Sửa đề: Chứng minh \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\)

Bài làm:

\(VT=\left(a+b\right)^3=\left(a+b\right)\left(a+b\right)\left(a+b\right)\\ =\left(a^2+ab+ab+b^2\right)\left(a+b\right)\\ =\left(a^2+2ab+b^2\right)\left(a+b\right)\\ =a^3+2a^2b+ab^2+a^2b+2ab^2+b^3\\ =a^3+3a^2b+3ab^2+b^3=VP\left(DPCM\right)\)

a) 

\(\left\{{}\begin{matrix}\dfrac{1}{x-1}+\dfrac{1}{y}=-1\\\dfrac{3}{x-1}-\dfrac{2}{y}=7\end{matrix}\right.\left(x\ne1;x\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-1}+\dfrac{2}{y}=-2\\\dfrac{3}{x-1}-\dfrac{2}{y}=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-1}+\dfrac{1}{y}=-1\\\dfrac{5}{x-1}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1+\dfrac{1}{y}=-1\\x-1=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=-1-1=-2\\x=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=2\end{matrix}\right.\left(tm\right)\)

b) 

\(\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{1}{y+1}=3\\\dfrac{4}{x-2}-\dfrac{3}{y+1}=1\end{matrix}\right.\left(x\ne2;y\ne-1\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-2}+\dfrac{2}{y+1}=6\\\dfrac{4}{x-2}-\dfrac{3}{y+1}=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y+1}=5\\\dfrac{2}{x-2}+\dfrac{1}{y+1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+1=1\\\dfrac{2}{x-2}+1=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=0\\\dfrac{2}{x-2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x-2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=0\\x=3\end{matrix}\right.\left(tm\right)\)

c)

\(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{y-1}=1\end{matrix}\right.\left(x\ne2;y\ne1\right) \Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{2}{y-1}=4\\\dfrac{2}{x-2}-\dfrac{3}{y-1}=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{y-1}=2\\\dfrac{5}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{3}{5}=2\\y-1=\dfrac{5}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}=2-\dfrac{3}{5}=\dfrac{7}{5}\\y=\dfrac{5}{3}+1=\dfrac{8}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=\dfrac{5}{7}\Leftrightarrow x=\dfrac{5}{7}+2=\dfrac{19}{7}\\y=\dfrac{8}{3}\end{matrix}\right.\left(tm\right)\)

a) 

\(\left\{{}\begin{matrix}\dfrac{1}{3x}+\dfrac{1}{3y}=\dfrac{1}{4}\\\dfrac{5}{6x}+\dfrac{1}{y}=\dfrac{2}{3}\end{matrix}\right.\left(x,y\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{4}\\\dfrac{5}{6x}+\dfrac{1}{y}=\dfrac{2}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{5}{6x}=\dfrac{3}{4}-\dfrac{2}{3}\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{6x}=\dfrac{1}{12}\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{4}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x=12\Leftrightarrow x=2\\\dfrac{1}{2}+\dfrac{1}{y}=\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\\dfrac{1}{y}=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\left(tm\right)\)

b) 

\(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=2\\2\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=16\end{matrix}\right.\left(x,y\ne0\right) \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=2\\\dfrac{1}{x}+\dfrac{1}{y}=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}=10\\\dfrac{1}{x}+\dfrac{1}{y}=8\end{matrix}\right. \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\5+\dfrac{1}{y}=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\\dfrac{1}{y}=8-5=3\end{matrix}\right. \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=\dfrac{1}{3}\end{matrix}\right.\left(tm\right)\)

$2x^2-5xy-3y^2$

$=2x^2-6xy+xy-3y^2$

$=2x(x-3y)+y(x-3y)$

$=(x-3y)(2x+y)$

y = (m-2)x+2m-1 (a = m-2 và b=2m-1) 

a) Đề hàm số là hàm số bậc nhất thì:

\(a\ne0\Rightarrow m-2\ne0\Leftrightarrow m\ne2\)

b) y=-2x+3 (a'=-2)

Để (d) song song với (d') thì: 

\(a=a'\\ \Rightarrow m-2=-2\Rightarrow m=0\) 

c) Để (d) cắt (d1) tại một điểm trên trục hoành thì: `y=0`

=> (d1) `y=x-2=0=>x=2` 

\(\left(d\right)y=\left(m-2\right)x+2m-1=0\Rightarrow\left(m-2\right)x=1-2m\Rightarrow x=\dfrac{1-2m}{m-2}\)

Mà: `x=2` nên:

\(2=\dfrac{1-2m}{m-2}\Leftrightarrow2\left(m-2\right)=1-2m\Leftrightarrow2m-4=1-2m\\ \Leftrightarrow2m+2m=1+4=5\\ \Leftrightarrow4m=5\\ \Leftrightarrow m=\dfrac{5}{4}\left(tm\right)\)

= 2x² - 4xy - xy + 2y²

⁼ 2x(x - 2y) - y(x - 2y)

= (2x - y)(x - 2y)

Đề yêu cầu gì thế bạn?

bạn ơi cho mik hỏi là đề là phân tích thành nhân tử à?

1) $-xyz^2-3xz.yz=-xyz^2-3xyz^2=-4xyz^2$

2) $-8x^2y-x.(xy)=-8x^2y-x^2y=-9x^2y$

3) $4xy^2.x-(-12x^2y^2)=4x^2y^2+12x^2y^2=16x^2y^2$

4) $\frac12 x^2y^3-\frac13 x^2y.y^2=\frac12 x^2y^3-\frac13 x^2y^3=\frac16 x^2y^3$

5) $3xy.(x^2y)-\frac56 x^3y^2=3x^3y^2-\frac56 x^3y^2=\frac{13}{6}x^3y^2$

6) $\frac34 x^4y-\frac16 xy.x^3=\frac34 x^4y-\frac16 x^4y=\frac{7}{12}x^4y$

7) $\frac45y^2x^5-x^3.x^2y^2=\frac45 x^5y^2-x^5y^2=-\frac15 x^5y^2$

8) $-xy^3-\frac27 y^2.xy=-xy^3-\frac27 xy^3==\frac97 xy^3$

9) $\frac56 xy^2z-\frac14 xyz.y=\frac56 xy^2z-\frac14 xy^2z=\frac{7}{12} xy^2z$

10) $15x^4+7x^4-20x^2.x^2$

$=22x^4-20x^4=2x^4$

11) $\frac12 x^5y-\frac34 x^5y+xy.x^4$

$=-\frac14 x^5y+x^5y=\frac34 x^5y$

12) $13x^2y^5-2x^2y^5+x^6$

$=11x^2y^5+x^6$

Bài 10:

1: \(-xyz^2-3xz\cdot yz=-xyz^2-3xyz^2=-4xyz^2\)

2: \(-8x^2y-x\cdot xy=-8x^2y-x^2y=-9x^2y\)

3: \(4xy^2\cdot x-\left(-12x^2y^2\right)=4x^2y^2+12x^2y^2=16x^2y^2\)

4: \(\dfrac{1}{2}x^2y^3-\dfrac{1}{3}x^2y\cdot y^2=\dfrac{1}{2}x^2y^3-\dfrac{1}{3}x^2y^3=\dfrac{1}{6}x^2y^3\)

5: \(3xy\cdot\left(x^2y\right)-\dfrac{5}{6}x^3y^2=3x^3y^2-\dfrac{5}{6}x^3y^2=\dfrac{13}{6}x^3y^2\)

6: \(\dfrac{3}{4}x^4y-\dfrac{1}{6}xy\cdot x^3=\dfrac{3}{4}x^4y-\dfrac{1}{6}x^4y=x^4y\left(\dfrac{3}{4}-\dfrac{1}{6}\right)=\dfrac{7}{12}x^4y\)

7: \(\dfrac{4}{5}x^5y^2-x^3\cdot x^2y^2=\dfrac{4}{5}x^5y^2-x^5y^2=-\dfrac{1}{5}x^5y^2\)

8: \(-xy^3-\dfrac{2}{7}\cdot y^2\cdot xy=-xy^3-\dfrac{2}{7}xy^3=-\dfrac{9}{7}xy^3\)

9: \(\dfrac{5}{6}xy^2z-\dfrac{1}{4}xyz\cdot y=\dfrac{5}{6}xy^2z-\dfrac{1}{4}xy^2z=xyz^2\left(\dfrac{5}{6}-\dfrac{1}{4}\right)=\dfrac{7}{12}xyz^2\)

10:

\(15x^4+7x^4-20x^2\cdot x^2=22x^4-20x^4=2x^4\)

11:

\(\dfrac{1}{2}x^5y-\dfrac{3}{4}x^5y+xy\cdot x^4\)

\(=\dfrac{1}{2}x^5y-\dfrac{3}{4}x^5y+x^5y\)

\(=x^5y\left(\dfrac{1}{2}-\dfrac{3}{4}+1\right)=\dfrac{3}{4}x^5y\)

12: \(13x^2y^5-2x^2y^5+x^6=x^2y^5\left(13-2\right)+x^6=x^6+11x^2y^5\)