A=10

B=5

C=2

G.tuỳ ý

C. Số tự nhiên

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

b) \(\left(3x-\frac{1}{8}y\right)^2=9x^2-\frac{3}{4}xy+\frac{1}{64}y^2\)

c) \(\left(-6x-\frac{2}{5}\right)^2=36x^2+\frac{24}{5}x+\frac{4}{25}\)

d) \(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-1\)

e) \(\left(x-y\right)^2\left(x+y\right)^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\)

f) \(\left(\frac{1}{2}x-\frac{1}{3}y-1\right)^2=\frac{1}{4}x^2+\frac{1}{9}y^2+1-\frac{1}{3}xy-x+\frac{2}{3}y\)

a, \(\left(x+2y\right)^2=x^2+4xy+4y^2\)

b, \(\left(3x-\frac{1}{8}y\right)^2=9x^2-\frac{3}{4}xy+\frac{1}{64}y^2\)

e, \(\left(x-y\right)^2\left(x+y\right)^2=x^4-2x^2y^2+y^4\)

Bài làm:

Ta có: \(3\left(3-2x^2\right)+3x\left(2x-1\right)=9\)

\(\Leftrightarrow9-6x^2+6x^2-3x=9\)

\(\Leftrightarrow3x=0\)

\(\Rightarrow x=0\)

\(3\left(2-2x^2\right)+3x\left(2x-1\right)=9\)

\(\Leftrightarrow3\left(2-2x^2+2x^2-x\right)=9\)

\(\Leftrightarrow2-x=3\)

\(\Leftrightarrow x=-1\)

mình nghĩ làm chọn đáp án là: C 

Nếu đúng thì k cho mình nha!

Chúc học tốt!

He said he ____________ to the party.

A. had  rather not go                                  B. would rather did not go

C. would rather not go                               D. had better not going

\(A=5-8x+x^2=-8x+x^2+6-11\)

\(=\left(x-4\right)^2-11\)

Vì \(\left(x-4\right)^2\ge0\forall x\)\(\Rightarrow\left(x-4\right)^2-11\ge-11\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)

Vậy Amin = - 11 <=> x = 4

\(B=\left(2-x\right)\left(x+4\right)=-x^2-2x+8\)

\(=-\left(x^2+2x+1\right)+9=-\left(x+1\right)^2+9\)

Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow-\left(x+1\right)^2+9\le9\)

Dấu "=" xảy ra \(\Leftrightarrow-\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy Bmax = 9 <=> x = - 1

16 phần tử

BĐT CẦN CM <=>   \(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge a+b+c\)

<=>   \(a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\ge a+b+c\)

<=>   \(2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\ge0\)

<=>   \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge0\)

THỰC TẾ LÀ    \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}>0\)    nhé do    \(a;b;c>0\)     mà !!!!!!

Bình phương 2 vế BĐT , ta có :

\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge a+b+c\)

\(\Leftrightarrow a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\ge a+b+c\)

\(\Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ac}>0\left(\forall a,b,c>0\right)\)

=) ĐPCM