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x²+2x³+x² =x^2*(1+2x+1)
x³-x+3x²y+3xy²+y³-y = (y+x-1)*(y+x)*(y+x+1)
5x²-10xy+5y²-20z² = -5*(2*z-y+x)*(2*z+y-x)
https://coccoc.com/search/math#query=Ph%C3%A2n+t%C3%ADch+%C4%91a+th%E1%BB%A9c+th%C3%A0nh+nh%C3%A2n+t%E1%BB%AD%3A+x%5E8%2B98x%5E4%2B1
x8 + 98x4 + 1 = (x8 + 2x4 + 1 ) + 96x4
= (x4 + 1)2 + 16x2(x4 + 1) + 64x4 - 16x2(x4 + 1) + 32x4
= (x4 + 1 + 8x2)2 – 16x2(x4 + 1 – 2x2) = (x4 + 8x2 + 1)2 - 16x2(x2 – 1)2
= (x4 + 8x2 + 1)2 - (4x3 – 4x )2
= (x4 + 4x3 + 8x2 – 4x + 1)(x4 - 4x3 + 8x2 + 4x + 1)
a) \(A=\left(4n+3\right)^2-5^2=\left(4n+3-5\right)\left(4n+3+5\right)=\left(4n-2\right)\left(4n+8\right)\)
\(=8\left(n-1\right)\left(n+2\right)\). Vì A chứa thừa số 8 nên A chia hết cho 8
b) \(B=\left(2n+3\right)^2-3^2=\left(2n+3-3\right)\left(2n+3+3\right)=2n\left(2n+6\right)=4n\left(n+3\right)\)
Vì B chứa thừa số 4 nên B chia hết cho 4
ĐKXĐ: \(x\ne2;-2\)
\(M=\left(\frac{x-2}{x+2}+\frac{x+2}{x-2}-\frac{4x^2}{4-x^2}\right):\frac{3x^2+4}{x+2}\)
\(=\left(\frac{\left(x-2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{4x^2}{x^2-4}\right):\frac{3x^2+4}{x+2}\)
\(=\left(\frac{x^2-2x+1}{\left(x+2\right)\left(x-2\right)}+\frac{x^2+2x+1}{\left(x+2\right)\left(x-2\right)}+\frac{4x^2}{\left(x+2\right)\left(x-2\right)}\right):\frac{3x^2+4}{x+2}\)
\(=\frac{6x^2+2}{\left(x+2\right)\left(x-2\right)}.\frac{x+2}{3x^2+4}\)
\(=\frac{6x^2+2}{\left(x-2\right)\left(3x^2+4\right)}\)
b) ta có để M <-1 thì \(\frac{6x^2+2}{\left(x-2\right)\left(3x^2+4\right)}\)\(< -1\)
\(< =>\frac{6x^2+2}{\left(x-2\right)\left(3x^2+4\right)}+1< 0\)
đến đây bạn quy đồng rồi giải nhé
a)\(36-4a^2+20ab-25b^2=6^2-\left(4a^2-20ab+25b^2\right)\)
\(=6^2-\left[\left(2a\right)^2-2.2a.5b+\left(5b\right)^2\right]\)
\(=6^2-\left(2a-5b\right)^2\)
\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)
b)\(a^3+3a^2+3a+1-27b^3=\left(a+1\right)^3-\left(3b\right)^3\)(chỗ này mình sửa 27b2 thành 27b3 vì mình nghĩ nhầm đề)
\(=\left(a+1-3b\right)\left[\left(a+1\right)^2+\left(a+1\right)3b+\left(3b\right)^2\right]\)
\(=\left(a+1-3b\right)\left(a^2+2a+1+3ab+3b+9b^2\right)\)
c)\(x^3+3x^2+3x+1-3x^2-3x=\left(x+1\right)^3-3x\left(x+1\right)\)
\(=\left(x+1\right)\left[\left(x+1\right)^2-3x\right]\)
\(=\left(x+1\right)\left(x^2+2x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)\)
\(\left(x^2+\frac{2}{5}y\right)\left(x^2-\frac{2}{5}y\right)\)
\(=x^4+\frac{2}{5}x^2y-\frac{2}{5}x^2y-\left(\frac{2}{5}y\right)^2\)
\(=x^4-\frac{4}{25}y^2\)
2) a) Không mất tính tổng quát, ta giả sử \(a\ge b\ge c>0\).Suy ra \(a+b\ge a+c\ge b+c\)
Ta có : \(\frac{b}{c+a}< \frac{b}{b+c}\); \(\frac{c}{a+b}< \frac{c}{b+c}\); \(\frac{a}{b+c}< 1\)
\(\Rightarrow\frac{b}{c+a}+\frac{c}{a+b}+\frac{a}{b+c}< \frac{b+c}{b+c}+1=2\)
b) Đặt \(x=b+c-a\); \(y=c+a-b\); \(z=a+b-c\);
Khi đó : \(2a=y+z\Rightarrow a=\frac{y+z}{2}\). \(b=\frac{x+z}{2}\); \(c=\frac{x+y}{2}\)
\(\Rightarrow\frac{\frac{y+z}{2}}{x}+\frac{\frac{x+z}{2}}{y}+\frac{\frac{x+y}{2}}{z}=\frac{1}{2}\left[\left(\frac{y}{x}+\frac{x}{y}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)\right]\)
Mặt khác ta có : \(\frac{x}{y}+\frac{y}{x}\ge2\); \(\frac{y}{z}+\frac{z}{y}\ge2\); \(\frac{x}{z}+\frac{z}{x}\ge2\)
\(\Rightarrow\frac{\frac{y+z}{2}}{x}+\frac{\frac{x+z}{2}}{y}+\frac{\frac{x+y}{2}}{z}\ge\frac{1}{2}\left(2+2+2\right)\)
hay \(\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\ge3\)(đpcm)