Sao đề bài... nó khó hiểu quá!

\(a,\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\\ \Rightarrow2x=-4\\ \Rightarrow x=-2\\ b,\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\\ \Rightarrow\left(2021x-1\right)\left(x-2020\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2020=0\\2021x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)

a) \(\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\)

\(\Rightarrow2x=-4\Rightarrow x=-2\)

b) \(\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\Rightarrow\left(x-2020\right)\left(2021x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)

Xét \(\left(x^2+2020\right)\left(x-10\right)=0\)

Vì \(x^2\ge0\forall x\)\(\Rightarrow x^2+2020\ge2020\forall x\)

\(\Rightarrow\left(x^2+2020\right)\left(x-10\right)=0\)\(\Leftrightarrow x-10=0\)\(\Leftrightarrow x=10\)

Ta thấy: trong biểu thức \(P=\left(x^2-1\right)\left(x^2-2\right)\left(x^2-3\right)......\left(x^2-2020\right)\)có chứa thừa số \(x^2-100\)

Thay \(x=10\)vào thừa số \(x^2-100\)ta được: \(10^2-100=100-100=0\)

\(\Rightarrow P=0\)

Vậy \(P=0\)

Theo đề bài, ta có: (x^2+2020)(x-10)=0

Vì x^2 luôn lớn hơn hoặc bằng 0 nên x^2+2020>0

=> x-10=0

Khi đó P=(x^2-1)(x^2-2)...(x^2-100)(x^2-101)...(x^2-2020)

 => P=(10^2-1)(10^2-2)...(10^2-100)(10^2-101)...(10^2-2020)

=> P=0 < Vì 10^2-100=0>

Vậy P=0

Lời giải:

$\frac{x+2}{2020}+\frac{x+2}{2020}=\frac{x+2019}{3}+\frac{x+2020}{2}$

$\frac{x+2}{2020}+1+\frac{x+2}{2020}+2=\frac{x+2019}{3}+1+\frac{x+2020}{2}+1$

$\frac{x+2022}{2020}+\frac{x+2022}{2020}=\frac{x+2022}{3}+\frac{x+2022}{2}$

$(x+2022)(\frac{1}{2020}+\frac{1}{2020}-\frac{1}{3}-\frac{1}{2})=0$

Dễ thấy $\frac{1}{2020}+\frac{1}{2020}-\frac{1}{3}-\frac{1}{2}<0$

Do đó: $x+2022=0$

$\Rightarrow x=-2022$

1, \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow x=0;x=\pm5\)

2, \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+9\right)=0\Leftrightarrow x=-9;x=1\)

3, \(6x\left(x-2\right)=x-2\Leftrightarrow\left(6x-1\right)\left(x-2\right)=0\Leftrightarrow x=\frac{1}{6};x=2\)

4, \(7\left(x-2020\right)^2-x+2020=0\Leftrightarrow7\left(x-2020\right)^2-\left(x-2020\right)=0\)

\(\Leftrightarrow\left(x-2020\right)\left[7\left(x-2020\right)-1\right]=0\Leftrightarrow x=2020;x=\frac{14141}{7}\)

5, \(x^2-10x=-25\Leftrightarrow x^2-10x+25=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)

6, \(x^2-2x-3=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\Leftrightarrow x=-1;x=3\)

\(1,\)

\(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x^2-25\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-25=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

\(2,\)

\(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow x^2-x+9x-9=0\)

\(\Leftrightarrow x\left(x-1\right)+9\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+9=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-9\\x=1\end{cases}}\)

\(3,\)

\(6x\left(x-2\right)=x-2\)

\(\Leftrightarrow6x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(6x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{6}\end{cases}}\)

\(4,\)

\(7\left(x-2020\right)^2-x+2020=0\)

\(\Leftrightarrow7\left(x-2020\right)^2-\left(x-2020\right)=0\)

\(\Leftrightarrow\left(x-2020\right)[7\left(x-2020\right)-1]=0\)

\(\Leftrightarrow\left(x-2020\right)[7x-14141]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\7x=14141\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=\frac{14141}{7}\end{cases}}\)

\(5,\)

\(x^2-10x=-25\)

\(\Leftrightarrow x^2-10x+25=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

\(6,\)

\(x^2-2x-3=0\)

\(\Leftrightarrow x^2-3x+x-3=0\)

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

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

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

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

pls help me mk đang cần vội :(

Bài 1:

\(a,=6x^2+19x-7-6x^3-4x^2+7x=-6x^3+2x^2+26x-7\\ b,B=26\cdot\left(63^2+63\cdot37+37^2\right):26+63\cdot37\\ =63^2+63\cdot37+37^2+63\cdot37\\ =\left(63+37\right)^2=100^2=10000\)

Bài 2:

\(a,=x\left(y^2-25\right)=x\left(y-5\right)\left(y+5\right)\\ b,=\left(x-y\right)\left(x+2\right)\\ c,=\left(x-3\right)\left(x^2-4\right)=\left(x-2\right)\left(x-3\right)\left(x+2\right)\)

Bài 2: 

b: \(=\left(x-y\right)\left(x+2\right)\)