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a)
\(A=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
Daaus = xayr ra khi: x = 2
b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)
Dấu = xảy ra khi x = 3
c) \(C=4x^2+2y^2-4xy-4y+1=\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3=\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)
Dấu = xảy ra khi
2x = y và y = 2
=> x = 1 và y = 2
a) A = \(-x^2+4x+3=-\left(x-2\right)^2+7\le7\)
Dấu "=" <=> x = 2
b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)
Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)
c) \(4x^2+2y^2-4xy-4y+1\)
= \(\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3\)
= \(\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)
Dấu "=" <=> \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
\(A=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow x=1\\ B=2\left(x^2-3x\right)=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ B=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ B_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\\ C=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\\ C_{max}=7\Leftrightarrow x=2\)
a,\(A=x^2-2x+5=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)
Dấu "=" \(\Leftrightarrow x=-1\)
b,\(B=2\left(x^2-3x\right)=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
Dấu "=" \(\Leftrightarrow x=\dfrac{3}{2}\)
c,\(=C=-\left(x^2-4x-3\right)=-\left[\left(x^2-4x+4\right)-7\right]=-\left(x-2\right)^2+7\le7\)
Dấu "=" \(\Leftrightarrow x=2\)
\(A=4x^2-4x+5\)
\(=\left(4x^2-4x+1\right)+4\)
\(=\left(2x-1\right)^2+4\ge4\)
\(\Rightarrow A\ge4\)
Dấu = khi \(\left(2x-1\right)^2=0\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\Leftrightarrow\)\(x=\frac{1}{2}\)
Vậy MinA=4 khi \(x=\frac{1}{2}\)
Ta có: 4x2-4x-9 = (4x2-4x+1)-10 = (2x-1)2-10 ≥ -10
Dấu "=" xảy ra ⇔ x=1/2
\(4x^2-4x-9=\left(2x-1\right)^2-10\)
Vì \(\left(2x-1\right)^2\ge0\Rightarrow\left(2x-1\right)^2-10\ge10\)
\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)
Ta có :
\(4x^2-4x+5\)
\(=\left(2x\right)^2-2.2x.1+1+4\)
\(=\left(2x-1\right)^2+4\)
Vì ( 2x - 1 ) ^ 2 \(\ge0\)
=> \(\left(2x-1\right)^2+4\ge4\)
Dấu " = " xảy ra khi x = 1 / 2
Vậy ..............
\(x^2-6x+11=x^2-2.3.x+9+2=\left(x-3\right)^2+2\ge2\)
dấu"=" xảy ra<=>x=3
\(4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-2.2x+4-7\right)\)
\(=-[\left(x-2\right)^2-7]\le7\) dấu"=" xay ra<=>x=2
a) Ta có: \(x^2-6x+11\)
\(=x^2-6x+9+2\)
\(=\left(x-3\right)^2+2\ge2\forall x\)
Dấu '=' xảy ra khi x=3
b) Ta có: \(-x^2+4x+3\)
\(=-\left(x^2-4x-3\right)\)
\(=-\left(x^2-4x+4-7\right)\)
\(=-\left(x-2\right)^2+7\le7\forall x\)
Dấu '=' xảy ra khi x=2
Bài 1:
a: \(M=x^2-10x+3\)
\(=x^2-10x+25-22\)
\(=\left(x^2-10x+25\right)-22\)
\(=\left(x-5\right)^2-22>=-22\forall x\)
Dấu '=' xảy ra khi x-5=0
=>x=5
b: \(N=x^2-x+2\)
\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi x-1/2=0
=>x=1/2
c: \(P=3x^2-12x\)
\(=3\left(x^2-4x\right)\)
\(=3\left(x^2-4x+4-4\right)\)
\(=3\left(x-2\right)^2-12>=-12\forall x\)
Dấu '=' xảy ra khi x-2=0
=>x=2
Ta có: \(E=4x^2+4x-5\)
\(=4x^2+4x+1-6\)
\(=\left(2x+1\right)^2-6\ge-6\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)
x có điều kiện gì ko bạn
\(2x^2-4x=2x^2-4x+2-2=2\left(x^2-2x+1\right)-2=2\left(x-1\right)^2-2\)
Vì \(2\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow2x^2-4x\ge-2\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)
Vậy GTNN của biểu thức là \(-2\)\(\Leftrightarrow x=1\)