\((3x-5)(x^2 -2x-3)\)

\(=3x.x^2+3x.\left(-2x\right)+3x.\left(-3\right)+\left(-5\right).x^2+\left(-5\right).\left(-2x\right)+\left(-5\right).\left(-3\right)\)

\(=3x^3-6x^2-9x-5x^2+10x+15\)

\(=3x^3-11x^2+x+15\)

Ta có: \((x + y + z)^2 = 0 \)

\(\Leftrightarrow\) \( x^2 + y^2 + z^2 + 2.(xy + yz + xz) = 0 \)

\(\Leftrightarrow\) \(1 + 2(xy + yz + xz) = 0 \)

\(\Leftrightarrow\) \(xy + yz + xz =\) \(-\dfrac{1}{2}\)

Lại có:

\(x^2y^2 + y^2z^2 + x^2z^2 \)

\(= (xy + yz + xz)^2 - 2xyz(x + y + z) \)

\(=\dfrac{1}{4}-0\)

\(=\dfrac{1}{4}\)

\(\Rightarrow\) \( x^4+y^4+z^4 = (x^2 + y^2 + z^2)^2 - 2(x^2y^2 + y^2z^2 + x^2z^2) \)

\(=\) \(1 - 2.\dfrac{1}{4} \)

\(=\dfrac{1}{2}\)

Vậy \(x^4+y^4+z^4=\dfrac{1}{2}\)

Ta có: \((a^2+b^2)(x^2+y^2)=(ax+by)^2 \)

\(\Leftrightarrow\) \(a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2 = a^2x^2 + 2abxy + b^2y^2 \)

\(\Leftrightarrow\) \(a^2y^2 + b^2x^2 = 2abxy \)

\(\Leftrightarrow\) \(a^2y^2 + b^2x^2 - 2abxy = 0 \)

\(\Leftrightarrow\) \((ay - bx)^2 = 0 \)

\(\Rightarrow\) \(ay - bx = 0 \)

\(\Rightarrow\) \(ay = bx \)

\(\Rightarrow\) \(\dfrac{a}{x}=\dfrac{b}{y}\)( Đpcm )

\(M=a^3 +b^3+3ab(a^2+b^2)+6a^2.b^2(a+b) \)

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

\(M=a^2-ab+b^2+3ab\left(a^2+b^2+2ab\right)\)

\(M=a^2-ab+b^2+3ab\left(a+b\right)^2\)

\(M=a^2-ab+b^2+3ab\)

\(M=a^2+b^2+2ab\)

\(M=\left(a+b\right)^2\)

\(M=1\)

Vì \(2^{2017}\) và \(5^{2017}\) có cùng số mũ nên ta có:

\(\left(2.5\right)^{2017}=10^{2017}\)

Vì \(2017\) là số mũ của 10

\(\Rightarrow\) Có \(2017\) chữ số \(0 + 1\) chữ số \(1 = 2018\)

Vậy hai số \(2^{2017}\) và \(5^{2017}\) khi viết liền nhau thì có \(2018\) chữ số.

Nốt bài 70 nhé :"v

Ta có: \( a^3 + b^3 + c^3 = 3abc \)

\(\Leftrightarrow a^3 + b^3 + c^3 - 3abc = 0 \)

\(\Leftrightarrow\)\( a^3 + b^3 + 3a^2b + 3ab^2 - 3a^2b - 3ab^2 + c^3 - 3abc = 0 \)

\(\Leftrightarrow(a+b)^3 - 3a^2b - 3ab^2 + c^3 - 3abc = 0 \)

\(\Leftrightarrow[(a+b)^3+c^3]-3ab(a+b+c)=0\)

\(\Leftrightarrow(a+b+c)[(a+b)^2 - c(a+b) + c^2] – 3ab(a+b+c) = 0 \)

\(\Leftrightarrow(a+b+c)(a^2+ 2ab + b^2 - ac – bc + c^2 - 3ab) = 0 \)

\(\Leftrightarrow(a+b+c)(a^2+b^2+c^2-bc-ab-ca)=0\)

\(\Leftrightarrow\dfrac{\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{2}=0\)

Vì \(a,b,c\) là số dương nên \(a+b+c>0\)

\(\Rightarrow\left[{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow a-b=b-c=c-a\)

\(\Rightarrow a=b=c\) ( Đpcm )

1. \(\dfrac{1}{x-1}-\dfrac{1}{x+1}\)

\(=\dfrac{1.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}-\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{x+1+\left(-x+1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{x+1-x+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{1}{x^2-1}\)

2. \(\dfrac{x}{x^2-1}-\dfrac{1}{x-1}\)

\(=\dfrac{x}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{x}{\left(x+1\right)\left(x-1\right)}+\dfrac{-\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{x+\left(-x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{-1}{x^2-1}\)

3. \(\dfrac{1}{x\left(x-y\right)}-\dfrac{1}{x\left(x-y\right)}\)

\(=\dfrac{1}{y\left(x-y\right)}+\dfrac{-1}{x\left(x-y\right)}\)

\(=\dfrac{1x}{y\left(x-y\right)x}+\dfrac{-1y}{x\left(x-y\right)y}\)

\(=\dfrac{x}{xy\left(x-y\right)}+\dfrac{-y}{xy\left(x-y\right)}\)

\(=\dfrac{x-y}{xy\left(x-y\right)}=\dfrac{1}{xy}\)

4. \(\dfrac{1}{x}-\dfrac{1}{x-1}\)

\(=\dfrac{1\left(x-1\right)}{x\left(x-1\right)}-\dfrac{1x}{\left(x-1\right)x}\)

\(=\dfrac{x-1}{x\left(x-1\right)}+\dfrac{-x}{x\left(x-1\right)}\)

\(=\dfrac{\left(x-1\right)-x}{x\left(x-1\right)}\)

\(=\dfrac{-1}{x\left(x-1\right)}\)

5. \(\dfrac{1}{x}-\dfrac{1}{x+1}\)

\(=\dfrac{1\left(x+1\right)}{x\left(x+1\right)}-\dfrac{1x}{\left(x+1\right)x}\)

\(=\dfrac{x+1}{x\left(x+1\right)}+\dfrac{-x}{x\left(x+1\right)}\)

\(=\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}\)

6. \(\dfrac{1}{2x^2-10x}-\dfrac{1}{x-5}\)

\(=\dfrac{1}{2x\left(x-5\right)}-\dfrac{1}{x-5}\)

\(=\dfrac{1}{2x\left(x-5\right)}-\dfrac{1.2x}{2x\left(x-5\right)}\)

\(=\dfrac{1}{2x\left(x-5\right)}+\dfrac{-2x}{2x\left(x-5\right)}\)

\(=\dfrac{1-2x}{2x\left(x-5\right)}\)

7. \(\dfrac{x-1}{x^2-1}.\dfrac{x+1}{x+3}\)

\(=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x^2-1\right)\left(x+3\right)}\)

\(=\dfrac{x^2-1}{\left(x^2-1\right)\left(x+3\right)}\)

8. \(\dfrac{2}{2x^2+10x}.\dfrac{x+5}{3x}\)

\(=\dfrac{2x\left(x+5\right)}{2x^2+10x.3x}\)

\(=\dfrac{2\left(x+5\right)}{2x\left(x+5\right)3x}\)

\(=\dfrac{2}{6x^2}=\dfrac{1}{3x^2}\)

1. \(\dfrac{4x^3y^2z^4}{8x^3yz^3}=\dfrac{4x^3yz^3.yz}{4x^3yz^2.2}=\dfrac{yz}{2}\)

2. \(\dfrac{x^2-9}{x-3}=\dfrac{\left(x+3\right).\left(x-3\right)}{x-3}=x+3\)

3. \(\dfrac{5x^2-10x}{x^2-4x+4}=\dfrac{5x.\left(x-2\right)}{\left(x-2\right)^2}=\dfrac{5x.\left(x-2\right)}{\left(x-2\right)\left(x-2\right)}=\dfrac{5x}{x-2}\)

1. \(\dfrac{2x^2+4x}{3x+6}=\dfrac{2x.\left(x+2\right)}{3.\left(x+2\right)}=\dfrac{2x}{3}\)

2. \(\dfrac{3x^2y^3}{6xy}=\dfrac{xy^23xy}{2.3xy}=\dfrac{xy^2}{2}\)

3. \(\dfrac{x^2+2x+1}{2x+2}=\dfrac{\left(x+1\right)^2}{2.\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x+1\right)}{2.\left(x+1\right)}=\dfrac{x+1}{2}\)

4. \(\dfrac{9-x}{x-9}=\dfrac{9-x}{-\left(9-x\right)}=\dfrac{1}{-1}=-1\)

5. \(\dfrac{-x^2+4}{2x-4}=\dfrac{-\left(x^2-4\right)}{2.\left(x-2\right)}=\dfrac{-\left(x-2\right).\left(x-2\right)}{2.\left(x-2\right)}=\dfrac{-\left(x+2\right)}{2}\)