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a, A = (x^2-3x)^2 - 1 >=-1
Dấu "=" xảy ra <=> x^2-3x = 0 <=>x.(x-3) = 0 <=> x=3 hoặc x=0
Vậy Min A = -1 <=> xz=3 hoặc x=0
b, Đề thiếu kìa bạn ơi
\(A=\left(y^2+2y\left(x+1\right)+\left(x+1\right)^2\right)+\left(2x^2-2x+2-\left(x+1\right)^2\right)\)
\(=\left(y+x+1\right)^2+\left(x-2\right)^2-3\ge-3\)
Min A=-3 khi x=2;y=-3
\(B=\left(x^2+x\left(y-3\right)+\frac{\left(y-3\right)^2}{4}\right)+\left(y^2-3y-\frac{\left(y-3\right)^2}{4}\right)\)
\(=\left(x+\frac{y-3}{2}\right)^2+\frac{3\left(y^2-2y+1\right)-12}{4}\)
\(=\left(....\right)^2+\frac{3}{4}\left(y-1\right)^2-3\ge3\)
Min B=-3 khi y=1;x=1
2)
\(A=2x^2+2x+y^2-2xy=x^2-2xy+y^2+x^2+2x+1-1\)
\(=\left(x-y\right)^2+\left(x+1\right)^2-1\ge-1\)
Dấu \(=\)khi \(\hept{\begin{cases}x-y=0\\x+1=0\end{cases}}\Leftrightarrow x=y=-1\).
Vậy GTNN của \(A\)là \(-1\)đạt tại \(x=y=-1\).
\(B=2a^2+b^2+c^2-ab+ac+bc\)
\(2B=4a^2+2b^2+2c^2-2ab+2ac+2bc\)
\(=a^2-2ab+b^2+a^2+2ac+c^2+b^2+2bc+c^2+2a^2\)
\(=\left(a-b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2+2a^2\ge0\)
Dấu \(=\)khi \(a=b=c=0\).
Vậy GTNN của \(B\)là \(0\)đạt tại \(a=b=c=0\).
1.
a) \(2x^2+2x+1=x^2+x^2+2x+1=x^2+\left(x+1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x+1=0\end{cases}}\)(vô nghiệm)
suy ra đpcm
b) \(x^2+y^2+2xy+2y+2x+2=\left(x+y\right)^2+2\left(x+y\right)+1+1=\left(x+y+1\right)^2+1>0\)
c) \(3x^2-2x+1+y^2-2xy+1=x^2-2xy+y^2+x^2-2x+1+x^2+1\)
\(=\left(x-y\right)^2+\left(x-1\right)^2+x^2+1>0\)
d) \(3x^2+y^2+10x-2xy+26=x^2-2xy+y^2+x^2+10x+25+x^2+1\)
\(=\left(x-y\right)^2+\left(x+5\right)^2+x^2+1>0\)
Câu a, b, c thì đơn giản òi. Câu d phải chú ý điểm rơi:v
d) Ta có: \(D=\left(x-\frac{1}{2}\right)^4+\frac{1}{2}\left(3x^2-3x+\frac{15}{8}\right)\)
\(=\left(x-\frac{1}{2}\right)^4+\frac{3}{2}\left(x-\frac{1}{2}\right)^2+\frac{9}{16}\ge\frac{9}{16}\)
Đẳng thức xảy ra khi x = 1/2
a)\(\left(3x+2\right)\left(2x-3\right)=6x^2-5x-6\)
b) viết lại đề nhin (dùng f(x) viết mới rõ ra dduocj) ko phải dùng {[(...)]} cho chuẩn vào
c) \(\left(x-2\right)^3-x^2.\left(x-6\right)=x^3-3.x.2\left(x-2\right)-8-x^3+6x^2\)
\(=x^3-6x^2+12x-8-x^3+6x^2=12x-8=4\Rightarrow x=1\)
a) \(3x^2-3y^2-x-y\)
\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)
\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(\Leftrightarrow3\left(x-y\right)\)
d) \(3x^2-7x+4\)
\(\Leftrightarrow3x^2-7x+7-3\)
\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)
\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)
e) \(-2x^2+3x-1\)
\(\Leftrightarrow\left(-2x^2-1^2\right)+3x\)
\(\Leftrightarrow\left(-2x-1\right)\left(-2x+1\right)+3x\)
f) \(x^2+2xy+y^2-2x-2y\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
k) \(2x^2+5x+3\)
\(\Leftrightarrow2x^2+2x+3x+3\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)
l) \(x^2-2x-y^2+1\)
\(\Leftrightarrow\left(x^2-2x+1\right)-y^2\)
\(\Leftrightarrow\left(x-1\right)^2-y^2\)
\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)
a) \(3x^2-3y^2-x-y\)
\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)
\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(\Leftrightarrow3\left(x-y\right)\)
d) \(3x^2-7x+4\)
\(\Leftrightarrow3x^2-7x+7-3\)
\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)
\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)
e) \(-2x^2+3x-1\)
\(\Leftrightarrow\left(-2x^2-1^2\right)+3x\)
\(\Leftrightarrow\left(-2x-1\right)\left(-2x+1\right)+3x\)
f) \(x^2+2xy+y^2-2x-2y\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
k) \(2x^2+5x+3\)
\(\Leftrightarrow2x^2+2x+3x+3\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)
l) \(x^2-2x-y^2+1\)
\(\Leftrightarrow\left(x^2-2x+1\right)-y^2\)
\(\Leftrightarrow\left(x-1\right)^2-y^2\)
\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)