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Ta có: \(7x^2+8xy+7y^2=10\)
\(\Rightarrow4x^2+8xy+4y^2+3x^2+3y^2=10\)
\(\Rightarrow4\left(x+y\right)^2+3\left(x^2+y^2\right)=10\)
\(\Rightarrow3\left(x^2+y^2\right)=10-4\left(x+y\right)^2\)
\(\Rightarrow S_{Max}=x^2+y^2=\dfrac{10-4\left(x+y\right)^2}{3}\le\dfrac{10}{3}\)
Đẳng thức xảy ra khi \(x=-y\)
Ta có: \(x^2+y^2\ge2xy\forall x,y\) đẳng thức xảy ra khi \(x=y\)
Thay vào \(7x^2+8xy+7y^2=10\) ta có:
\(7x^2+8x^2+7x^2=10\)
\(\Rightarrow22x^2=10\Rightarrow x^2=\dfrac{10}{22}\Rightarrow y^2=\dfrac{10}{22}\)
Khi đó \(S_{Min}=\dfrac{10}{22}+\dfrac{10}{22}=\dfrac{10}{11}\)
Đẳng thức xảy ra khi \(x=y\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(7x^2+8xy+7y^2=10\)
\(\Rightarrow\left(x^2+2xy+y^2\right)+6\left(x^2+y^2\right)=10\)
\(\Rightarrow6\left(x^2+y^2\right)=10-\left(x+y\right)^2\)
\(\Rightarrow x^2+y^2=\frac{10-\left(x+y\right)^2}{6}=\frac{5}{3}-\frac{\left(x+y\right)^2}{6}\)
Vì \(\left(x+y\right)^2\ge0\forall x,y\)\(\Rightarrow\frac{\left(x+y\right)^2}{6}\ge0\)
\(\Rightarrow x^2+y^2\le\frac{5}{3}\)
Dấu \("="\)xảy ra \(\Leftrightarrow\left(x+y\right)^2=0\)
\(\Leftrightarrow x+y=0\)
\(\Leftrightarrow x=-y\)
\(\Leftrightarrow7x^2-8x^2+7x^2=10\)
\(\Leftrightarrow6x^2=10\)
\(\Leftrightarrow x^2=\frac{5}{3}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{5}{3}\end{cases}}\)
hoặc \(\hept{\begin{cases}x=-\frac{5}{3}\\y=\frac{5}{3}\end{cases}}\)
Ta dễ dàng chứng minh được : \(2xy\le x^2+y^2\forall x,y\)
\(\Rightarrow8xy\le4\left(x^2+y^2\right)\)
Ta có :\(7x^2+8xy+7y^2=7\left(x^2+y^2\right)+8xy=10\)
\(\Rightarrow7\left(x^2+y^2\right)=10-8xy\ge10-4\left(x^2+y^2\right)\)
\(\Rightarrow11\left(x^2+y^2\right)\ge10\)
\(\Rightarrow x^2+y^2\ge\frac{10}{11}\)
Dấu \("="\)xảy ra \(\Leftrightarrow x=y\)
\(\Leftrightarrow7x^2+8x^2+7x^2=10\)
\(\Leftrightarrow22x^2=10\)
\(\Leftrightarrow x^2=\frac{5}{11}\)
\(\Leftrightarrow\orbr{\begin{cases}x=y=\sqrt{\frac{5}{11}}\\x=y=-\sqrt{\frac{5}{11}}\end{cases}}\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
a) =(5x)^2-2*5x+1+3
=(5x-1)^2+3
suy ra min=3
b) = -(x^2-2x+1)-1
=-(x^2-1)^2-1
suy ra Max=-1
c)=(x^2-2x+1)+(y^2-4y+4)+1
=(x^2-1)^2+(y^2-2)^2+1
suy ra Min=1
# mk ko chắc lắm đâu
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(A=x^2-6x+11\)
\(=x^2-6x+9+2\)
\(=\left(x^2-6x+9\right)+2\)
\(=\left(x-3\right)^2+2\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-3\right)^2+2\ge2\forall x\)
Dấu '=' xảy ra khi
\(\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: GTNN của đa thức \(A=x^2-6x+11\) là 2 khi x=3
b) Ta có: \(B=x^2-4x+3\)
\(=x^2-4x+4-1\)
\(=\left(x^2-4x+4\right)-1\)
\(=\left(x-2\right)^2-1\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-2\right)^2-1\ge-1\forall x\)
Dấu '=' xảy ra khi
\(\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy: GTNN của đa thức \(B=x^2-4x+3\) là -1 khi x=2
c) Ta có: \(C=x^2+5x\)
\(=x^2+2\cdot x\cdot\frac{5}{2}+\frac{25}{4}-\frac{25}{4}\)
\(=\left(x^2+2\cdot x\cdot\frac{5}{2}+\frac{25}{4}\right)-\frac{25}{4}\)
\(=\left(x+\frac{5}{2}\right)^2-\frac{25}{4}\)
Ta có: \(\left(x+\frac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{5}{2}\right)^2-\frac{25}{4}\ge\frac{-25}{4}\forall x\)
Dấu '=' xảy ra khi
\(\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=\frac{-5}{2}\)
Vậy: GTNN của đa thức \(C=x^2+5x\) là \(\frac{-25}{4}\) khi \(x=\frac{-5}{2}\)
d) Ta có: \(D=x^2+x+1\)
\(=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
Dấu '=' xảy ra khi
\(\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=\frac{-1}{2}\)
Vậy: GTNN của đa thức \(D=x^2+x+1\) là \(\frac{3}{4}\) khi \(x=\frac{-1}{2}\)
e) Ta có: \(E=4x^2+4x-2\)
\(=\left(2x\right)^2+2\cdot2x\cdot1+1-3\)
\(=\left[\left(2x\right)^2+2\cdot2x\cdot1+1\right]-3\)
\(=\left(2x+1\right)^2-3\)
Ta có: \(\left(2x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(2x+1\right)^2-3\ge-3\forall x\)
Dấu '='xảy ra khi
\(\left(2x+1\right)^2=0\Leftrightarrow2x+1=0\Leftrightarrow2x=-1\Leftrightarrow x=\frac{-1}{2}\)
Vậy: GTNN của đa thức \(E=4x^2+4x-2\) là -3 khi \(x=\frac{-1}{2}\)
g) Ta có: \(G=x^2-7x\)
\(=x^2-2\cdot x\cdot\frac{7}{2}+\frac{49}{14}-\frac{49}{14}\)
\(=\left(x^2-2\cdot x\cdot\frac{7}{2}+\frac{49}{4}\right)-\frac{49}{4}\)
\(=\left(x-\frac{7}{2}\right)^2-\frac{49}{4}\)
Ta có: \(\left(x-\frac{7}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{7}{2}\right)^2-\frac{49}{4}\ge\frac{-49}{4}\forall x\)
Dấu '=' xảy ra khi
\(\left(x-\frac{7}{2}\right)^2=0\Leftrightarrow x-\frac{7}{2}=0\Leftrightarrow x=\frac{7}{2}\)
Vậy: GTNN của đa thức \(G=x^2-7x\) là \(\frac{-49}{4}\) khi \(x=\frac{7}{2}\)
\(A=x^2-6x+11\)
\(A=x^2-2.x.3+3^2-3^2+11\)
\(A=\left(x^2-6x+3^2\right)-3^2+11\)
\(A=\left(x-3\right)^2+2\)
Vì \(\left(x-3\right)^2\ge0\forall x\)
=>\(\left(x-3\right)^2\ge0\ge2\forall x\)
Min A = 2 khi \(\left(x-3\right)^2=0\)
=> \(x-3=0hayx=3\)
Vậy Min A = 2 khi x = 3
\(B=x^2-4x+3\)
\(B=x^2-2.x.2+2^2-2^2+3\)
\(B=\left(x^2-4x+2^2\right)-4+3\)
\(B=\left(x-2\right)^2-1\)
=> \(\left(x-2\right)^2-1\ge0\forall x\)
MIn B = -1 khi \(\left(x-2\right)^2=0\)
=>\(\left(x-2\right)=0hayx=2\)
Vậy Min B = -1 khi x= 2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
7x2+8xy+7y2=0 (*)
=>4x2+8xy+4y2+3x2+3y2=0
=>4(x+y)2+3(x2+y2)=0
=>3(x2+y2)=0-4(x+y)2
=>x2 + y2 =0-4(x+y)2/3
Vậy A lớn nhất khi (x+y)2=0=>x=-y
=>Amax=0/3
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(A=\left(x+y\right)^2+4xy=x^2+2xy+y^2+4xy=x^2+6xy+y^2\)
2) \(B=\left(6x-2\right)^2+4\left(3x-1\right)\left(2+y\right)+\left(y+2\right)^2\)
\(=\left(6x-2\right)^2+2\left(6x-2\right)\left(y+2\right)+\left(y+2\right)^2\)
\(=\left(6x-2+y+2\right)^2=\left(6x+y\right)^2=36x^2+12xy+y^2\)
3) \(C=\left(x-y\right)^2+2\left(x^2-y^2\right)+\left(x+y\right)^2\)
\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x-y+x+y\right)^2=\left(2x\right)^2=4x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: 7x2+8xy+7y2=10 (*)
=>4x2+8xy+4y2+3x2+3y2=10
=>4(x+y)2+3(x2+y2)=10
=>3(x2+y2)=10-4(x+y)2
Vậy A lớn nhất khi (x+y)2=0=>x=-y
Amax=10/3
Áp dụng bất đẳng thức Cosy cho 2 số dương ta có:
A=x2+y22xy,
=> Amin khi x=y
Thay vào (*) ta được:
7x2+8x2+7x2=10
=>22x2=10
=>x2=10/22
=> y2=10/22
=>Amin=10/22+10/22=10/11.
Vậy Amin=10/3<=> x=-y
Amax=10/11<=>x=y.
k cho mình nhé mọi người