\(A=6x-x^2+10\)

\(-A=x^2-6x+10\)

\(-A=\left(x^2-6x+9\right)+1\)

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

Mà  \(\left(x-3\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge1\Leftrightarrow A\le-1\)

Dấu "=" xảy ra khi :  \(x-3=0\Leftrightarrow x=3\)

Vậy  \(A_{Max}=-1\Leftrightarrow x=3\)

Bạn tự vẽ hình ik nha.

a) tg abd= tg aed: AB=AE; BAD=EAD; AD: cạnh chung

b) tg DBM= tg DEC: BẠN TỰ LÀM NHA DỄ LẮM 

Toán lớp 7 nha!

\(\frac{11}{13}-\left(\frac{5}{42}-x\right)=-\left(\frac{15}{28}-\frac{11}{13}\right)\)

<=> \(\frac{11}{13}-\frac{5}{42}+x=-\left(-\frac{113}{364}\right)\)

<=> \(\frac{397}{546}+x=\frac{113}{364}\)

<=> \(x=\frac{-5}{12}\)

học tốt

I love you

--học tốt--

 là :I love you 

\(A=4x^4+4x^2-3\)

\(A=\left[\left(2x^2\right)^2+2.2x^2.1+1^2\right]-4\)

\(A=\left(2x+1\right)^2-4\)

Ta có: \(\left(2x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(2x+1\right)^2-4\ge-4\forall x\)

\(A=-4\Leftrightarrow\left(2x+1\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy \(A_{min}=-4\Leftrightarrow x=-\frac{1}{2}\)

S = 2² + 4² + .........+ 20²

S = 2² . 1 + 2² . 2² + ..... + 2² . 10²

S = 2² . (1 + 2² + ... + 10²)

S = 4 . S

S = 4 . 485

S = 1540

S = 2² + 4² + 6² +......... +20²

 Trả lời : 

S = 2² + 4² + .........+ 20²

S = 2² . 1 + 2² . 2² + ..... + 2² . 10²

S = 2² . (1 + 2² + ... + 10²)

S = 4 . S

S = 4 . 485

S = 1940