\(C=-x^2+10x-5=-\left(x^2-10+5\right)\)

\(=-\left(x^2-10x+25-20\right)\)

\(=-\left[\left(x-5\right)^2-20\right]\)

\(=-\left(x-5\right)^2+20\le20\)

Vậy \(C_{max}=20\Leftrightarrow x-5=0\Leftrightarrow x=5\)

Ta có: \(x^2\ge0\)

\(\Leftrightarrow2x^2\ge0\)

\(\Leftrightarrow-2x^2\le0\)

\(\Leftrightarrow4-2x^2\le4\)

Vậy \(B_{max}=4\Leftrightarrow x=0\)

Có 4-2x²=-2x²+4

Vì x² > hoặc = 0

=)2x² > hoặc = 0

=)-2x² < hoặc = 0

=)-2x²+4 < hoặc = 4

Dấu '=' xảy ra khi -2x²+4=4

=)Bn lm nốt 

Mình ko chắc nx

Ta có : Q(x) = -(x+1)(x+2019) + 2020

                   = - (x²+2019x+x+2019) + 2020

                   = -x² - 2020x - 2019 +2020

                   = -x² - 2020x + 1

                   = - (x²+2020x + 1020100) + 1020101

                   = - (x+1010)²+1020101

Vì (x+1010)²  \(\ge\) 0 \(\forall x\) nên - (x+1010)² \(\le0\forall x\)

=>  - (x+1010)²+1020101 \(\le\)1020101 với mọi x

=> Q(x) \(\le\)1020101 với mọi x

Ta thấy Q(x) = 1020101 khi (x+1010)² = 0 => x+1010 = 0 => x = -1010

Vậy Q(x) đạt GTLN là 1020101 khi x = -1010

=\(\left(2-1\right)\left(2+1\right)\left(2^2-1\right)....\left(2^{20}-1\right)\) +1

=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{20}+1\right)+1\)

=\(\left(2^4-1\right)\left(2^4+1\right)....\left(2^{20}+1\right)+1\)

=.....

=\(\left(2^{20}-1\right)\left(2^{20}+1\right)+1\)

=\(2^{40}-1+1\)

=\(2^{40}\)

Chuc ban hoc tot

Sai rồi, nếu mũ là 32 thì bài này làm thế đc chứ mũ 20 thì ko làm như này được

\(F=\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)\)

\(=\left(x^2-7x+10\right)\left(x^2-7x-10\right)\)

Đặt \(x^2-7x=a\)

\(F=\left(a-10\right)\left(a+10\right)=a^2-100\ge-100\)

\(\Rightarrow F_{min}=-100\Leftrightarrow x^2-7x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=7\end{cases}}\)

\(a,25-x^4=5^2-\left(x^2\right)^2=\left(5-x^2\right)\left(5+x^2\right)\)

\(b,\left(3x+y\right)^2=\left(3x\right)^2+2.3x.y+y^2=9x^2+6xy+y^2\)

\(c,\left(x+1\right)^2=x^2+2x+1\)

\(d,\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)

\(e,\left(2+3x\right)^2=2^2+2.2.3x+\left(3x\right)^2=4+12x+9x^2\)

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

\(g,\left(x-1\right)^3=x^3-3.x^2.1+3.x.1^2-1^3=x^3-3x^2+3x-1\)

\(h,x^3-8=\left(x-2\right)\left(x^2-2x+4\right)\)

a) \(25-x^4\)

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

\(=\left(\sqrt{5}-x\right)\left(\sqrt{5}+x\right)\left(5+x^2\right)\)

đặt a+b-c=x;b+c=y;c+a-b=z có x+y+z=a+b+c

có:(x+y+z)^3-x^3-y^3-z^3=3(x+y)(y+z)(x+z)

tách đoạn đầu đi mình đánh mỏi tay lắm :v

=24abc nhé :v

<=> \(\frac{b+c-a}{2a}+1+\frac{a-b+c}{2b}+1+\frac{a+b-c}{2c}+1\ge\frac{3}{2}+3\)

<=> \(\frac{a+b+c}{2c}+\frac{a+b+c}{2b}+\frac{a+b+c}{2c}\ge\frac{9}{2}\)

<=> \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

<=> \(\frac{a}{a}+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{b}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+\frac{c}{c}\ge9\)

<=> \(\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge6\)

Ap dung bdt  \(\frac{a}{b}+\frac{b}{a}\ge2\)

Suy ra ve trai >= 2.3=6=ve phai

=> DPCM

Dau = xay ra <=> a=b=c

mik phai di hoc nen tra loi tat mong ban thong cam

cảm ơn nhiều ạ