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\(A=2x^2+5x-3=2\left(x^2+\frac{5}{2}x-\frac{2}{3}\right)\)
\(=2\left(x^2+2.\frac{5}{4}x+\frac{25}{16}-\frac{107}{48}\right)\)
\(=2\left[\left(x+\frac{5}{4}\right)^2-\frac{107}{48}\right]\)
\(=2\left[\left(x+\frac{5}{4}\right)^2\right]-\frac{107}{24}\ge\frac{-107}{24}\)
Vậy \(A_{min}=\frac{-107}{24}\Leftrightarrow x+\frac{5}{4}=0\Leftrightarrow x=-\frac{5}{4}\)
<=> \(^{x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10}\)
<=> \(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-10\)
<=> 12x - 4 = -10
<=> 12x =-6
<=> x= \(\frac{-6}{12}=\frac{-1}{2}\)
1/
a,\(A=x-x^2=-x^2+x=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\)
Vì \(-\left(x-\frac{1}{2}\right)^2\le0\Rightarrow A=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Dấu "=" xảy ra <=>x=1/2
Vậy Amax=1/4 khi x=1/2
b, \(B=2x-2x^2-5=-2x^2+2x-5\)
\(\Rightarrow2B=-4x^2+4x-10=-\left(4x^2-4x+1\right)-9=-\left(2x-1\right)^2-9\)
Vì \(-\left(2x-1\right)^2\le0\Rightarrow2B=-\left(2x-1\right)^2-9\le-9\Rightarrow B\le\frac{-9}{2}\)
Dấu "=" xảy ra <=>x=1/2
Vậy Bmax=-9/2 khi x=1/2
2/
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)
Bài 1 : hđt bạn tự làm nhé
Bài 2 :
\(\left(x-1\right)\left(x^2+x+1\right)-\left(x-4\right)^2x\)
\(=x^3-1-x\left(x^2-8x+16\right)=x^3-1-x^3+8x^2-16x\)
\(=8x^2-16x-1\)
\(\left(x+7\right)\left(x^2-7x+49\right)-\left(5-x\right)\left(5+x\right)\left(x-1\right)\)
\(=x^3+343-\left(25-x^2\right)\left(x-1\right)=x^3+343-\left(25x-25-x^3+x^2\right)\)
\(=x^3+343+x^3-x^2-25x+25=2x^3-x^2-25x+368\)
Đáp án :
x = 1 ; x = -1
# Hok tốt !
bị ngốc