y = 0                     và    y = -3

Rồi tự thay vào r tính A đi eiu :)