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4) Ta có : A=(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)
=> (a+d)2 - (b+c)2= (a-d)2 - (c-b)2
=> a2+ d2+ 2ad - b2- c2- 2bc=a2 + d2 - 2ad - c2-b2+2bc
Rút gọn ta được: 4ad = 4bc => ad = bc =>\(\dfrac{a}{c}=\dfrac{b}{d}\)
1) a2+b2+c2+3=2(a+b+c) =>(a-1)2+(b-1)2+(c-1)2=0
=> a-1=b-1=c-1=0 => a=b=c=1 =>đpcm
Câu 1:
Ta có: \(\left(\dfrac{a+b}{2}\right)^2\ge ab\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{2^2}-ab\ge0\)
\(\Leftrightarrow\dfrac{a^2+2ab+b^2-4ab}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\ge ab\) (1)
Ta có: \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}-\dfrac{\left(a+b\right)^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{2a^2-2b^2-a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\) (2)
Từ (1) và (2) \(\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\)
5 , a3+b3+c3\(\ge\) 3abc
\(\Leftrightarrow\) a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc\(\ge\) 0
\(\Leftrightarrow\) (a+b)3+c3-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+b2+c2-ab-bc-ca)\(\ge0\) (1)
ta co : a,b,c>0 \(\Rightarrow\)a+b+c>0 (2)
(a-b)2+(b-c)2+(c-a)2\(\ge0\)
<=> 2a2+2b2+2c2-2ac-2cb-2ab\(\ge0\)
<=>a2+b2+c2-ab-bc-ac\(\ge\) 0 (3)
Từ (1)(2)(3)=> pt luôn đúng
\(1,x^3-7x+6\)
\(=x^3+3x^2-3x^2-9x+2x+6\)
\(=x^2\left(x+3\right)-3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-3x+2\right)\)
\(=\left(x+3\right)\left(x^2-2x-x+2\right)\)
\(=\left(x+3\right)\left(x-2\right)\left(x-1\right)\)
\(2,x^3-9x^2+6x+16\)
\(=x^3+x^2-10x^2-10x+16x+16\)
\(=x^2\left(x+1\right)-10x\left(x+1\right)+16\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left(x^2-2x-8x+16\right)\)
\(=\left(x+1\right)\left(x-8\right)\left(x-2\right)\)
mk ms lm hai câu thôi mà đã mệt r , bh mk lm bt mai đi học ,lúc khác lm đ cko bn
Câu 1:
- Chứng minh a3+b3+c3=3abc thì a+b+c=0
\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)
\(\Rightarrow\left[\left(a+b\right)^3+c^3\right]-3abc\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow0=0\) Đúng (Đpcm)
- Chứng minh a3+b3+c3=3abc thì a=b=c
Áp dụng Bđt Cô si 3 số ta có:
\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc\)
Dấu = khi a=b=c (Đpcm)
Câu 2
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3\cdot\frac{1}{abc}\)
Ta có:
\(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3}+\frac{abc}{b^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(=abc\cdot3\cdot\frac{1}{abc}=3\)
a/ \(\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)-24\)
\(=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
b/ \(\left(x^2-2x+4\right)\left(x^2+3x+4\right)-14x^2\)
\(=x^2\left[\left(x-2+\frac{4}{x}\right)\left(x+3+\frac{4}{x}\right)-14\right]\)
\(=x^2\left[\left(x-2+\frac{4}{x}\right)^2+5\left(x-2+\frac{4}{x}\right)-14\right]\)
\(=x^2\left(x-2+\frac{4}{x}-2\right)\left(x-2+\frac{4}{x}+7\right)\)
\(=x^2\left(x-4+\frac{4}{x}\right)\left(x+5+\frac{4}{x}\right)\)
\(=\left(x^2-4x+4\right)\left(x^2+5x+4\right)\)
\(=\left(x-2\right)^2\left(x+1\right)\left(x+4\right)\)
c/ \(a^2b-a^2c+b^2\left(a-c\right)+c^2a-c^2b\)
\(=b\left(a^2-c^2\right)-ac\left(a-c\right)+b^2\left(a-c\right)\)
\(=\left(a-c\right)\left(ab+bc-ac+b^2\right)\)
d/ \(a^3+b^3+3ab\left(a+b\right)+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(\left(a+b\right)^2-\left(a+b\right)c+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
a,Câu hỏi của Cỏ dại - Toán lớp 8 - Học toán với OnlineMath
b,Câu hỏi của Ngô Đức Duy - Toán lớp 8 - Học toán với OnlineMath
c,Câu hỏi của Cỏ dại - Toán lớp 8 - Học toán với OnlineMath