a) Ta thấy:

\(\left(x-3\right)^2\ge0\)

\(\left(y+2\right)^2\ge0\)

\(\Rightarrow\left(x-3\right)^2+\left(y+2\right)^2\ge0\)

Để \(\left(x-3\right)^2+\left(y+2\right)^2=0\)

\(\Rightarrow\begin{cases}\left(x-3\right)^2=0\\\left(y+3\right)^2=0\end{cases}\)

\(\Rightarrow\begin{cases}x-3=0\\y+3=0\end{cases}\)

\(\Rightarrow\begin{cases}x=3\\y=-3\end{cases}\)

Vậy \(\begin{cases}x=3\\y=-3\end{cases}\)

c) Ta thấy:

\(\left(x-12+y\right)^{200}\ge0\)

\(\left(x-4-y\right)^{200}\ge0\)

\(\Rightarrow\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}\ge0\)

Để \(\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\)

\(\Rightarrow\begin{cases}\left(x-12+y\right)^{200}=0\\\left(x-4-y\right)^{200}=0\end{cases}\)

\(\Rightarrow\begin{cases}x-12+y=0\\x-4-y=0\end{cases}\)

\(\Rightarrow\begin{cases}x+y=12\\x-y=4\end{cases}\)

\(\Rightarrow\begin{cases}x=\left(12+4\right):2\\y=\left(12-4\right):2\end{cases}\)

\(\Rightarrow\begin{cases}x=8\\y=4\end{cases}\)

Vậy \(\begin{cases}x=8\\y=4\end{cases}\)

Để biểu thức a = 0 thì x - 3 = 0 và y + 2 = 0

Xét x-3=0

=>x=3

Xét y+2=0

=>y=-2

Vậy x=0;y=-2

a) (y + 5) x 2020 = 205 x 2020

(y + 5) x 2020 = 414100

y + 5 = 414100 : 2020 = 205

y = 205 - 5 = 200

a) y= 200

b) y-45600 = 1600 × 4 ×25

y-45600=160000

y= 114400

b) y - 45600 = (2019 - 419) x 4 x 25

y - 45600 = 1600 x 100

y - 45600 = 160000

y = 160000 + 45600 = 205 600

Bài 1:

a)\(\begin{cases}\left(x-3\right)^2+\left(y+2\right)^2=0\\\begin{cases}\left(x-3\right)^2\ge0\\\left(y+2\right)^2\ge0\end{cases}\end{cases}\)

\(\Rightarrow\begin{cases}\left(x-3\right)^2=0\\\left(y+2\right)^2=0\end{cases}\)\(\Rightarrow\begin{cases}x=3\\y=-2\end{cases}\)

b) tương tự 

b) (x-12+y)²⁰⁰+(x-4-y)²⁰⁰= 0

\(\begin{cases}\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\\\begin{cases}\left(x-12+y\right)^{200}\ge0\\\left(x-4-y\right)^{200}\ge0\end{cases}\end{cases}\)

\(\Rightarrow\begin{cases}\left(x-12+y\right)^{200}=0\\\left(x-4-y\right)^{200}=0\end{cases}\)\(\Rightarrow\begin{cases}x-12+y=0\\x-4-y=0\end{cases}\)\(\Rightarrow\begin{cases}x+y=12\left(1\right)\\x-y=4\left(2\right)\end{cases}\)

Trừ theo vế của (1) và (2) ta được:

\(2y=8\Rightarrow y=4\)\(\Rightarrow\begin{cases}x+4=12\\x-4=4\end{cases}\)\(\Rightarrow x=8\)

Vậy x=8; y=4

Các bạn ơi chỉ cần trả lời câu 1 thôi cũng được mình làm được câu 2 rồi

\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)

\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)

\(-18x+13=0\)

\(x=\dfrac{13}{18}\)

Vậy \(S=\left\{\dfrac{13}{18}\right\}\)

\(b.\left(x-1\right)^3-125=0\)

\(\left(x-1\right)^3=125\)

\(x-1=5\)

\(x=6\)

Vậy \(S=\left\{6\right\}\)

\(c.\left(x-1\right)^2+\left(y +2\right)^2=0\)

\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)

Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)

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

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

Vậy \(S=\left\{1;-2\right\}\)

\(d.x^2-4x+4+x^2-2xy+y^2=0\)

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

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

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

Vậy \(S=\left\{2;2\right\}\)

\(1)|5-2x|=|x+4|\)

\(\Leftrightarrow\orbr{\begin{cases}5-2x=x+4\\5-2x=-x-4\end{cases}\Leftrightarrow\orbr{\begin{cases}-2x-x=4-5\\-2x+x=-4-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}-3x=-1\\-x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=9\end{cases}}}\)

Vậy \(x=\frac{1}{3};x=9\)

\(2)|x-1|=|2x+5|\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=2x+5\\x-1=-2x-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x-2x=5+1\\x+2x=-5+1\end{cases}\Leftrightarrow}\orbr{\begin{cases}-x=4\\3x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\x=-\frac{4}{3}\end{cases}}}\)

Vậy \(x=-4;x=-\frac{4}{3}\)

\(3)|x+1|+|x+2|+|x+3|=0\left(1\right)\)

Ta có: \(|x+1|\ge0\forall x;|x+2|\ge0\forall x;|x+3|\ge0\forall x\)

\(\Leftrightarrow|x+1|+|x+2|+|x+3|\ge0\forall x\)

\(\left(1\right)\Leftrightarrow|x+1|+|x+2|+|x+3|=0\)

\(\Leftrightarrow\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=0\)

\(\Leftrightarrow x+1+x+2+x+3=0\)

\(\Leftrightarrow\left(x+x+x\right)+\left(1+2+3\right)=0\)

\(\Leftrightarrow3x+6=0\)

\(\Leftrightarrow3x=-6\)

\(\Leftrightarrow x=-6:3\)

\(\Leftrightarrow x=-2\)

Vậy x=-2

1: x = 1/3 , x=9

2: x = 4 , x = -4/3

3: x=2

(x - 12 + y)²⁰⁰ + (x - 4 + y)²⁰⁰ = 0

\(\Leftrightarrow\hept{\begin{cases}\left(x-12+y\right)^{200}=0\\\left(x-4+y\right)^{200}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=12\\x+y=4\end{cases}}\)  Vô lý 

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