Ta có: \(\dfrac{x}{2\cdot3}+\dfrac{x}{3\cdot4}+...+\dfrac{x}{99\cdot100}=-1\)

\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=-1\)

\(\Leftrightarrow x\cdot\dfrac{49}{100}=-1\)

hay \(x=-\dfrac{100}{49}\)

$\dfrac{x}{2.3}+\dfrac{x}{3.4}+...+\dfrac{x}{99.100}=1$

`<=>x/2 - x/3 +x/3-x/4+...+x/(99)-x/(100)=1`

`<=>x/2-x/(100)=1`

`<=>(50x)/(100)-x/(100)=(100)/(100)`

`<=>50x-x=100`

`<=>49x=100`

`<=>x=(100)/(49)`

Vậy `x=(100)/(49)`

Ta có:\(\dfrac{x}{2.3}+\dfrac{x}{3.4}+.....+\dfrac{x}{99.100}=-1\)

     ⇔\(x\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{99.100}\right)=-1\)

     ⇔\(x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{99}-\dfrac{1}{100}\right)=-1\)

    ⇔\(x\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=-1\)

    \(\Leftrightarrow x.\dfrac{49}{100}=-1\)

     \(\Leftrightarrow x=-\dfrac{100}{49}\)

Vậy phương trình có nghiệm x=\(-\dfrac{100}{49}\)

a) Ta có: \(x^4+64\)

\(=x^4+16x^2+64-16x^2\)

\(=\left(x^2+8\right)^2-\left(4x\right)^2\)

\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)

b) Ta có: \(81x^4+4y^4\)

\(=81x^4+36x^2y^2+4y^4-36x^2y^2\)

\(=\left(9x^2+2y^2\right)^2-\left(6xy\right)^2\)

\(=\left(9x^2-6xy+2y^2\right)\left(9x^2+6xy+2y^2\right)\)

c) Ta có: \(x^5+x+1\)

\(=x^5+x^2-x^2+x-1\)

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

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

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

a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)

b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)

\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)

a. 9x² - 6x - 3 = 0

<=> 3(3x² - 2x - 1) = 0

<=> 3(3x² - 3x + x - 1) = 0

<=> \(3\left[3x\left(x-1\right)+\left(x-1\right)\right]=0\)

<=> 3(3x + 1)(x - 1) = 0

<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)

b. (2x + 1)² - 4(x + 2)² = 9

<=> (2x + 1)² - \(\left[2\left(x+2\right)\right]^2=9\)

<=> (2x + 1 - 2x - 4)(2x + 1 + 2x + 4) = 9

<=> -3(4x + 5) = 9

<=> 4x + 5 = -3

<=> 5 + 3 = -4x

<=> -4x = 8

<=> -x = 2

<=> x = -2

a) \(\Leftrightarrow\left(9x^2-6x+1\right)-4=0\)

\(\Leftrightarrow\left(3x-1\right)^2-4=0\)

\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b) \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)

\(\Leftrightarrow12x=-24\Leftrightarrow x=-2\)

c) \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)

\(\Leftrightarrow9x=18\Leftrightarrow x=2\)

d) \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)

\(\Leftrightarrow2x=-40\Leftrightarrow x=-20\)

(1)-a)Với mọi x, ta luôn có: \(\left(x+1\right)^2+3>0\Leftrightarrow x^2+1+2x+3>0\Leftrightarrow x^2+2x+4>0\)

            \(\sqrt{x^2+2x+4}=2\Leftrightarrow x^2+2x+4=2^2=4\)

                                           \(\Leftrightarrow x^2+2x=0\\\Leftrightarrow\left(x+2\right)x=0\\\Leftrightarrow\left[{}\begin{matrix}x+2=0\Leftrightarrow x=-2\\x=0\end{matrix}\right. \)

        ➤\(x\in\left\{-2;0\right\}\)

b) \(\left\{{}\begin{matrix}x+2y-1=0\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2y=1\\4x+2y=10\end{matrix}\right.\)

                                  \(\Leftrightarrow\left\{{}\begin{matrix}2y=1-x\\3x=9\Leftrightarrow x=\dfrac{9}{3}=3\end{matrix}\right.\)

Do \(x=3\Leftrightarrow1-x=1-3=-2\) nên ta có: \(2y=1-x=-2\Leftrightarrow y=\dfrac{-2}{2}=-1\)

➤\(\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

(2): +)ĐK để 2 hàm số cắt nhau là: \(2a\ne1\Leftrightarrow a\ne\dfrac{1}{2}\Leftrightarrow a\ne0,5\) 

Ta có hệ phương trình sau: \(\left\{{}\begin{matrix}y=2ax+a+1\\y=x+2\end{matrix}\right.\)

➢Do đó, ta có: \(2ax+a+1=x+2\Leftrightarrow2ax+a-x=2-1=1\)

a: Ta có: \(A=\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)

\(=8x^3+27-8x^3+2\)

=29

b: Ta có: \(B=\left(64x^3-1\right)-\left(4x-3\right)\left(16x^2+3\right)\)

\(=64x^3-1-64x^3-12x-48x^2+9\)

\(=-12x+8\)

c: Ta có: \(2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)

\(=2\left(x^2+xy+y^2\right)-3\left(-2xy\right)\)

\(=2x^2+2xy+2y^2+6xy\)

\(=2x^2+8xy+2y^2\)

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

b: Phương trình hoành độ giao điểm là:

-x+3=-2x+1

\(\Leftrightarrow x=-2\)

Thay x=-2 vào y=-x+3, ta được;
y=2+3=5

Thay x=-2 và y=5 vào (d), ta được:

\(-2\left(2-m\right)+2m-1=5\)

\(\Leftrightarrow2m-4+2m-1=5\)

\(\Leftrightarrow4m=10\)

hay \(m=\dfrac{5}{2}\)

\(3xy-4x+2y=1\Rightarrow x\left(3y-4\right)=1-2y\Rightarrow x=\dfrac{1-2y}{3y-4}\)

-Vì x,y nguyên nên \(\left(1-2y\right)⋮\left(3y-4\right)\)

\(\Rightarrow\left(3-6y\right)⋮\left(3y-4\right)\)

\(\Rightarrow\left(-6y+8-5\right)⋮\left(3y-4\right)\)

\(\Rightarrow-5⋮\left(3y-4\right)\)

\(\Rightarrow3y-4\inƯ\left\{-5\right\}\)

\(\Rightarrow3y-4\in\left\{1;5;-1;-5\right\}\)

\(\Rightarrow y\in\left\{3;1\right\}\)

*\(y=1\Rightarrow x=\dfrac{1-2.1}{3.1-4}=1\)

*\(y=3\Rightarrow x==\dfrac{1-2.3}{3.3-4}=-1\)