Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
D = 4x2 - 4x
D = (2x)2 - 2 . 2x . 1 + 12 - 12
D = (2x - 1)2 - 1
(2x - 1)2 lớn hơn hoặc bằng 0
(2x - 1)2 - 1 lớn hoặc bằng 1
Vậy Min D = -1 khi 2x - 1 = 0 <=> x = 1/2
Ta có: \(\left(2x+5\right)\left(2x-7\right)-\left(-4x-3\right)^2=16\)
\(\Leftrightarrow4x^2-14x+10x-35-\left[-\left(4x+3\right)\right]^2=16\)
\(\Leftrightarrow4x^2-4x-35-\left(-1\right)^2\cdot\left(4x+3\right)^2-16=0\)
\(\Leftrightarrow4x^2-4x-35-\left(4x+3\right)^2-16=0\)
\(\Leftrightarrow4x^2-4x-35-\left(16x^2+24x+9\right)-16=0\)
\(\Leftrightarrow4x^2-4x-35-16x^2-24x-9-16=0\)
\(\Leftrightarrow-12x^2-28x-60=0\)
\(\Leftrightarrow-12\left(x^2+\frac{28}{12}x+5\right)=0\)
\(\Leftrightarrow x^2+\frac{28}{12}x+5=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\frac{7}{6}+\frac{49}{36}+\frac{131}{36}=0\)
\(\Leftrightarrow\left(x+\frac{7}{6}\right)^2+\frac{131}{36}=0\)(vô lý)
Vậy: \(S=\varnothing\)
a) A = 5x2 - 20x + 2020 = 5(x2 - 4x + 4) + 2000 = 5(x - 2)2 + 2000 \(\ge\)2000 \(\forall\)x
Dấu "=" xảy ra <=> x - 2 = 0 <=> x = 2
Vậy MinA = 2000 khi x = 2+
b) B = -3x2 - 6x + 15 = -3(x2 + 2x + 1) + 18 = -3(x + 1)2 + 18 \(\le\)18 \(\forall\)x
Dấu "=" xảy ra <=> x + 1 = 0 <=> x = -1
Vậy MaxB = 18 khi x = -1
c) C = 9x2 + 2x + 7 = (9x2 + 2x + 1/9) + 62/9 = (3x + 1/3)2 + 62/9 \(\ge\)62/9 \(\forall\)x
Dấu "=" xảy ra <=> 3x + 1/3 = 0 <=> x = -1/9
Vậy MinC = 62/9 khi x = -1/9
d) D = 16 - 2x2 - 8x = -2(x2 + 4x + 4) + 24 = -2(x + 2)2 + 24 \(\le\) 24 \(\forall\)x
Dấu "=" xảy ra <=> x + 2 = 0 <=> x = -2
Vậy MaxD = 24 khi x = -2
\(A=x^2+4x+100\)
\(A=x^2+2.x.2+2^2+96\)
\(A=\left(x+2\right)^2+96\)
\(\left(x+2\right)^2+96\le0\)
\(\left(x+2\right)^2+96\le96\)
\(\Leftrightarrow A\le96\)
\(A_{min}\Leftrightarrow A=10\)
Dấu "=" xảy ra : \(\left(x+2\right)^20\)
\(x+2=0\)
\(x=-2\)
\(A=1\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(=...\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(=2^{64}-1-2^{64}=-1\)
\(M=x^2-2xy+4y^2+12xy+22\)
\(M=\left(x^2-2xy+y^2\right)+\left(3y^2+12y+12\right)+10\)
\(M=\left(x-y\right)^2+3\left(x+2\right)^2+10\ge10\forall x;y\)
Dấu " = " xảy ra \(\Leftrightarrow x=y=-2\)
( Chỗ \(M=\left(x-y\right)^2+3\left(x+2\right)^2+10\ge10\forall x;y\) bạn phân tích từng cái đã nhá, mình làm tắt )
M=(x2+6x+9)-10
=(x+3)2-10≥-10
Dấu = khi x+3=0 khi x=-3
Vậy GTNN của M =-10 khi x=-3