Ta có: \(\left(x-1\right)^{2020}\ge0\forall x\)

\(\left|y-3\right|\ge0\forall y\)

Do đó: \(\left(x-1\right)^{2020}+\left|y-3\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-1=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)

Vậy: (x,y)=(1;3)

a. (2x + 1)² - 4x² + 2x² - 2 = 0

<=> (2x + 1 - 2x)(2x + 1 + 2x) + 2(x² - 1) = 0

<=> (4x + 1) + 2x² - 2 = 0

<=> 4x + 1 + 2x² - 2 = 0

<=> 2x² + 4x - 2 + 1 = 0

<=> 2x² + 4x - 1 = 0

<=> 2x² + 4x = 1

<=> 2x(x + 2) = 1

Vì 1 chỉ có tích là 1 . 1 nên:

<=> \(\left[{}\begin{matrix}2x=1\\x+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)

\(a,\Leftrightarrow4x^2+4x+1-4x^2+2x^2-2=0\\ \Leftrightarrow2x^2+4x-1=0\\ \Leftrightarrow2\left(x^2+2x+1\right)-3=0\\ \Leftrightarrow2\left(x+1\right)^2-3=0\\ \Leftrightarrow\left(x+1\right)^2=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{3}{2}}\\x+1=-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{6}}{2}\\x=\dfrac{-2+\sqrt{6}}{2}\end{matrix}\right.\)

\(b,\left(x-2\right)\left(x+2\right)-\left(x+3\right)^2-2x-5=0\\ \Leftrightarrow x^2-4-x^2-6x-9-2x-5=0\\ \Leftrightarrow-8x=18\\ \Leftrightarrow x=-\dfrac{9}{4}\)

a,\(x\in\left\{5;1,5;\dfrac{-4}{3}\right\}\)

1: Ta có: \(\left(3-x\right)^2+\left(2x+1\right)^2-\left(2-x\right)^2-\left(2x+1\right)^2=0\)

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

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

\(\Leftrightarrow x=\dfrac{5}{2}\)

2: Ta có: \(\left(1-2x\right)^2-3\left(x-1\right)^2+\left(x+1\right)^2-\left(x-1\right)^2-\left(x-1\right)^2=0\)

\(\Leftrightarrow4x^2-4x+1-3x^2+6x-3+\left(x+1\right)^2-2\left(x-1\right)^2=0\)

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

\(\Leftrightarrow2x^2+4x+1-2x^2+4x-2=0\)

\(\Leftrightarrow x=\dfrac{1}{8}\)

a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)

b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)

c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Xem lại đề câu d 

a) x - 452 = 77 + 48

x - 452 = 125

x= 125 + 452 

x= 577

b) x + 58 = 64 + 58

x + 58 = 122

x= 122 - 58

x= 64

c) x - 1 - 2 - 3 - 4 = 0

x= 0 + 4 + 3 + 2 +1

x= 10

a)x-452=77+48

x-452=125

x=125+452

x=577

b)x+58=64+58

x+58=122

x=122-58

x=64

c)x-1-2-3-4=0

x=0+4+3+2+1

x=10

Nhớ tick cho mk nha^^

\(a)\left(x-2\right)\left(x^2+2x-3\right)\ge0.\)

Đặt \(f\left(x\right)=\left(x-2\right)\left(x^2+2x-3\right).\)

Ta có: \(x-2=0.\Leftrightarrow x=2.\\ x^2+2x-3=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=-3.\end{matrix}\right.\)

Bảng xét dấu:

x                   \(-\infty\)       -3       1       2     \(+\infty\)

\(x-2\)                    -      |    -   |   -   0   +

\(x^2+2x-3\)         +     0    -   0  +   |    +

\(f\left(x\right)\)                     -     0    +  0   -  0   +

Vậy \(f\left(x\right)\ge0.\Leftrightarrow x\in\left[-3;1\right]\cup[2;+\infty).\)

\(b)\dfrac{x^2-9}{-x+5}< 0.\)

Đặt \(g\left(x\right)=\dfrac{x^2-9}{-x+5}.\)

Ta có: \(x^2-9=0.\Leftrightarrow\left[{}\begin{matrix}x=3.\\x=-3.\end{matrix}\right.\)

\(-x+5=0.\Leftrightarrow x=5.\)

Bảng xét dấu:

x            \(-\infty\)      -3       3        5       \(+\infty\)

\(x^2-9\)            +   0   -   0   +   |    +

\(-x+5\)          +    |   +   |    +  0    -

\(g\left(x\right)\)              +    0   -   0   +  ||    -

Vậy \(g\left(x\right)< 0.\Leftrightarrow x\in\left(-3;3\right)\cup\left(5;+\infty\right).\)

b) (x+1)^3-x(x-2)^2+x-1=0

 ⇔x^3+3x^2+3x+1-(x^3-4x^2+4x)=0

⇔ x^3+3x^2+3x+1-x^3+4x^2-4x+x-1=0

⇔7x^2-2=0

⇔7x^2=2

⇔7x^2=-2⇔x=-3

⇔7x^2=2⇔x=-căn 5

a) PT \(\Leftrightarrow\left(x+1\right)^4+\sqrt{\left(x+1\right)^2+9}=3\).

Ta có \(\left(x+1\right)^4+\sqrt{\left(x+1\right)^2+9}\ge\sqrt{9}=3\).

Đẳng thức xảy ra khi và chỉ khi x = -1.

Vậy..

b) \(x^2=\sqrt{x^3-x^2}+\sqrt{x^2-x}\)

Đk: \(\left\{{}\begin{matrix}x^3-x^2\ge0\\x^2-x\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2\left(x-1\right)\ge0\\x\left(x-1\right)\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x=0\end{matrix}\right.\\\left[{}\begin{matrix}x\ge1\\x\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x\ge1\end{matrix}\right.\)

Thay x=0 vào pt thấy thỏa mãn => x=0 là một nghiệm của pt

Xét \(x\ge1\) 

Pt \(\Leftrightarrow x^4=\left(\sqrt{x^3-x^2}+\sqrt{x^2-x}\right)^2\le2\left(x^3-x\right)\) (Theo bđt bunhiacopxki)

\(\Leftrightarrow x^4\le2x\left(x^2-1\right)\le\left(x^2+1\right)\left(x^2-1\right)=x^4-1\)

\(\Leftrightarrow0\le-1\) (vô lí)

Vậy x=0

c) \(\sqrt{x-1}+\sqrt{3-x}+x^2+2x-3-\sqrt{2}=0\)  (đk: \(1\le x\le3\))

Xét x-1=0 <=> x=1 thay vào pt thấy thỏa mãn => x=1 là một nghiệm của pt

Xét \(x\ne1\)

Pt\(\Leftrightarrow\dfrac{x-1}{\sqrt{x-1}}+\dfrac{1-x}{\sqrt{3-x}+\sqrt{2}}+\left(x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{x-1}}-\dfrac{1}{\sqrt{3-x}+\sqrt{2}}+x+3\right)=0\) (1)

Xét \(\dfrac{1}{\sqrt{x-1}}-\dfrac{1}{\sqrt{3-x}+\sqrt{2}}+x+3\)

Có \(\sqrt{3-x}+\sqrt{2}\ge\sqrt{2}\) 

\(\Leftrightarrow\dfrac{-1}{\sqrt{3-x}+\sqrt{2}}\ge-\dfrac{1}{\sqrt{2}}\)

Có \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-1}}>0\\x+3\ge4\end{matrix}\right.\)  \(\Rightarrow\dfrac{1}{\sqrt{x-1}}-\dfrac{1}{\sqrt{3-x}+\sqrt{2}}+x+3>0-\dfrac{1}{\sqrt{2}}+4>0\)

Từ (1) => x-1=0 <=> x=1

Vậy pt có nghiệm duy nhất x=1