e) 16x <1284
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a: \(x^2-8x+21=x^2-8x+16+5=\left(x-4\right)^2+5>=5\)
Dấu '=' xảy ra khi x=4
b: \(16x^2+16x-30\)
\(=16x^2+2\cdot4x\cdot2+4-34\)
\(=\left(4x+2\right)^2-34>=-34\)
Dấu '=' xảy ra khi x=-1/2
d: \(-x^2+12x+34\)
\(=-\left(x^2-12x-34\right)\)
\(=-\left(x^2-12x+36-70\right)\)
\(=-\left(x-6\right)^2+70< =70\)
Dấu '=' xảy ra khi x=6
a) 2x3 + 3x - 5
= 2x3 - 2x + 5x - 5
= (2x3 - 2x) + (5x - 5)
= 2x(x2-1) + 5(x-1)
= 2x(x+1)(x-1) + 5(x-1)
= (2x2+2x)(x-1) + 5(x-1)
= (x-1)(2x2+2x+5)
b) 16x - 5x^2 - 3 = - 5^2 + 16x - 3 = -5^2+15x +x - 3 = 5x(3-x) - (3-x) = (3-x)(5x-1);
c) x^2 +4x +3 = x^2 +x + 3x +3 = x(x+1) +3(x+1) = (x+3)(x+1);
d) x^2 -4x -5 = x^2 + x - 5x - 5 = x(x+1) -5 (x+1) = (x-5)(x+1)
e) 16x - 5x^2 - 3 => tương tự câu b
\(E=-16x^2+3x-3=-\left(4x-\frac{3}{8}\right)^2-\frac{183}{64}\le\frac{-183}{64}\)
Vậy \(MaxE=\frac{-183}{64}\) khi \(x=\frac{3}{32}\)
Bạn xem lại đề phần \(F\) nhé.
\(G=-3x^2-9x+2=-3\left(x^2+3x-\frac{2}{3}\right)=-3[x^2+2x.\frac{3}{2}+\left(\frac{3}{2}\right)^2]+\frac{35}{4}\)
\(=-3\left(x+\frac{3}{2}\right)^2+\frac{35}{4}\le\frac{35}{4}\forall x\)
Vậy \(MaxG=\frac{35}{4}\) khi: \(\left(x+\frac{3}{2}\right)^2=0\Rightarrow x=\frac{-3}{2}\)
\(H=-7x^2+14x-3=-7\left(x^2-2x+\frac{3}{7}\right)=-7\left(x^2-2x+1\right)+4=-7\left(x-1\right)^2+4\le4\forall x\)
Vậy \(MaxH=4\) khi: \(\left(x-1\right)^2=0\Rightarrow x=1\)
1) \(16x\left(2-x\right)-\left(4x-5\right)^2=0\)
\(32x-16x^2-16x^2+40x-25=0\)
\(72x-16x^2-25=0\)
Đề sai ko bạn
2) \(\left(x-7\right)^2+3=\left(x-2\right)\left(x+2\right)\)
\(\left(x^2-14x+7\right)+3-\left(x-2\right)\left(x+2\right)=0\)
\(x^2-14x+7+3-x^2+4=0\)
\(-14x+14=0\)
\(x=1\)
3) \(\left(2x-3\right)^2-\left(7x-2x\right)^2=2\)
\(\left(2x-3\right)^2-\left(5x\right)^2=2\)
\(\left(2x-3-5x\right)\left(2x-3+5x\right)=2\)
\(\left(-3x-3\right)\left(7x-3\right)=2\)
=> lập bảng tìm x
4) \(\left(5x-7\right)^2-\left(1-3x\right)^2=16x\left(x-3\right)\)
\(25x^2-70x+49-9x^2+6x-1-16x^2+48x=0\)
\(-16x+48=0\)
\(x=3\)
a) -5x2+x+15x-3 = \(-5x\left(x-\frac{1}{5}\right)+15\left(x-\frac{1}{5}\right)\)=(3-x)(5x-1)
b)x2+x-6x-6 = x(x+1)-6(x+1) = (x-6)(x+1)
c) x2-x-6x+6 = x(x-1)-6(x-1) = (x-6)(x-1)
a: \(A=-3\left(x^2-2x+\dfrac{2}{3}\right)\)
\(=-3\left(x^2-2x+1-\dfrac{1}{3}\right)\)
\(=-3\left(x-1\right)^2+1< =1\)
Dấu '=' xảy ra khi x=1
b: \(B=-\left(16x^2+8x-4\right)\)
\(=-\left(16x^2+8x+1-5\right)\)
\(=-\left(4x+1\right)^2+5< =5\)
Dấu '=' xảy ra khi x=-1/4
d: \(x^2+2x+3=\left(x+1\right)^2+2>=2\)
=>E<=1/2
Dấu '=' xảy ra khi x=-1
`16^{x} < 128^{4}`
`(2^{4})^{x} < (2^{7})^{4}`
`2^{4.x} < 2^{28}`
`4.x < 28`
`x < 28:4`
`x < 7`
đề ??