bài 1 Tìm GTNN của biểu thức
A= 25x2+3y2-10x+11
B= (x-3)2+(x-11)2
C= (x+1).(x-2).(x-3).(x-6)
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BÀi 1
D = 4x - 10 - x2= - (x2 - 4x +10) = - (x - 2 )2 - 6
Vì - (x - 2 )2 \(\le0\)nên - (x - 2 )2 - 6 \(\le-6< 0\)
Vậy D = 4x - 10 - x2 luôn âm (dpcm)
\(a,\\ A=25x^2-10x+11\\ =\left(5x\right)^2-2.5x.1+1^2+10\\ =\left(5x+1\right)^2+10\ge10\forall x\in R\\ Vậy:min_A=10.khi.5x+1=0\Leftrightarrow x=-\dfrac{1}{5}\\ B=\left(x-3\right)^2+\left(11-x\right)^2\\ =\left(x^2-6x+9\right)+\left(121-22x+x^2\right)\\ =x^2+x^2-6x-22x+9+121=2x^2-28x+130\\ =2\left(x^2-14x+49\right)+32\\ =2\left(x-7\right)^2+32\\ Vì:2\left(x-7\right)^2\ge0\forall x\in R\\ Nên:2\left(x-7\right)^2+32\ge32\forall x\in R\\ Vậy:min_B=32.khi.\left(x-7\right)=0\Leftrightarrow x=7\\Tương.tự.cho.biểu.thức.C\)
b:
\(D=-25x^2+10x-1-10\)
\(=-\left(25x^2-10x+1\right)-10\)
\(=-\left(5x-1\right)^2-10< =-10\)
Dấu = xảy ra khi x=1/5
\(E=-9x^2-6x-1+20\)
\(=-\left(9x^2+6x+1\right)+20\)
\(=-\left(3x+1\right)^2+20< =20\)
Dấu = xảy ra khi x=-1/3
\(F=-x^2+2x-1+1\)
\(=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1< =1\)
Dấu = xảy ra khi x=1
A=-(11+5x+5x+x^2)
A=-((5x+x^2)+(5x+20)-9)
A=-(x(5+x)+5(5+x)-9)
A=-((5+x)^2-9)
A=(5+x)^2+9
Max A= 9 Khi x=-5
Theo bài ra ta có:
|x+\(\frac{1}{2}\)|\(\ge\)0
|x+\(\frac{1}{6}\)|\(\ge\)0
............................
|x+\(\frac{1}{110}\)|\(\ge\)0
\(\Rightarrow\)|x+\(\frac{1}{2}\)|+|x+\(\frac{1}{6}\)|+...+|x+\(\frac{1}{110}\)|\(\ge\)0
\(\Rightarrow\)11.x\(\ge\)0
\(\Rightarrow\)x\(\ge\)0
\(\Rightarrow\)x dương.
Khi đó:|x+\(\frac{1}{2}\)|+|x+\(\frac{1}{6}\)|+...+|x+\(\frac{1}{110}\)|=11.x
\(\Rightarrow\)x+\(\frac{1}{2}\)+x+\(\frac{1}{6}\)+...+x+\(\frac{1}{110}\)=11.x
\(\Rightarrow\)27.x+\(\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)=11x
\(\Rightarrow\)\(\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)=-16x
\(\Rightarrow\)\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)=-16x
\(\Rightarrow\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)=-16x
\(\Rightarrow\)\(\frac{10}{11}\)=-16x
\(\Rightarrow\)\(\frac{10}{-176}=x\)
Vậy \(x=\frac{10}{-176}\).
bài4 tìm x
a (x+7)^2=7
b (x- căn 2)^2=2
c (x+ căn 7)^2=7
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C=[(x+1)(x-6)][(x-2)(x-3)]
=(x2-5x-6)(x2-5x+6)
=(x2-5x)2-36>=-36
GTNN cua C=-36 tai x2-5x=0=>x(x-5)=0=>x=0 hoac x=5
B=(x-3)2+(x-11)2
=x2-6x+9+x2-22x+121
=2x2-28x+130
=2(x2-14x+65)
=2(x2-2.7x+72-72+65)
=2[(x-7)2-49+65]
=2(x-7)2+32
=> vì 2(x-7)2 >= 0
=>2(x-7)2+32 >= 32
=> GTNN của B=32. Khi x=7