Mình đang cần gấp mấy bạn có thể làm giúp mình không ạ ?

a) |3x - 2| - 1 = x

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

ĐK : x + 1 \(\ge0\Rightarrow x\ge-1\)

Khi đó (1) <=> \(\orbr{\begin{cases}3x-2=x+1\\3x-1=-x-1\end{cases}}\Rightarrow\orbr{\begin{cases}2x=3\\4x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1,5\\x=0\end{cases}}\)(tm)

Vậy \(x\in\left\{1,5;0\right\}\)là giá trị cần tìm

b) |2x - 1| + 1 = x

=> |2x - 1| = x - 1 (1)

Đk : \(x-1\ge0\Rightarrow x\ge1\)

Khi đó (1) <=> \(\orbr{\begin{cases}2x-1=x-1\\2x-1=-x+1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\3x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\left(\text{loại}\right)\)

Vậy \(x\in\varnothing\)

​5/6 : x = 20 : 3 

5/6 : x = 20/3

        x = 20/3 : 5/6

        x =        8

Vậy x bằng 8

CHÚC BẠN HOK TỐT !!!!!!!!!!!!!………!!!!!!!!

\(\left(x-2020\right)^2+\left|y-2021\right|^5=0\)

Vì \(\left(x-2020\right)^2\ge0\forall x\)

     \(\left|y-2021\right|\ge0\forall y\)\(\Rightarrow\left|y-2021\right|^5\ge0\forall y\)

\(\Rightarrow\left(x-2020\right)^2+\left|y-2021\right|^5\ge0\forall x,y\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2020=0\\y-2021=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2020\\y=2021\end{cases}}\)

Vậy \(x=2020\)và \(y=2021\)

Ta có \(\hept{\begin{cases}\left(x-2020\right)^2\ge0\forall x\\\left|y-2021\right|^5\ge0\forall y\end{cases}}\Rightarrow\left(x-2020\right)^2+\left|y-2021\right|^5\ge0\forall x;y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2020=0\\y-2021=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2020\\y=2021\end{cases}}\)

Vậy x = 2020 ; y = 2021 là giá trị cần tìm 

\(A=1.2+2.3+3.4+...+n\left(n+1\right)\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+.....+n\left(n+1\right).3\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+......+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)

\(=1.2.3+2.3.4-1.2.3+......+n\left(n+1\right)\left(n+2\right)-n\left(n-1\right)\left(n+1\right)\)

\(=n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

A = 1.2 + 2.3 + 3.4 + .... + n(n + 1)

=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + .... + n(n + 1).3

=> 3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + .... + n(n + 1).[(n + 2) - (n- 1)]

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + n(n + 1)(n + 2) - (n - 1)n(n + 1)

=> 3A = n(n + 1)(n + 2)

=> A = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

A=232+252+272+...+220172<5041009A=232+252+272+...+220172<5041009

A=23.3+25.5+27.7+...+22017.2017<5041009A=23.3+25.5+27.7+...+22017.2017<5041009

A=23.5+25.7+27.9+...+22017.2019<5041009A=23.5+25.7+27.9+...+22017.2019<5041009                                                  

A=13−15+15−17+17−19+...+12017−12019<5041009A=13-15+15-17+17-19+...+12017-12019<5041009                                         

A=13−12019<5041009A=13-12019<5041009                                                                        

A=224673<5041009A=224673<5041009

Vậy A<5041009

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