Trả lời :

Mk giúp bn câu a ) thôi mà sai thì thôi nhé :)))

a, \(\left|x\right|+\left|y\right|=0\)

\(\Leftrightarrow x=0;y=0\) \(\Rightarrow\left|x\right|+\left|y\right|=0\)

Vậy x = 0 ; y = 0

_Học tốt

câu a,b,c dạng tương tự nhau nha nên mình làm câu a

a)\(\left|x\right|+\left|y\right|=0\left(1\right)\)

Ta có: \(\hept{\begin{cases}\left|x\right|\ge0;\forall x,y\\\left|y\right|\ge0;\forall x,y\end{cases}\Rightarrow}\left|x\right|+\left|y\right|\ge0;\forall x,y\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\hept{\begin{cases}\left|x\right|=0\\\left|y\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=0\end{cases}}}\)

Vậy \(\left(x,y\right)=\left(0;0\right)\)

d) \(\left|x^2+1\right|=12\left(1\right)\)

Ta thấy \(x^2\ge0;\forall x\)

\(\Rightarrow x^2+1\ge1>0;\forall x\left(2\right)\)

Từ (1) và (2) \(\Rightarrow x^2+1=12\)

                      \(\Leftrightarrow x^2=11\)

                      \(\Leftrightarrow x=\pm\sqrt{11}\)

Vậy \(x=\pm\sqrt{11}\)

a) /x/+/y/ là số nguyên dương vì /x/,/y/ bao giờ cũng là số nguyên dương

mà /x/+/y/=0 suy ra /x/ , /y/ đều =0

b) x-1=0

    y+2=0 suy ra x=1. y=-2

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

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

Mà \(\hept{\begin{cases}\left|x-1\right|\ge0\\-\left|x^2+3\right|\le0\end{cases}\left(\forall x\right)}\)

Dấu "=" xảy ra khi: \(\left|x-1\right|=-\left|x^2+3\right|=0\)

\(\Rightarrow x^2=-3\) => vô lý

Vậy PT vô nghiệm

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

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

Mà \(\hept{\begin{cases}\left|x-1\right|\ge0\\-\left|x^2-1\right|\le0\end{cases}\left(\forall x\right)}\)

Dấu "=" xảy ra khi: \(\left|x-1\right|=-\left|x^2-1\right|=0\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\x^2=1\end{cases}}\Rightarrow x=1\)

Vậy x = 1

a. 

\(=\left(x+1\right)\left(x+2\right)\left(x-2\right)\left(x-3\right)\)

b. 

\(=\left(x+1\right)\left(x+1\right)\left(x^2+x+1\right)\)

c. 

\(1,\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)

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

\(2,\)

\(3x\left(x+5\right)-6\left(x+5\right)=0\)

\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

\(3,\)

\(x^4-x^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

\(4,\)

\(x^2-2x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(5,\)

\(x\left(x+6\right)-10\left(x-6\right)=0\)

\(\Leftrightarrow x^2+6x-10x+60=0\)

\(\Leftrightarrow x^2-4x+60=0\)

\(\Leftrightarrow x^2-4x+4+56=0\)

\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)

=> Phương trình vô nghiệm

Ta có : \(\hept{\begin{cases}\left|x^2+x-2\right|\ge0\forall x\\\left|x^2-1\right|\ge0\forall x\end{cases}}\Rightarrow\left|x^2+x-2\right|+\left|x^2-1\right|\ge0\forall x\)

Đẳng thức |x² + x - 2| + |x² - 1| = 0 xảy ra 

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

+) Nếu : (x + 2)(x - 1) = 0

=> \(\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)

+) Nếu x² = 1

=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Vậy  x = 1 

1/ a) TH1: x-2 = 0 => x= 0+2 = 2

       TH2: 5-x= 0 => x= 5-0 = 5

b)???

duyệt đi

\(a,\Delta=4\left(m-1\right)^2-4\left(-2m-3\right)=4m^2-8m+4+8m+12\\ \Delta=4m^2+16>0\left(đpcm\right)\\ b,\Delta=\left(2m-1\right)^2-4\left(2m-2\right)=4m^2-4m+1-8m+8\\ \Delta=4m^2-12m+9=\left(2m-3\right)^2\ge0\left(đpcm\right)\\ c,Sửa:x^2-2\left(m+1\right)x+2m-2=0\\ \Delta=4\left(m+1\right)^2-4\left(2m-2\right)=4m^2+8m+4-8m+8\\ \Delta=4m^2+12>0\left(đpcm\right)\\ d,\Delta=4\left(m+1\right)^2-4\cdot2m=4m^2+8m+4-8m\\ \Delta=4m^2+4>0\left(đpcm\right)\\ e,\Delta=4m^2-4\left(m+7\right)=4m^2-4m+7=\left(2m-1\right)^2+6>0\left(đpcm\right)\\ f,\Delta=4\left(m-1\right)^2-4\left(-3-m\right)=4m^2-8m+4+12+4m\\ \Delta=4m^2-4m+16=\left(2m-1\right)^2+15>0\left(đpcm\right)\)

1/2 + 1/6+1/12 + 1/20 +....+ 1/x(x+1) = 2021/2022

1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +...+ 1/x. (x+1) = 2021/2020

1 - 1/2 + 1/2 - 1/3 + 1/3- 1/4 + 1/4 - 1/5 +...+ 1/x - 1/(x+1) = 2021/2020

1 - 1/(x+1)        = 2021/2020

     1/(x+1)         = 1 - 2021/2020

     1/(x+1)         = -1/2020

     1/(x+1)         = 1/-2020

       x + 1           = - 2020

        x                 = -2020 - 1

        x                 = -2021

Giải:

1/2+1/6+1/12+1/20+...+1/x.(x+1)=2021/2022

1/1.2+1/2.3+1/3.4+1/4.5+...+1/x.(x+1)=2021/2022

1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=2021/2022

1/1-1/x+1                                    =2021/2022

      1/x+1                                    =1/1-2021/2022

      1/x+1                                    =1/2022

⇒x+1=2022

       x=2022-1

       x=2021

Chúc bạn học tốt!

1. (-2x - 1)(x² - x - 3) - (x + 2)(x + 1)²

= -2x³ + 2x² + 6x - x² + x + 3 - (x + 2)(x² + 2x + 1)

= -2x³ + x² + 7x + 3 - x³ - 2x² - x - 2x² - 2x - 2

= -3x³ - 3x² + 4x + 1

2. (x + 2)(x - 1) - (x - 3)(x + 2) = 3

=> (x + 2)(x - 1 - x + 3) = 3

=> (x + 2).0 = 3

...(xem lại đề)

\(\left(x+2\right)\left(x-1\right)-\left(x-3\right)\left(x+2\right)=3\)

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

\(\Leftrightarrow2\left(x+2\right)=3\)

\(\Leftrightarrow x+2=\frac{3}{2}\)

\(\Leftrightarrow x=\frac{3}{2}-2\)

\(\Leftrightarrow x=-\frac{1}{2}\)

\(x^2+11x-13=0\)

Ta có: \(\Delta=11^2+4.13=173\)

Vậy pt có 2 nghiệm phân biệt:

\(x_1=\frac{-11+\sqrt{173}}{2}\);\(x_2=\frac{-11-\sqrt{173}}{2}\)

\(2x^2-5x+2=0\)

Ta có: \(\Delta=5^2-4.2.2=9,\sqrt{\Delta}=3\)

Vậy pt có 2 nghiệm phân biệt:

\(x_1=\frac{5+3}{4}=2\);\(x_2=\frac{5-3}{4}=\frac{1}{2}\)