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a) \(-x^2+6x-16=-\left(x^2-6x+9\right)-7=-\left(x-3\right)^2-7< 0\)
b) \(-5x^2+20x-49=-5\left(x^2-4x+4\right)-29=-5\left(x-2\right)^2-29< 0\)
c) \(-1+x-x^2=-\left(x^2-x+1\right)=-\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{3}{4}=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}< 0\)
a. \(2x^2-4x+10=x^2-2x+1+x^2-2x+1+8=\left(x-1\right)^2+\left(x-1\right)^2+8=2\left(x-1\right)^2+8\)
Vì \(2\left(x-1\right)^2\ge0\Rightarrow2\left(x-1\right)^2+8\ge8\)
Vậy...
b. \(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy..
c. \(2x^2-6x+5=x^2-4x+4+x^2-2x+1=\left(x-2\right)^2+\left(x-1\right)^2\)
Vì \(\hept{\begin{cases}\left(x-2\right)^2\ge0\\\left(x-1\right)^2\ge0\end{cases}}\Rightarrow\left(x-2\right)^2+\left(x-1\right)^2\ge0\)
Vậy...
\(a,P=5x\left(2-x\right)-\left(x+1\right)\left(x+9\right)\)
\(=10x-5x^2-\left(x^2+x+9x+9\right)\)
\(=10x-5x^2-x^2-x-9x-9\)
\(=\left(10x-x-9x\right)+\left(-5x^2-x^2\right)-9\)
\(=-6x^2-9\)
Ta thấy: \(x^2\ge0\forall x\)
\(\Rightarrow-6x^2\le0\forall x\)
\(\Rightarrow-6x^2-9\le-9< 0\forall x\)
hay \(P\) luôn nhận giá trị âm với mọi giá trị của biến \(x\).
\(b,Q=3x^2+x\left(x-4y\right)-2x\left(6-2y\right)+12x+1\)
\(=3x^2+x^2-4xy-12x+4xy+12x+1\)
\(=\left(3x^2+x^2\right)+\left(-4xy+4xy\right)+\left(-12x+12x\right)+1\)
\(=4x^2+1\)
Ta thấy: \(x^2\ge0\forall x\)
\(\Rightarrow4x^2\ge0\forall x\)
\(\Rightarrow4x^2+1\ge1>0\forall x\)
hay \(Q\) luôn nhận giá trị dương với mọi giá trị của biến \(x\) và \(y\).
#\(Toru\)
a : x2 + 4x + 7 = (x + 2)2 + 3 > 0
b : 4x2 - 4x + 5 = (2x - 1)2 + 4 > 0
c : x2 + 2y2 + 2xy - 2y + 3 = (x + y)2 + (y - 1)2 + 2 > 0
d : 2x2 - 4x + 10 = 2(x - 1)2 + 8 > 0
e : x2 + x + 1 = (x + 0,5)2 + 0,75 > 0
f : 2x2 - 6x + 5 = 2(x - 1,5)2 + 0,5 > 0
a,\(-\left(x^2-3x+4\right)\)
\(-\left[\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\right]\)
\(\Leftrightarrow-\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\le-\frac{7}{4}\)(luôn âm)
b\(-2\left(x^2-5x+\frac{15}{2}\right)\)
\(-2\left[\left(x-\frac{5}{2}\right)^2+\frac{5}{4}\right]\)
\(-2\left(x-\frac{5}{4}\right)^2-\frac{5}{2}\le-\frac{5}{2}\)(luôn âm)
c,\(-\left[\left(4x^2-4x+1\right)+\left(2y^2-6y+5\right)\right]\)
\(=-\left[\left(2x-1\right)^2+2\left(y^2-3y+\frac{5}{2}\right)\right]\)
\(=-\left[\left(2x-1\right)^2+2\left(y-\frac{3}{2}\right)^2+\frac{1}{4}\right]\)
\(=-\left[\left(2x-1\right)^2+2\left(y-\frac{3}{2}\right)^2\right]-\frac{1}{4}\le-\frac{1}{4}\)(luôn âm)
\(4x^2-12x+11=\left(2x\right)^2-2.x.6+36-\) \(25\)
= \(\left(2x-6\right)^2-25>=-25\)
A đạt GTNN = -25 <=> \(\left(2x-6\right)^2=0\)
<=> \(x=3\)
các câu còn lại tương tự
TÌM GIÁ TRỊ NHỎ NHẤT, LỚN NHẤT CỦA BIỂU THỨC
\(a,A=4x^2-12x+11\)
\(A=4x^2-12x+9+2\)
\(A=\left(2x-3\right)^2+2\)
Nhận xét: \(\left(2x-3\right)^2\ge0\forall x\)
\(\Rightarrow\left(2x-3\right)^2+2\ge2\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(2x-3\right)^2=0\Rightarrow2x=3\Rightarrow x=\frac{3}{2}\)
Vậy \(minA=2\Leftrightarrow x=\frac{3}{2}\)
\(b,B=x^2-x+1\)
\(B=x^2-2x.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1\)
\(B=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}+1\)
\(B=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
Nhận xét: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\Rightarrow x=\frac{1}{2}\)
Vậy \(minB=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)
\(c,C=-x^2+6x-15\)
\(C=-\left(x^2-6x+15\right)\)
\(C=-\left(x^2-6x+4+11\right)\)
\(C=-\left[\left(x-2\right)^2+11\right]\)
\(C=-\left(x-2\right)^2-11\)
Nhận xét: \(-\left(x-2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-2\right)^2-11\le-11\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow-\left(x-2\right)^2=0\Rightarrow x=2\)
Vậy \(maxC=-11\Leftrightarrow x=2\)
\(d,D=\left(x-3\right)\left(1-x\right)-2\)
\(D=x-x^2-3+3x-2\)
\(D=-x^2+4x-5\)
\(D=-\left(x^2-4x+5\right)\)
\(D=-\left(x^2-4x+4+1\right)\)
\(D=-\left[\left(x-2\right)^2+1\right]\)
\(D=-\left(x-2\right)^2-1\)
Nhận xét: \(-\left(x-2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-2\right)^2-1\le-1\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow-\left(x-2\right)^2=0\Rightarrow x=2\)
Vậy \(maxD=-1\Leftrightarrow x=2\)
\(D=-x^2-y^2+2x+2y-3\)
\(D=-\left(x^2-2x+1\right)-\left(y^2-2y+1\right)-1\)
\(D=-\left(x-1\right)^2-\left(y-1\right)^2-1\)
Ta thấy \(-\left(x-1\right)^2< 0;-\left(y-1\right)^2< 0\forall x;y\). Mà -1 < 0
\(\Rightarrow-\left(x-1\right)^2-\left(y-1\right)^2-1< 0\forall x;y\)\(\Rightarrow D< 0\forall x;y\)(đpcm).
a) Đặt \(A=x^2+4x+7\)
\(A=\left(x^2+4x+4\right)+3\)
\(A=\left(x+2\right)^2+3\)
Mà \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow A\ge3>0\)
b) Đặt \(B=4x^2-4x+5\)
\(B=\left(4x^2-4x+1\right)+4\)
\(B=\left(2x-1\right)^2+4\)
Mà \(\left(2x-1\right)^2\ge0\forall x\)
\(\Rightarrow B\ge4>0\)
c) Đặt \(C=x^2+2y^2+2xy-2y+3\)
\(C=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)+2\)
\(C=\left(x+y\right)^2+\left(y-1\right)^2+2\)
Mà \(\left(x+y\right)^2\ge0\forall x;y\)
\(\left(y-1\right)^2\ge0\forall y\)
\(\Rightarrow C\ge2>0\)
\(a,-x^2+6x-16\)
\(=-x^2+3x+3x-9-5\)
\(=-x\left(x-3\right)+3\left(x-3\right)-5\)
\(=\left(3-x\right)\left(x-3\right)-5\)
\(=-\left(x-3\right)^2-5\le-5\)=>Luôn âm
\(c,-1+x-x^2\)
\(=-x^2+x-1\)
\(=-\left(x^2-x+\frac{1}{2}+\frac{1}{2}\right)\)
\(=-\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\le\frac{-1}{2}\)=>Luôn âm