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2) \(A=\frac{x^3-27}{x-3}+5x\)
\(=\frac{\left(x-3\right).\left(x^2+3x+9\right)}{x-3}+5x\)
\(=x^2+3x+9+5x=x^2+8x+9\)
\(=\left(x+\text{4}\right)^2-7\ge-7\)
Vậy \(A_{min}=-7\)
4) Số đỉnh của đa giác có tổng các góc trong bằng \(1080^o\)là 8
P/s cn mấy cái kia kh bk =))
1) \(B=5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)+2\left(5-3x\right)^2\)
\(=5\left(4x^2-4x+1\right)+\left(4x-4\right)\cdot\left(x+3\right)+2\left(25-30x+9x^2\right)\)
\(=20x^2-20x+5+4x^2+12x-4x-12+50-60+18x^2\)
\(=42x^2-72x+43\)
2) \(C=\left(2a^2+2a+1\right)\left(2a^2-2a+1\right)-\left(2a+1\right)^2\)
\(=4a^4-4a^3+2a^2+4a^3-4a^2+2a+2a^2-2a+1-\left(4a^2+4a+1\right)\)
\(=4a^4+2a^2-4a^2+2a^2+1-4a^2-4a-1\)
\(=4a^4-4a^2-4a\)
3) Sky Sơn Tùng làm đúng rồi nhé.
4) \(E=\left(x^2-5x+1\right)^2+2\left(5x-1\right)\left(x^2-5x+1\right)\left(5x-1\right)^2\)
\(=x^4+27x^2+1-10x^3+250x^5-1400x^4+1030x^3-302x^2+40x-2\)
\(=-1399x^4-275x^2-1+1020x^3+250x^5+40x\)
5) \(F=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2\)
\(=\left[a^2+b^2-c^2-\left(a^2-b^2+c^2\right)\right]\cdot\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\)
\(=\left(a^2+b^2-c^2-a^2+b^2-c^2\right)\cdot2a^2\)
\(=\left(2b^2-2c^2\right)\cdot2a^2\)
\(=2\left(b^2-c^2\right)\cdot2a^2\)
\(=2\left(b-c\right)\left(b+c\right)\cdot2a^2\)
\(=2\cdot2a^2\cdot\left(b-c\right)\left(b+c\right)\)
\(=4a^2\cdot\left(b-c\right)\left(b+c\right)\)
6) \(G=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+\left(-c\right)^2+2ab-2ac-2bc-2\left(a^2+2ab+b^2\right)\)
\(=a^2+b^2+c^2+2ab+a^2+b^2+\left(-c\right)^2+2ab-2a^2-4ab-2b^2\)
\(=0+0+c^2+0+c^2\)
\(=2c^2\)
7) \(H=\left(a+c\right)\left(a-c\right)-\left(a-b-c\right)\left(a-b+c\right)+b\left(b-2x\right)\)
\(=a^2-c^2-\left[\left(a-b\right)^2-c^2\right]+b^2-2bx\)
\(=a^2-c^2-\left(a^2-2ab+b^2-c^2\right)+b^2-2bx\)
\(=a^2-b^2-a^2+2ab-b^2+c^2+b^2-2bx\)
\(=2ab-2bx\)
\(D=\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)=\left(9x-1+1-5x\right)^2=\left(4x\right)^2=16x^2\)
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\)\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\)\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow\)\(a^2+b^2+c^2-ab-bc-ca=0\) (do a+b+c # 0)
\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\) (bn nhân với 2 rồi tách, nhóm lại nhé)
\(\Leftrightarrow\)\(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\)\(\Leftrightarrow\)\(a=b=c\)
\(D=\left(\frac{a}{2b}\right)^2+\left(\frac{b}{2c}\right)^2+\left(\frac{c}{2a}\right)^2\)
\(=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2\)\(=\frac{3}{4}\)
a)\(\left(a+b+c\right)^2-\left(a+b\right)^2-c^2\\ =\left(a+b\right)^2+2\left(a+b\right)c+c^2-\left(a+b\right)^2-c^2\\ =2\left(a+b\right)c\)
b)\(\left(a+b+c\right)^2-\left(b+c\right)^2-2a\left(b+c\right)\\ =a^2+2a\left(b+c\right)+\left(b+c\right)^2-\left(b+c\right)^2-2a\left(b+c\right)\\ =a^2\)
c)\(\left(3a+1\right)^2-2\left(2a+5\right)\left(3a+1\right)+\left(2a+5\right)^2\\ =\left(3a+1-2a-5\right)^2\\ =\left(a-4\right)^2\)
a2 +b2 +c2 +42 = 2a+8b+610c
a2 -2a+1 + b2-8b+16 +c2 -10c + 24 =0
(a-1)2 +(b-4)2+(c-5)2=0
suy ra a= 1 ;b= 4; c= 5
vậy a+b+c = 10