Chọn A

A

A nha!!!

a) \(x^2+10x=0\)

\(x\left(x+10\right)=0\)

\(\left[{}\begin{matrix}x=0\\x=-10\end{matrix}\right.\)

b) \(\left(x-7\right)^3=x-7\)

\(\left(x-7\right)^3-\left(x-7\right)=0\)

\(\left(x-7\right)\left[\left(x-7\right)^2-1\right]=0\)

\(\left(x-7\right)\left(x-7-1\right)\left(x-7+1\right)=0\)

\(\left(x-7\right)\left(x-8\right)\left(x-6\right)=0\)

\(\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)

c) \(x^2-20x+100=0\)

\(x^2-10x-10x+100=0\)

\(x\left(x-10\right)-10\left(x-10\right)=0\)

\(\left(x-10\right)\left(x-10\right)=0\)

\(\left(x-10\right)^2=0\)

=> x = 10

a) \(x^2+10x=0\)

\(\Leftrightarrow x\left(x+10\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+10=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-10\end{matrix}\right.\)

Vậy..

b) \(\left(x-7\right)^3=\left(x-7\right)\)

\(\Leftrightarrow\left(x-7\right)^3-\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left[\left(x-7\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-8\right)\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-7=0\\x-8=0\\x-6=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)

Vậy..

c) \(x^2-20x+100=0\)

\(\Leftrightarrow\left(x-10\right)^2=0\)

\(\Rightarrow x-10=0\)

\(\Rightarrow x=10\)

a, \(x^2+10x=0\)

\(\Leftrightarrow x\left(x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-10\end{matrix}\right.\)

Vậy x = 0 hoặc x = -10

b, \(\left(x-7\right)^3=\left(x-7\right)\)

\(\Leftrightarrow\left(x-7\right)^3-\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left[\left(x-7\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\\left(x-7\right)^2-1=0\end{matrix}\right.\)

+) \(x-7=0\Leftrightarrow x=7\)

+) \(\left(x-7\right)^2-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=6\end{matrix}\right.\)

Vậy x = 7 hoặc x = 8 hoặc x = 6

c, \(x^2-20x+100=0\)

\(\Leftrightarrow\left(x-10\right)^2=0\)

\(\Leftrightarrow x=10\)

Vậy x = 10

A = x² - 20x + 100 = x² - 2.x.10 + 10² = ( x - 10 )² = ( 20 - 10 )² = 10² = 100

B = ( x + 2 )² + 2( x + 2 )( 3x + 5 ) + ( 3x + 5 )² ( vầy thôi nhỉ ? )

= [ ( x + 2 ) + ( 3x + 5 ) ]²

= ( x + 2 + 3x + 5 )²

= ( 4x + 7 )²

= \(\left[4\cdot\left(-\frac{1}{4}\right)+7\right]^2\)

= ( -1 + 7 )² = 6² = 36

Để  \(B\)lớn nhất thì  \(\frac{1}{B}\) nhỏ nhất 

Ta có:   \(\frac{1}{B}=\frac{x^2+20x+100}{x}=x+\frac{100}{x}+20\)

Áp dụng BĐT Cô-si ta có:   \(\frac{1}{B}=x+\frac{100}{x}+20\ge2\sqrt{x.\frac{100}{x}}+20=2.\sqrt{100}+20=40\)

Dấu :'=" xảy ra   \(\Leftrightarrow\)\(x=\frac{100}{x}\)\(\Leftrightarrow\)\(x=10\)

Min  \(\frac{1}{B}=40\)  \(\Rightarrow\) Max  \(B=\frac{1}{40}\) \(\Leftrightarrow\)\(x=10\)

P/s:  tham khảo nhé, nếu có sai đâu m.n chỉ  mk nhé  (yếu nhất cực trị)

a) x*(1+2+3+...+20)=2100

                   x*210 =2100

                   x        =10

dấu * là nhân hả bạn

đúng rồi bạn BÌNH NGUYỄN

\(C=16x^2-8x+2024\)

\(\Rightarrow C=16x^2-8x+1+2023\)

\(\Rightarrow C=\left(4x-1\right)^2+2023\ge2023\left(\left(4x-1\right)^2\ge0\right)\)

\(\Rightarrow Min\left(C\right)=2023\)

\(D=-25x^2+50x-2023\)

\(\Rightarrow D=-\left(25x^2-50x+25\right)-1998\)

\(\Rightarrow D=-\left(5x-5\right)^2-1998\le1998\left(-\left(5x-5\right)^2\le0\right)\)

\(\Rightarrow Max\left(D\right)=1998\)

\(B=-x^2+20x+100=-\left(x^2-20x+100\right)+200=-\left(x-10\right)^2+200\le200\left(-\left(x-10\right)^2\le0\right)\)

\(\Rightarrow Max\left(B\right)=200\)

\(E=\left(2x-1\right)^2-\left(3x+2\right)\left(x-5\right)\)

\(\Rightarrow E=4x^2-4x+1-\left(3x^2-13x-10\right)\)

\(\Rightarrow E=4x^2-4x+1-3x^2+13x+10\)

\(\Rightarrow E=x^2+9x+11=x^2+9x+\dfrac{81}{4}-\dfrac{81}{4}+11\)

\(\Rightarrow E=\left(x+\dfrac{9}{2}\right)^2-\dfrac{37}{4}\ge-\dfrac{37}{4}\left(\left(x+\dfrac{9}{2}\right)^2\ge0\right)\)

\(\Rightarrow Min\left(E\right)=-\dfrac{37}{4}\)

\(F=\left(3x-5\right)^2-\left(3x+2\right)\left(4x-1\right)\)

\(\Rightarrow F=9x^2-30x+25-\left(12x^2+3x-2\right)\)

\(\Rightarrow F=-3x^2-33x+27=-3\left(x^2-10x+9\right)\)

\(\Rightarrow F=-3\left(x^2-10x+25\right)+48=-3\left(x-5\right)^2+48\le48\left(-3\left(x-5\right)^2\le0\right)\)

\(\Rightarrow Max\left(F\right)=48\)

đề của bạn là (x+10)² à

Nếu đề là (x+10)² thì đáp án là D nhé

nhớ like nha

D

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a) pt a <=> 3x+1=0     hoặc        x-2000=0         hoặc      3x+6000=0

           <=> x=-1/3      hoặc        x=2000            hoặc      x=-2000