\(x^2-2\left(m+1\right)x+4m-3=0\)

Theo Vi - ét, ta có :

\(\left\{{}\begin{matrix}x_`+x_2=-\dfrac{b}{a}=2\left(m+1\right)=2m+2\\x_1x_2=\dfrac{c}{a}=4m-3\end{matrix}\right.\)

Ta có :

\(x_1^2x_2+x_1x_2^2=4\)

\(\Leftrightarrow x_1x_2\left(x_1+x_2\right)-4=0\)

\(\Leftrightarrow\left(4m-3\right)\left(2m+2\right)-4=0\)

\(\Leftrightarrow8m^2+8m-6m-6-4=0\)

\(\Leftrightarrow8m^2+2m-10=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=1\\m=-\dfrac{5}{4}\end{matrix}\right.\)

1.Cho  \(\frac{x^2-4x+4}{x^2-4}< 2\)

<=>\(\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}< 2\)

<=>\(\frac{x-2}{x+2}< 2\)

<=>\(\frac{x-2}{x+2}-2< 0\)

<=>\(\frac{x-2}{x+2}-\frac{2\left(x+2\right)}{x+2}< 0\)

<=>\(\frac{x-2-2\left(x+2\right)}{x+2}< 0\)

<=>\(\frac{x-2-2x-4}{x+2}< 0\)

<=>\(\frac{-x-6}{x+2}< 0\)

<=>\(\orbr{\begin{cases}\hept{\begin{cases}-x-6< 0\\x+2>0\end{cases}}\\\hept{\begin{cases}-x-6>0\\x+2< 0\end{cases}}\end{cases}}\)

<=>\(\orbr{\begin{cases}\hept{\begin{cases}x< -6\\x< -2\end{cases}}\\\hept{\begin{cases}x>-6\\x>-2\end{cases}}\end{cases}}\)

<=>\(\orbr{\begin{cases}x< -2\\x>-6\end{cases}}\)

Vậy -6 < x < -2

1) \(\frac{x^2-4x+4}{x^2-4}=\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x-2}{x+2}\)

\(\frac{x-2}{x+2}< 2\)

\(\Leftrightarrow\frac{x-2}{x+2}-2< 0\)

\(\Leftrightarrow\frac{x-2}{x+2}-\frac{2\left(x+2\right)}{x+2}< 0\)

\(\Leftrightarrow\frac{x-2-2x-4}{x+2}< 0\)

\(\Leftrightarrow\frac{-x-6}{x+2}< 0\)

\(\Leftrightarrow-x-6< 0\)

\(\Leftrightarrow-x< 6\)

\(\Leftrightarrow x>-6\)

vậy \(x>-6\)thì giá trị của phân thức \(>2\)

2) \(\frac{2x^2-4x+8}{x^3+8}\)

\(=\frac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)

\(=\frac{2}{x+2}\)\(\left(x\ne-2\right)\)

khi đó \(\frac{2}{x+2}>2\)

\(\Leftrightarrow\frac{2}{x+2}-2>0\)

\(\Leftrightarrow\frac{2}{x+2}-\frac{2\left(x+2\right)}{x+2}>0\)

\(\Leftrightarrow\frac{2-2x-4}{x+2}>0\)

\(\Leftrightarrow\frac{-2x-2}{x+2}>0\)

\(\Leftrightarrow-2x-2>0\)

\(\Leftrightarrow-2x>2\)

\(\Leftrightarrow x< -1\)

\(P=\left(x+1\right)^3+3\left(x+1\right)^2y+3\left(x+1\right)y^2+y^3=\left(x+1+y\right)^3=10^3=1000\\ \)

1) Ta có: 
x³ + y³ + z³ - 3xyz = (x+y)³ - 3xy(x-y) + z³ - 3xyz 
= [(x+y)³ + z³] - 3xy(x+y+z) 
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z) 
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy] 
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy) 
= (x+y+z)(x² + y² + z² - xy - xz - yz). 

Câu 2:

\(\frac{x^2-y^2+6x+9}{x+y+3}\)

\(=\frac{x^2-y^2+x^2+6x+9-x^2}{x+y+3}\)

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

\(=\frac{\left(x-y+3\right)\left(x+y+3\right)}{x+y+3}\)

\(=x-y+3\)

1, phân tích thành nhân tử

x^3+y^3+z^3-3xyz

2, tính chia

(x^2-y^2+6x+9):((x+y+3)

3.Tìm gtri nguyên của n để x^4-x^3+6x^2+n chia hết cho x^2-x+5

4, CMR :a(2^a-3)-2a(a+1) chia hết cho 5

5, C=x2x−2 +x²⁺¹2−2x² 

tìm x để gtri phân thức C>0

mk chua hoc

mk moi hoc lop6

y xđ khi \(x^2-3x+1\ne0\)

       \(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ne0\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ne\dfrac{3+\sqrt{5}}{2}\\x\ne\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)

\(Q=\left(\dfrac{3-\sqrt{5}}{2}\right)^3+\left(\dfrac{3+\sqrt{5}}{2}\right)^3+3\left(\dfrac{3-\sqrt{5}}{2}+\dfrac{3+\sqrt{5}}{2}\right)\left(\dfrac{3-\sqrt{5}}{2}.\dfrac{3+\sqrt{5}}{2}\right)-13\left(\dfrac{3-\sqrt{5}}{2}.\dfrac{3+\sqrt{5}}{2}\right)\) \(=\left(\dfrac{3-\sqrt{5}}{2}+\dfrac{3+\sqrt{5}}{2}\right)^3-13\)

\(=3^3-13=27-13=14\)

1) \(Q=-x\) khi:

\(\dfrac{x-3}{x+1}=-x\)

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

2) \(Q< 1\) khi:

\(\dfrac{x-3}{x+1}< 1\)

\(\Leftrightarrow x-3< x+1\)

\(\Leftrightarrow x-x< 1+3\)

\(\Leftrightarrow0< 4\) (luôn đúng) 

Vậy \(Q< 0\) với mọi x 

3) \(Q=m\) khi:

\(\dfrac{x-3}{x+1}=m\)

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

\(\Leftrightarrow x-3=mx+m\)

\(\Leftrightarrow x-mx=m+3\)

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

\(\Leftrightarrow1-m\ne0\)

\(\Leftrightarrow m\ne1\)