áp dụng BĐT : \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) , ta đc:

\(Q=\left|x+1,7\right|+\left|4,3-x\right|\ge\left|x+1,7+4,3-x\right|=6\)

đẳng thức xảy ra khi :\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1,7\ge0\\4,3-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1,7\le0\\4,3-x\le0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}1,7\le x\le4,3\left(nhận\right)\\1,7\le x\ge4,3\left(loại\right)\end{matrix}\right.\)

vậy MIN Q=6 tại \(1,7\le x\le4,3\)

goi 3canh cua TG đó la : a , b ,c 

Ta có : a/2= b/3=c/4           và a+b+c= 45 

4-x

\(\cdot\dfrac{10}{x}< \dfrac{x}{11}\Rightarrow x^2< 110\Rightarrow-\sqrt{110}< x< \sqrt{110}\\ \cdot\dfrac{x}{11}< \dfrac{12}{x}\Rightarrow x^2< 132\Rightarrow-\sqrt{132}< x< \sqrt{132}\)\(\Rightarrow-\sqrt{132}< x< \sqrt{132}\)

\(D=x^2+2x+y^2-4y+7\\ D=x^2+2x+1+y^2-4y+4+2\\ D=\left(x+1\right)^2+\left(y-2\right)^2+2\ge2\)

đẳng thức xảy ra khi x=-1 và y=2

vậy MIN D=2 tại x=-1 và y=2

\(C=x^2+2xy+y^2+4y^2=\left(x+y\right)^2+4y^2\ge0\)

đẳng thức xảy ra khi x=y=0

vậy MIN C=0 tại x=y=0

áp dụng công thức này mà lm câu a,b,e nhá:

\(A=ax^2+bx+c=a\left(x+\dfrac{b}{2a}\right)^2+\dfrac{4ac-b^2}{4a}\\ \left[{}\begin{matrix}A\ge\dfrac{4ac-b^2}{4a}\left(với\text{ }\text{ }\text{ }a\ge0\right)\\A\le\dfrac{4ac-b^2}{4a}\left(với\text{ }a< 0\right)\end{matrix}\right.\)

bài 1:

a)

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

b)\(x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Rightarrow x=0\)

c)\(x^2-10x=-25\\ \Leftrightarrow x^2-2.5.x+25=0\\ \Leftrightarrow\left(x-5\right)^2=0\\ \Rightarrow x=5\)

bài 2:

a)

\(x^6-y^6=\left(x^3+y^3\right)\left(x^3-y^3\right)\\ =\left(x+y\right)\left(x^2+y^2+xy\right)\left(x-y\right)\left(x^2+y^2-xy\right)\)

b)

\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x+y\right)-3xzy\\ =\left(x+y+z\right)^3-\left(3xz+3zy\right)\left(x+y+z\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2zy-3xz-3zy-3xy\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-zx-zy\right)\)

sửa đề: \(\left(\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{2}{\sqrt{2}+\sqrt{3}}\right).\sqrt{21-12\sqrt{3}}\)

\(\left(\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{2}{\sqrt{2}+\sqrt{3}}\right).\sqrt{21-12\sqrt{3}}\\ =\left(\dfrac{\sqrt{2}+\sqrt{3}+2\sqrt{2}-2\sqrt{3}}{2-3}\right).\sqrt{3}.\sqrt{7-4\sqrt{3}}\\ =\dfrac{3\sqrt{2}-\sqrt{3}}{-1}.\sqrt{3}.\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left(-3\sqrt{2}+\sqrt{3}\right).\sqrt{3}.\left(2-\sqrt{3}\right)\\ =\left(\sqrt{3}-3\sqrt{2}\right)\left(2\sqrt{3}-3\right)\\ =6-3\sqrt{3}-6\sqrt{6}+9\sqrt{2}\)

tới đó thì ko bt còn hay hết