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Áp dụng Bunyakovsky, ta có :
\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)
=> \(\left(x^2+y^2\right)\ge\frac{1}{2}\)
=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)
Mấy cái kia tương tự
\(A=\left(x^2+4y^2+1-4xy+2x-4y\right)+\left(x^2-4x+4\right)-3\)
\(A=\left(x-2y+1\right)^2+\left(x-2\right)^2-3\ge-3\)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(2;\dfrac{3}{2}\right)\)
\(M=\dfrac{1}{2}\left(4x^2+y^2+1-4xy+4x-2y\right)+\dfrac{9}{2}y^2+3y-\dfrac{1}{2}\)
\(M=\dfrac{1}{2}\left(2x-y+1\right)^2+\dfrac{9}{2}\left(y+\dfrac{1}{3}\right)^2-1\ge-1\)
\(M_{min}=-1\) khi \(\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
b) Ta có: \(B=x^2+2x+y^2-4y+6\)
\(=x^2+2x+1+y^2-4y+4+1\)
\(=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Vậy: \(B_{min}=1\) khi (x,y)=(-1;2)
c) Ta có: \(C=4x^2+4x+9y^2-6y-5\)
\(=4x^2+4x+1+9y^2-6y+1-7\)
\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(C_{min}=-7\) khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
\(A=2x^2+x=2\left(x^2+\dfrac{1}{2}x\right)=2\left(x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}-\dfrac{1}{16}\right)\)
\(=2\left[\left(x+\dfrac{1}{4}\right)^2-\dfrac{1}{16}\right]\ge-\dfrac{1}{8}\) dấu"=' xảy ra<=>x=\(-\dfrac{1}{4}\)
\(B=x^2+2x+y^2-4y+6\)
\(=x^2+2x+1+y^2-4y+4+1=\left(x+1\right)^2+\left(y-2\right)^2+1\)
\(\ge1\) dấu"=" xảy ra<=>x=-1;y=2
\(C=4x^2+4x+9y^2-6y-5\)
\(=4x^2+4x+1+9y^2-6y+1-7\)
\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)
dấu"=" xảy ra<=>x=\(-\dfrac{1}{2},y=\dfrac{1}{3}\)
\(D=\left(2+x\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)^2\)
=\(x^2+6x+8-\left(x-1\right)\left(x+3\right)^2\)
\(=\left(x+3\right)^2-1-\left(x-1\right)\left(x+3\right)^2\)
\(=\left(x+3\right)^2\left(2-x\right)-1\ge-1\)
dấu"=" xảy ra\(< =>\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
a: Ta có: \(A=2x^2-8x+1\)
\(=2\left(x^2-4x+\dfrac{1}{2}\right)\)
\(=2\left(x^2-4x+4-\dfrac{7}{2}\right)\)
\(=2\left(x-2\right)^2-7\ge-7\forall x\)
Dấu '=' xảy ra khi x=2
`A=2x^2-2xy-6x+y^2+10`
`A=x^2-2xy+y^2+x^2-6x+10`
`A=(x-y)^2+x^2-6x+9+1`
`A=(x-y)^2+(x-3)^2+1`
Vì `(x-y)^2+(x-3)^2>=0=>A>=1`
Dấu "=" xảy ra khi `{(x-y=0),(x-3=0):}<=>x=y=3`
a) Ta có: x - 1 ≠ 0 ⇒ x ≠ 1
x2 - 1 = (x + 1)(x - 1) ≠ 0 ⇔ x ≠ -1 và x ≠ 1
x2 - 2x + 1 = (x - 1)2 ≠ 0 ⇔ x - 1 ≠ 0 ⇔ x ≠ 1
ĐKXĐ: x ≠ -1 và x ≠ 1
\(A=2x^2+2x+1=x^2+x^2+x+x+\frac{1}{4}+\frac{1}{4}+\frac{1}{2}\)
\(A=\left(x^2+x+\frac{1}{4}\right)+\left(x^2+x+\frac{1}{4}\right)+\frac{1}{2}\)
\(A=\left(x+\frac{1}{2}\right)^2+\left(x+\frac{1}{2}\right)^2+\frac{1}{2}\)
\(A=2\left(x+\frac{1}{2}\right)^2+\frac{1}{2}\)
Vì \(2\left(x+\frac{1}{2}\right)^2\ge0\forall x\in R\)nên \(Min\left(A\right)=\frac{1}{2}\)
\(\Rightarrow2\left(x+\frac{1}{2}\right)^2=0\Rightarrow x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)
Vậy giá trị nhỏ nhất của \(A=\frac{1}{2}\equiv x=-\frac{1}{2}\)
\(\equiv\)là tại nhé
k cho minh nha