Question

[Nguyenthithuytien]

bài 1: phân tích đa thức thành nhân tử

a) 3x - 3y +x²-2xy+y²

d)x²+2xy+y²-2x-2y +1

bài 2: tìm x

a)(x+1).(x+1)= (x+2).(x+5)

b)(x+3).(x+5)=0

c)4x²-9=0

d) (4x-5)²-(3x-4)²=0

Bài 3 tìm m sao cho A(x) chia hết B(x)

A(x)= 8x²- 26x +m chia hết B(x)= 2x-3

giúp mk vs ạ

Answer 1

[B.Thị Anh Thơ]

Bài 1

d, \(x^2+2xy+y^2-2x-2y+1\)

\(\Rightarrow x^2+y^2=1+2xy-2y-2x\)

\(\Rightarrow\left(x+y-1\right)^2\)

Bài 2:

a, \(\left(x+1\right)\left(x+1\right)=\left(x+2\right)\left(x+5\right)\)

\(\Leftrightarrow\left(x+1\right)^2=x^2+5x+2x+10\)

\(\Leftrightarrow x^2+2x+1=x^2=5x+2x+10\)

\(\Leftrightarrow-5x=9\)

\(\Leftrightarrow x=-\frac{9}{5}\)

b,\(\left(x+3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

c, \(4x^2-9=0\)

\(\Leftrightarrow4x^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\\frac{3}{2}\end{matrix}\right.\)

d,\(\left(4x-5\right)^2-\left(3x-4\right)^2=0\)

\(\Leftrightarrow16x^2-40x+25-\left(9x^2-24x+16\right)=0\)

\(\Leftrightarrow16x^2-40x+25-9x^2+24x-16=0\)

\(\Leftrightarrow7x^2-16x+9=0\)

\(\Leftrightarrow x=\frac{-\left(-16\right)\pm\sqrt{\left(-16\right)^2-4.7.9}}{14}\)

\(\Leftrightarrow x=\frac{16\pm\sqrt{256-252}}{14}\)

\(\Leftrightarrow x=\frac{16\pm\sqrt{4}}{14}\)

\(\Leftrightarrow x=\frac{16\pm2}{14}\)

\(\Leftrightarrow x=\left[{}\begin{matrix}\frac{16+2}{14}\\\frac{16-2}{14}\end{matrix}\right.\)

\(\Leftrightarrow x=\left[{}\begin{matrix}\frac{9}{7}\\1\end{matrix}\right.\)

Answer 2

[@Nk>↑@]

1.a)\(3x-3y+x^2-2xy+y^2\)

\(=3\left(x-y\right)+\left(x-y\right)^2\)

\(=\left(x-y\right)\left(3+x-y\right)\)

d)\(x^2+2xy+y^2-2x-2y+1\)

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

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

2.a)\(\left(x+1\right)\left(x+1\right)=\left(x+2\right)\left(x+5\right)\)

\(\Leftrightarrow\left(x+1\right)^2=x^2+5x+2x+10\)

\(\Leftrightarrow x^2+2x+1-x^2-7x-10=0\)

\(\Leftrightarrow-5x-9=0\)

\(\Leftrightarrow-5x=9\)

\(\Leftrightarrow x=-\frac{9}{5}\). Vậy \(S=\left\{-\frac{9}{5}\right\}\)

b)\(\left(x+3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\).Vậy \(S=\left\{-3;-5\right\}\)

c)\(4x^2-9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{3}{2}\end{matrix}\right.\). Vậy \(S=\left\{\pm\frac{3}{2}\right\}\)

d)\(\left(4x-5\right)^2-\left(3x-4\right)^2=0\)

\(\Leftrightarrow\left(4x-5+3x-4\right)\left(4x-5-3x+4\right)=0\)

\(\Leftrightarrow\left(7x-9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-9=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{9}{7}\\x=1\end{matrix}\right.\). Vậy \(S=\left\{1;\frac{9}{7}\right\}\)

3.Ta có:

Để \(A\left(x\right)⋮B\left(x\right)\) thì: \(m+21⋮2x-3\)

\(\Rightarrow m+21=0\)

\(\Rightarrow m=-21\)

Vậy...!