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\(4x^2+4x+2022=4x^2+4x+1+2021=\left(2x+1\right)^2+2021\ge2021\)
dấu "=" xảy ra \(< =>2x+1=0< =>x=\dfrac{-1}{2}\)
Đặt \(-6x^2+3x+3=0\)
\(\Leftrightarrow-6x^2+6x-3x+3=0\)
\(\Leftrightarrow-6x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) dễ tự làm
b) A(x) có bậc 6
hệ số: -1 ; 5 ; 6 ; 9 ; 4 ; 3
B(x) có bậc 6
hệ số: 2 ; -5 ; 3 ; 4 ; 7
c) bó tay
d) cx bó tay
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=x^2-3x+5\)
\(=x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\)
Vì \(\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\Rightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\)
\(\Rightarrow C\ge\dfrac{11}{4}\forall x\)
Dấu "=" xảy ra khi \(\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(MIN_C=\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{2}.\)
\(D=3x^2-6x-1\)
\(=3\left(x^2-3x-\dfrac{1}{3}\right)\)
\(=3\left(x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{31}{12}\right)\)
\(=3\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{31}{12}\right]\)
\(=3\left(x-\dfrac{3}{2}\right)^2-\dfrac{31}{4}\)
.......
Vậy \(MIN_D=\dfrac{-31}{4}\) khi \(x=\dfrac{3}{2}.\)
\(E=2x^2-6x\)
\(=2\left(x^2-3x\right)\)
\(=2\left[\left(x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}\right)-\dfrac{9}{4}\right]\)
\(=2\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\right]\)
\(=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)
.....
Vậy \(MIN_E=\dfrac{-9}{2}\) khi \(x=\dfrac{3}{2}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A\left(x\right)=-1+5^6-6x^2-5-9x^6+4x^4-3x^2\)
\(=-9x^6+4x^4-\left(3x^2+6x^2\right)+\left(5^6-1-5\right)\)
\(=-9x^6+4x^4-9x^2+\left(5^6-1-5\right)-15619\)
\(B\left(x\right)=2-5x^2+3x^4-4x^2+3x+x^4-4x^6-7x\)
\(=-4x^6+\left(3x^4+x^4\right)-\left(5x^2+4x^2\right)+\left(3x-7x\right)+2\)
\(=-4x^6+4x^4-9x^2-4x+2\)
\(A\left(x\right)-B\left(x\right)\)
\(=\left(-9x^6+4x^4-9x^2-15619\right)-\left(-4x^6+4x^4-9x^2-4x+2\right)\)
\(=-9x^6+4x^4-9x^2-15619+4x^6-4x^4+9x^2+4x-2\)
\(=-5x^6+4x-15621\)
Hình như C(x) vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4x^2+4x+6\)
\(=\left(2x\right)^2+2.2x.1+1+5\)
\(=\left(2x+1\right)^2+5\ge5\)
\(Min=5\Leftrightarrow2x+1=0\Rightarrow x=\frac{-1}{2}\)
\(x^2+6x+11\)
\(=x^2+2.x.3+9+2\)
\(=\left(x+3\right)^2+2\ge2\)
\(Min=2\Leftrightarrow x+3=0\Rightarrow x-3\)
\(x^2-3x+1\)
\(=x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{5}{4}\)
\(=\left(x+\frac{3}{2}\right)^2-\frac{5}{4}\le\frac{-5}{4}\)
\(MIn=\frac{-5}{4}\Leftrightarrow x+\frac{3}{2}=0\Rightarrow x=\frac{-3}{2}\)
B = 4x2 + 4x - 6 = (2x)2 + 2.2.x + 1 - 7 = (2x + 1)2 - 7 \(\ge\)-7
Vậy MinB = -7 khi 2x + 1 = 0 => x = -1/2
C = x2 + 6x + 11 = x2 + 2.3.x + 9 + 2 = (x + 3)2 + 2 \(\ge\)2
Vậy MinC = 2 khi x + 3 = 0 => x = -3
D = x2 - 3x + 1 \(=x^2-2.\frac{3}{2}.x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+1=\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)
Vậy MinD = -5/4 khi x - 3/2 = 0 => x = 3/2
Xe máy thứ nhất 1 giờ đi được 1/4 quảng đường
Xe máy thứ hai 1 giờ đi được 1/3 quảng đường
Sau 1,5 giờ 2 xe đi được:(1/4+1/3)x1,5=7/12x3/2=7/8(quảng đường)
quảng đường AB là:
15x8=120(km)