a ) \(\left(m+1\right)x+\left(m^2+m\right)=0.x+0\)

Đồng nhất hệ số 2 đa thức :

\(\Rightarrow\left\{{}\begin{matrix}m+1=0\\m^2+m=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=-1\\m\left(m+1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=-1\\\left[{}\begin{matrix}m=0\\m=-1\end{matrix}\right.\end{matrix}\right.\)

Vậy m = -1 thì pt bằng 0

Câu 4 :

Xét tứ giác ABCD có : \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360\)

=> x + 130 + 90 + 70 = 360

=.> x  =  70

Vậy x = 70

câu 6 : 

Q = \(\dfrac{2x^2+2}{\left(x+1\right)^2}\)

Đặt x + 1 = t

=> x = t - 1

\(\Rightarrow Q=\dfrac{2\left(t-1\right)^2+2}{t^2}\)

\(\Rightarrow Q=\dfrac{2t^2-4t+2+2}{t^2}\)

\(\Rightarrow Q=\dfrac{2t^2-4t+4}{t^2}=\dfrac{2t^2}{t^2}-\dfrac{4t}{t^2}+\dfrac{4}{t^2}\)

\(\Rightarrow Q=2-\dfrac{4}{t}+\dfrac{4}{t^2}=1+\left(1-\dfrac{4}{t}+\dfrac{4}{t^2}\right)\)

\(\Rightarrow Q=1+\left(1-\dfrac{2}{t}\right)^2\ge1\) Vì \(\left(1-\dfrac{2}{t}\right)^2\ge0\)

Vậy GTNN của Q là 1

Dấu = xảy ra  :\(\Leftrightarrow\left(1-\dfrac{2}{t}\right)^2=0\Leftrightarrow\left(1-\dfrac{2}{x+1}\right)=0\Leftrightarrow\dfrac{2}{x+1}=1\Leftrightarrow x+1=2\Leftrightarrow x=1\)

Ta có : \(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b-c}=-\left(\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)\\\dfrac{b}{c-a}=-\left(\dfrac{a}{b-c}+\dfrac{c}{a-b}\right)\\\dfrac{c}{a-b}=-\left(\dfrac{a}{b-c}+\dfrac{b}{c-a}\right)\end{matrix}\right.\)

⇒ \(\dfrac{a}{\left(b-c\right)^2}+\dfrac{b}{\left(c-a\right)^2}+\dfrac{c}{\left(a-b\right)^2}\)

\(=-\left(\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)\dfrac{1}{b-c}-\left(\dfrac{a}{b-c}+\dfrac{c}{a-b}\right)\dfrac{1}{c-a}-\left(\dfrac{a}{b-c}+\dfrac{b}{c-a}\right)\dfrac{1}{a-b}\)

\(=\dfrac{b}{\left(c-a\right)\left(c-b\right)}+\dfrac{c}{\left(a-b\right)\left(c-b\right)}+\dfrac{a}{\left(b-c\right)\left(a-c\right)}+\dfrac{c}{\left(a-b\right)\left(a-c\right)}+\dfrac{a}{\left(b-c\right)\left(b-a\right)}+\dfrac{b}{\left(c-a\right)\left(b-a\right)}\)

\(=\dfrac{a+b}{\left(a-c\right)\left(b-c\right)}-\dfrac{a+c}{\left(a-b\right)\left(b-c\right)}+\dfrac{b+c}{\left(a-b\right)\left(a-c\right)}\)

\(=\dfrac{\left(a+b\right)\left(a-b\right)-\left(a+c\right)\left(a-c\right)+\left(b+c\right)\left(b-c\right)}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)

\(=\dfrac{a^2-b^2-a^2+c^2+b^2-c^2}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}=0\)

➤ĐPCM

cũng đúng nhưng hơi dài mk còn cách khác tí mk làm cho