toán lop may z

toan vong may z

x-1+x-2+x-3+.......+x-99+x-100=2500

=>x+x+...+x-(1+2+......+100)=2500

=>100x-5050=2500

=>100x=7550

=>x=75,5

Yêu cầu là gì v c.

`@` `\text {Ans}`

`\downarrow`

`1)`

`x^3 - 2x - 1`

`= x^3 - x^2 - x + x^2 - x - 1`

`= (x^3 - x^2 - x) + (x^2 - x - 1)`

`= x(x^2 - x - 1) + (x^2 - x - 1)`

`= (x+1)(x^2 - x - 1)`

`2)`

`x^3 +3x^2 -4`

`= x^3 + 4x^2 + 4x - x^2 - 4x - 4`

`= (x^3 + 4x^2 + 4x) - (x^2 + 4x + 4)`

`= x(x^2 + 4x + 4) - (x^2 + 4x + 4)`

`= (x-1)(x^2 + 4x + 4)`

`3)`

`x^3y^3 + x^2y^2 + 4`

`= x^3 y^3 - x^2y^2 + 2x^2y^2 + 2xy - 2xy + 4`

`= (x^3y^3 - x^2y^2 + 2xy) + (2x^2y^2 - 2xy + 4)`

`= xy(x^2y^2 - xy + 2) + 2(x^2y^2 - xy + 2)`

`= (xy+2)(x^2y^2-xy+2)`

`4)`

`x^3 + 5x^2 -9xy^2+5y^3`

Bạn xem lại đề ;-;

`5)`

` x^4 + x^3 + 6x^2 + 5x + 5`

`= x^4 + x^3 + 5x^2 + x^2 + 5x + 5`

`= (x^4 + x^3 + x^2) + (5x^2 + 5x + 5)`

`= x^2(x^2 + x + 1) + 5(x^2 + x + 1)`

`= (x^2 + 5)(x^2 + x + 1)`

Điều kiện \(\left\{{}\begin{matrix}\dfrac{4x-3x^2y-9xy^2}{x+3y}\ge0\\x+3y\ne0\end{matrix}\right.\)

Với \(3y\ge x\), hệ tương đương:

\(\left\{{}\begin{matrix}\left(x^4-2x^2+4\right)\left(x^2+2\right)=6x^5y\\\left(3y-x\right)^2=\dfrac{4x}{x+3y}-3xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^6+8=6x^5y\left(1\right)\\x^3+27y^3=4x\end{matrix}\right.\left(I\right)\)

Vì \(x=0\) thì hệ vô nghiệm nên \(x\ne0\), khi đó:

\(\left(I\right)\Leftrightarrow\left\{{}\begin{matrix}1+\dfrac{8}{x^6}=\dfrac{6y}{x}\\1+\dfrac{27y^3}{x^3}=\dfrac{4}{x^2}\end{matrix}\right.\)

Đặt \(\dfrac{3y}{x}=a,\dfrac{2}{x^2}=b\) ta được hệ:

\(\Leftrightarrow\left\{{}\begin{matrix}1+a^3=2b\\1+b^3=2a\end{matrix}\right.\)

Giải hệ này ta được \(a=b\Leftrightarrow\dfrac{3y}{x}=\dfrac{2}{x^2}\Leftrightarrow y=\dfrac{2}{3x}\)

\(\left(1\right)\Leftrightarrow x^6-4x^4+8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\\x=\sqrt{1+\sqrt{5}}\\x=-\sqrt{1+\sqrt{5}}\end{matrix}\right.\)

TH1: \(x=\sqrt{2}\Rightarrow y=\dfrac{\sqrt{2}}{3}\)

TH2: \(x=-\sqrt{2}\Rightarrow y=-\dfrac{\sqrt{2}}{3}\)

TH3: \(x=\sqrt{1+\sqrt{5}}\Rightarrow y=\dfrac{2}{3\sqrt{1+\sqrt{5}}}\)

TH4: \(x=-\sqrt{1+\sqrt{5}}\Rightarrow y=-\dfrac{2}{3\sqrt{1+\sqrt{5}}}\)

Đối chiếu với các điều kiện ta được \(\left(x;y\right)=\left(-\sqrt{1+\sqrt{5}};-\dfrac{2}{3\sqrt{1+\sqrt{5}}}\right)\)

a)x+(x+1)+(x+2)+(x+3)+...+(x+99)+(x+100)=5555

=> 101x +5050 = 5555

=> 101x = 505

=> x = 505 : 101 = 5

Vậy, x = 5

b)1+2+3+4+...+x=820

=> ( x+1) x :2 = 820

=> (x+1)x = 1640

Mà 1640 = 40 . 41

=> x = 40 ( vì {x+1} - x = 1)

Vậy, x = 40

c) 3^x+1 = 9.27=243

=> 3^x+1 = 3⁵

=>x + 1 = 5

=> x = 4

Vậy, x=4

d) x+2x+3x+...+99x+100x=15150

=> [( 100 + 1) x 100 :2 ] x = 15150

=> 5050x = 15150

=> x = 15150:5050 = 3

Vậy, x =3

e)(x+1)+(x+2)+(x+3)+...+(x+100)=205550

=> 100x + 5050 = 205550

=> 100x =  205550 - 5050= 200500

=> x =  200500 : 100 = 2005

Vậy, x = 2005

f)3^x+3^x+1+3^x+2=351

=> 3^x + 3^x . 3 + 3^x x 9 = 351

=> 3^x ( 1+3+9) = 351

=> 3^x  . 13 = 351

=> 3^x  = 351 :13=27 mà 27 = 3³

=> x=3

Vậy, x=3

mình đg cần gấp á

a) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5555\)

\(\Rightarrow x+x+1+x+2+x+3+...+x+100=5555\)

\(\Rightarrow101\cdot x+5050=5555\)

\(\Rightarrow101\cdot x=5555-5050\)

\(\Rightarrow101\cdot x=505\)

\(\Rightarrow x=505:101\)

\(\Rightarrow x=5\)

b) \(1+2+3+4+...+x=820\)

\(\Rightarrow\left(x+1\right)\cdot\left[\left(x-1\right):1+1\right]:2=820\)

\(\Rightarrow\left(x+1\right)\cdot\left(x+1-1\right):2=820\)

\(\Rightarrow\left(x+1\right)\cdot x:2=820\)

\(\Rightarrow x\cdot\left(x+1\right)=820\cdot2\)

\(\Rightarrow x\cdot\left(x+1\right)=1640\)

Ta thấy: \(40\cdot41=1640\)

Vậy: \(x=40\)