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mình sẽ giải câu 3 cho bạn nhé
đề bài=> \(\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}+\frac{1}{x^2+6x+7x+42}=\frac{1}{18}\)
\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-...-\frac{1}{x+7}=\frac{1}{18}\)
\(\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(18\left(x+7\right)-18\left(x+4\right)=\left(x+7\right)\left(x+4\right)\)
\(\left(x+13\right)\left(x-2\right)=0\)
\(\orbr{\begin{cases}x=-13\\x=2\end{cases}}\)
nhớ thank mk nhé
câu 5 nà
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
<=>\(1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\ge9\)
<=>\(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge9\)
<=>\(3+2+2+2\ge9\)(bất đẳng thức luôn đúng)
=> điều phải chứng minh
\(9x^2-6x+2=9x^2-6x+1+1=\left(3x-1\right)^2+1>0\Rightarrowđpcm\)
\(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\left(đpcm\right)\)
\(25x^2-20x+7=25x^2-20x+4+3=\left(5x-2\right)^2+3>0\left(đpcm\right)\)
\(9x^2-6xy+2y^2+1=\left(9x^2+6xy+y^2\right)+y^2+1=\left(3x+y\right)^2+y^2+1>0\left(đpcm\right)\)
\(\Leftrightarrow x^2+y^2\ge xy;x^2+y^2\ge2\sqrt{x^2y^2}=2\left|xy\right|\ge\left|xy\right|\ge xy\Rightarrowđpcm\)
\(a,A=4x-x^2+3\)
\(=-\left(x^2-4x+4\right)+7\)
\(=-\left(x-2\right)^2+7\le7\forall x\)
Dấu"=" xảy ra<=> \(-\left(x-2\right)^2=0\Leftrightarrow x=2\)
Vậy......
\(b,B=4-x^2+2x\)
\(=-\left(x^2-2x+1\right)+5\)
\(=-\left(x-1\right)^2+5\le5\forall x\)
Dấu"=" xảy ra<=> \(-\left(x-1\right)^2=0\Leftrightarrow x=1\)
Vậy......
B2:
a) ta có: \(a^2+b^2-2ab\ge0\)
\(\Rightarrow\left(a-b\right)^2\ge0\forall a;b\) (luôn đúng)
\(\Rightarrowđpcm\)
b) Ta có: \(a^2+b^2\ge-2ab\)
\(\Rightarrow\left(a+b\right)^2\ge0\forall a;b\) (luôn đúng)
\(\Rightarrowđpcm\)
\(A=x^2-5x+7\)
\(=x^2-5x+\dfrac{25}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\)
Với mọi x ta có :
\(\left(x-\dfrac{5}{2}\right)^2\ge0\)
\(\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}>0\)
\(\Leftrightarrow A>0\)
Vậy..
Em thử nhé !
Bài 1 :
a) \(A=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-2.x.2+2^2\right)+7\)
\(=-\left(x-2\right)^2+7\le7\)
Dấu "=" xảy ra \(\Leftrightarrow-\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy : \(A_{max}=7\Leftrightarrow x=2\)
b) \(B=4-x^2+2x=-\left(x^2-2x-4\right)=-\left(x^2-2.x.1+1^2\right)+5\)
\(\Leftrightarrow B=-\left(x-1\right)^2+5\le5\)
Dấu "=" xảy ra \(\Leftrightarrow-\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy : \(B_{max}=5\Leftrightarrow x=1\)
Câu b:
Ta có: \(x^2 + 4y^2 + z^2 - 2x - 6z + 8y + 15\)
\(= (x^2 - 2x +1) + (4y^2 - 8y + 4) + (z^2 - 6z +9) +1\)
\(= (x-1)^2 + (2y-2)^2 + (z-3)^2 + 1\)
Mà \((x-1)^2 \geq 0; (2y-2)^2 \geq 0; (z-3)^2\geq 0\)
\(\implies\) \((x-1)^2+(2y-2)^2 +(z-3)^2\geq 0\)
\(\implies\)\((x-1)^2+(2y-2)^2 +(z-3)^2+1> 0\)
\(x^4-6x^2+15\)
\(=\left(x^2\right)^2-2.3.x^2+9+6\)
\(=\left(x^2-3\right)^2+6\ge6\forall x\)
Vậy \(x^4-6x^2+15\) luôn dương.
x4-6x2+15 = (x2)2 - 2.3.x2 +32+6
= (x2-3)2 +6
vì (x2-3)2 ≥ 0 ,∀ x.
⇒ (x2-3)2 +6 ≥ 6
Vậy x4-6x2+15 ≥ 6 hay x4-6x2+15 luôn dương