\(A=\frac{35^3+13^3}{48}-35.13\)

\(=\frac{\left(35+13\right)\left(35^2-35.13+13^2\right)}{48}-35.13\)

\(=\frac{48.\left(35^2-35.13+13^2\right)}{48}-35.13\)

\(=35^2-35.13+13^2-35.13\)

\(=\left(35-13\right)^2\)

\(=22^2=484\)

Ơ bài này dễ mà

Ta có: G đối xứng với D qua M \(\Rightarrow\)M là trung điểm của GD

Tứ giác BGCD có 2 đường chéo GD và BC cắt nhau tại trung điểm M của mỗi đường \(\Rightarrow\)BGCD là hình bình hành

Xét \(a^3+b^3+c^3=1\)

\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3=1\)

\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)=1\)

\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)=1\)

\(\Rightarrow\left(a+b+c\right)^3-3\left(a+b\right)c\left(a+b+c\right)-3ab\left(a+b\right)=1\)

\(\Rightarrow\left(a+b+c\right)^3-3\left(a+b\right)\left[ac+cb+c^2-ab\right]=1\)

Tìm a+b+c rồi thay vô P(a+b+c) ở dưới

\(a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{a}+\frac{1}{c}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(=a\left(P-\frac{1}{a}\right)+b\left(P-\frac{1}{b}\right)+c\left(P-\frac{1}{c}\right)\)

\(=P\left(a+b+c\right)-3\)

Đặt \(\frac{a}{b^2}=x;\frac{b}{c^2}=y;\frac{c}{a^2}=z\) thì \(\frac{b^2}{a}=\frac{1}{x};\frac{a^2}{c}=\frac{1}{y};\frac{c^2}{b}=\frac{1}{z}\)

\(\Rightarrow xyz=\frac{a}{b^2}\cdot\frac{b}{c^2}\cdot\frac{c}{a^2}=1\)

Ta có: \(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{xy+yz+zx}{xyz}=xy+yz+zx\)

Lại có: \(\left(x-1\right)\left(y-1\right)\left(z-1\right)=xyz-xy-yz-zx+x+y+z-1=1-x-y-z+x+y+z-1=0\)(vì xyz=1, xy+yz+zx=x+y+z)

=>x-1=0 hoặc y-1=0 hoặc z-1=0

=>x=1 hoặc y=1 hoặc z=1

=>a/b²=1 hoặc b/c²=1 hoặc c/a²=1

=>a=b² hoặc b=c² hoặc c=a² (ĐPCM)

Cách khác

Ta có: \(\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}=\frac{b^2}{a}+\frac{a^2}{c}+\frac{c^2}{b}\)

<=>\(a^2b^2c^2\left(\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\right)=abc\left(\frac{b^2}{a}+\frac{a^2}{c}+\frac{c^2}{b}\right)\) (a²b²c²=abc=1)

<=>\(\frac{a^3b^2c^2}{b^2}+\frac{a^2b^3c^2}{c^2}+\frac{a^2b^2c^3}{a^2}=\frac{ab^3c}{a}+\frac{a^3bc}{c}+\frac{abc^3}{b}\)

<=>\(a^3c^2+b^3a^2+c^3b^2=b^3c+a^3b+c^3a\)

<=>\(a^3c^2+b^3a^2+c^3b^2-b^3c-a^3b-c^3a-a^2b^2c^2+abc=0\) (a²b²c²=abc=1)

<=>\(\left(a^3c^2-a^2b^2c^2\right)+\left(b^3a^2-a^3b\right)+\left(c^3b^2-c^3a\right)+\left(abc-b^3c\right)=0\)

<=>\(-a^2c^2\left(b^2-a\right)+a^2b\left(b^2-a\right)+c^3\left(b^2-a\right)-bc\left(b^2-a\right)=0\)

<=>\(\left(b^2-a\right)\left(-a^2c^2+a^2b+c^3-bc\right)=0\)

<=>\(\left(b^2-a\right)\left[c^2\left(c-a^2\right)-b\left(c-a^2\right)\right]=0\)

<=>\(\left(b^2-a\right)\left(c^2-b\right)\left(c-a^2\right)=0\) 

Đến đây dễ rồi

Mk chỉ lm 2 phần đầu thôi ,bn tham khảo nha!!!

\(a,\left(3x-1\right)^2-16=\left(3x-1+4\right)\left(3x-1-4\right)=\left(3x+3\right)\left(3x-5\right)=3\left(x+1\right)\left(3x-5\right)\)

\(b,\left(5x-4\right)^2-49x^2=\left(5x-4+7x\right)\left(5x-4-7x\right)\)

\(=\left(12x-4\right)\left(-2x-4\right)\)

\(=4\left(3x-1\right)\left(-2\right)\left(x+2\right)\)

\(=-8\left(3x-1\right)\left(x+2\right)\)

=.= hok tốt!!

\(\left(3x-1\right)^2-16\)

\(=\left(3x-1\right)^2-4^2\)

\(=\left(3x-1-4\right)\left(3x-1+4\right)\)

\(=\left(3x-5\right)\left(3x+3\right)\)

\(=3\left(x+1\right)\left(3x-5\right)\)

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