a) \(A=\left(x+1\right)\left(2x-1\right)\)

\(A=2x^2+x-1\)

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

\(A=2\left[x^2+2\cdot x\cdot\frac{1}{4}+\left(\frac{1}{4}\right)^2-\frac{9}{16}\right]\)

\(A=2\left[\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\right]\)

\(A=2\left(x+\frac{1}{4}\right)^2-\frac{9}{8}\ge\frac{-9}{8}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x+\frac{1}{4}=0\Leftrightarrow x=\frac{-1}{4}\)

Vậy A_min = -9/8 khi và chỉ khi x = -1/4

b) \(B=4x^2-4xy+2y^2+1\)

\(B=\left(2x\right)^2-2\cdot2x\cdot y+y^2+y^2+1\)

\(B=\left(2x-y\right)^2+y^2+1\ge1\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2x-y=0\\y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=0\end{cases}\Rightarrow}}x=y=0\)

Vậy B_min = 1 khi và chỉ khi x = y = 0

Câu hỏi:

17+9+2005=?

Trả lời

17+9+2005=2031

bn sinh ngày 17/9/2005 à

#Koo#

Hello,hello,hello ,hello 

\(\text{Ta có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0.\)

\(\Leftrightarrow bc+ac+ab=0\Rightarrow\hept{\begin{cases}bc=-ac-ab\\ac=-bc-ab\\ab=-bc-ac\end{cases}}\)

\(\Rightarrow BT\text{hức}=\frac{bc}{a^2+2bc}+\frac{ac}{b^2+2ac}+\frac{ab}{c^2+2ab}\)

\(=\frac{bc}{a^2-ac-ab+bc}+\frac{ac}{b^2-bc-ab+ac}+\frac{ab}{c^2-bc-ac+ab}\)

\(=\frac{bc}{a\left(a-b\right)-c\left(a-b\right)}+\frac{ac}{b\left(b-a\right)-c\left(b-a\right)}+\frac{ab}{c\left(c-a\right)-b\left(c-a\right)}\)

\(=\frac{bc}{\left(a-c\right)\left(a-b\right)}-\frac{ac}{\left(b-c\right)\left(a-b\right)}+\frac{ab}{\left(a-c\right)\left(b-c\right)}\)

\(=\frac{bc\left(b-c\right)-ac\left(a-c\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{b^2c-bc^2-a^2c+ac^2+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{c\left(b^2-a^2\right)-c^2\left(b-a\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+ab\left(a+b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left(c^2-ac-bc+ab\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{c\left(c-b\right)-a\left(c-b\right)}{\left(b-c\right)\left(a-c\right)}=\frac{\left(a-c\right)\left(b-c\right)}{....}=1\)

Lâu ko lm đổi dấu hơi thừa ra!! ko hiểu chỗ nào thì ib mk giải thích cho

\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right).\)

\(\Rightarrow\left(2x^2+x-2013\right)^2-4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)+4\left(x^2-5x-2012\right)^2=0\)

\(\Leftrightarrow\left[\left(2x^2+x-2013\right)-2\left(x^2-5x-2012\right)\right]^2=0\)(Hằng đẳng thức)

\(\Leftrightarrow2x^2+x-2013-2x^2+10x+4024=0\)

\(\Leftrightarrow11x=-2011\)

\(\Leftrightarrow x=\frac{-2011}{11}\)

quy đồng nha bạn