chứ kh lẽ chăm:)

a) \(5x\left(1-2x\right)-3x\left(x+18\right)=0\Rightarrow5x-10x^2-3x^2-54x=0\Rightarrow-13x^2-49x=0\Rightarrow-x\left(13x+49\right)=0\Rightarrow\left[{}\begin{matrix}-x=0\\13x+49=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-49}{13}\end{matrix}\right.\)

Vậy x=0; x=-49/13

b) \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\Rightarrow5x-3\left[4x-2\left(4x-15x+6\right)\right]=182\Rightarrow5x-3\left[4x-2\left(-11x+6\right)\right]=182\Rightarrow5x-3\left(4x+22x-12\right)=182\Rightarrow5x-3\left(26x-12\right)=182\Rightarrow5x-78x+36=182\Rightarrow-73x=146\Rightarrow x=-2\)

Vậy x=-2

c) \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\Rightarrow x^2+5x+6-\left(x^2+3x-10\right)=0\Rightarrow x^2+5x+6-x^2-3x-10=0\Rightarrow2x=4\Rightarrow x=2\)

Vậy x=2

d) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\Rightarrow24x^2+16x-9x-6-\left(4x^2+16x+7x+28\right)=10x^2-2x+5x-1-33\Rightarrow24x^2+7x-6-4x^2-23x-28=10x^2+3x-34\Rightarrow10x^2-19x=0\Rightarrow x\left(10x-19\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\10x-19=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{10}\end{matrix}\right.\)

Vậy x=0; x=19/10

#h24cfs_646 mở đi ạ, mình cũng muốn viết 😇

Ý 2 của câu b

a) \(P\left(x\right)=2x^4+3x^3+3x^2-x^4-4x+2-2x^2+6x=\left(2x^4-x^4\right)+3x^3+\left(3x^2-2x^2\right)+\left(-4x+6x\right)+2=x^4+3x^3+x^2+2x+2\)\(Q\left(x\right)=x^4+3x^2+5x-1-x^2-3x+2+x^3=x^4+x^3+\left(3x^2-x^2\right)+\left(5x-3x\right)+\left(-1+2\right)=x^4+x^3+2x^2+2x+1\)

b) \(P\left(x\right)+Q\left(x\right)=x^4+3x^3+x^2+2x+2+x^4+x^3+2x^2+2x+1=\left(x^4+x^4\right)+\left(3x^3+x^3\right)+\left(x^2+2x^2\right)+\left(2x+2x\right)+\left(2+1\right)=2x^4+4x^3+3x^2+4x+3\)

\(P\left(x\right)-Q\left(x\right)=x^4+3x^3+x^2+2x+2-\left(x^4+x^3+2x^2+2x+1\right)=x^4^{ }+3x^3+x^2+2x+2-x^4-x^3-2x^2-2x-1=\left(x^4-x^4\right)+\left(3x^3-x^3\right)+\left(x^2-2x^2\right)+\left(2x-2x\right)+\left(2-1\right)=2x^3+x^2+1\)

a) Công thức tính chu vi hình vuông có độ dài 1 cạnh là x cm: 4x (cm)

b) Sắp xếp đa thức theo số mũ giảm dần của biến: \(B\left(x\right)=-2021x^3+2022x^2-3\)

Bậc của đa thức là 3

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

Giá trị của đa thức tại x=-1 là: \(-\left(-1\right)^3-3=-\left(-1\right)-3=1-3=-2\)

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