ta có

1+2+3+.........+x=5050

=>\(\frac{x.\left(x+1\right)}{2}=5050\)

=>x.(x+1)=5050.2

=>x.(x+1)=10100

=>x.(x+1)=100.101

=>x=100

giúp mình với ,cần gấp

giúp mình với

c) (3x-4)(x-1)³=0

\(\Leftrightarrow\orbr{\begin{cases}3x-4=0\\\left(x-1\right)^3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=4\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=1\end{cases}}}\)

d) (x-4)(x-3)=0

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=3\end{cases}}}\)

e) (x+3)(2-x)>0

=> x+3 và 2-x cung dấu

TH1: Cùng âm

\(\hept{\begin{cases}x+3< 0\\2-x< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -3\\x>2\end{cases}}}\)(loại)

TH2L cùng dương

\(\hept{\begin{cases}x+3>0\\2-x>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-3\\x< 2\end{cases}\Leftrightarrow}-3< x< 2}\)

f) (x+1)+(x+2)+(x+3)+....+(x+100)=7450

<=> (x+x+x+....+x)+(1+2+3++.....+100)=7450

<=> 100x+\(\frac{\left(100+1\right)\cdot100}{2}=7450\)

<=> 100x+5050=7450

<=> 100x=2400

<=> x=24

Toán lớp 6 mà có mấy cái bài này á hả (xàm) -.-

1. 

$(3^2-2^3)x+3^2.2^2=4^2.3$

$\Leftrightarrow x+36=48$

$\Leftrightarrow x=48-36=12$

2.

$x^5-x^3=0$

$\Leftrightarrow x^3(x^2-1)=0$

$\Leftrightarrow x^3(x-1)(x+1)=0$

$\Leftrightarrow x^3=0$ hoặc $x-1=0$ hoặc $x+1=0$

$\Leftrightarrow x=0$ hoặc $x=\pm 1$
3.

$(x-1)^2+(-3)^2=5^2(-1)^{100}$

$\Leftrightarrow (x-1)^2+9=25$

$\Leftrightarrow (x-1)^2=25-9=16=4^2=(-4)^2$

$\Rightarrow x-1=4$ hoặc $x-1=-4$

$\Leftrightarrow x=5$ hoặc $x=-3$

4.

$(2x-1)^2-(2x-1)=0$

$\Leftrightarrow (2x-1)(2x-1-1)=0$

$\Leftrightarrow (2x-1)(2x-2)=0$

$\Leftrightarrow 2x-1=0$ hoặc $2x-2=0$

$\Leftrightarrow x=\frac{1}{2}$ hoặc $x=1$

$\Lef

`@` `\text {Ans}`

`\downarrow`

\((3^2-2^3)x+3^2.2^2=4^2.3\)

`=> x + (3*2)^2 = 48`

`=> x+6^2 = 48`

`=> x + 36 = 48`

`=> x = 48 - 36`

`=> x=12`

Vậy, `x=12`

\(x^5-x^3=0\)

`=> x^3(x^2 - 1)=0`

`=>`\(\left[{}\begin{matrix}x^3=0\\x^2-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

Vậy, `x \in {0; +- 1 }`

\(\left(x-1\right)^2+\left(-3\right)^2=5^2\cdot\left(-1\right)^{100}\)

`=> (x-1)^2 + 9 = 25*1`

`=> (x-1)^2 + 9 = 25`

`=> (x-1)^2 = 25 - 9`

`=> (x-1)^2 = 16`

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

`=>`\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4+1\\x=-4+1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy, `x \in {5; -3}`

\((2x-1)^2-(2x-1)=0\)

`=> (2x-1)(2x-1) - (2x-1)=0`

`=> (2x-1)(2x-1-1)=0`

`=>`\(\left[{}\begin{matrix}2x-1=0\\2x-2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

Vậy, `x \in {1; 1/2}`

x(3-x)=0

Th1: x=0

Th2: 3-x=0     =>       x=3

Vậy x=0 và x=3

(x+2)(4x-8)=0

Th1: x+2= 0        =>         x=-2

Th2: 4x-8 =0           =>             4x =8

                                                x= 2

Vậy x= +- 2 (cộng trừ 2 nhé)

(x+1)+(x+2)+(x+3)+....+(x+199)=5750

199x +(1+2+3+...199) =5750

199x+ {(199+1)* [(199-1)+1] : 2} =5750

199x + 19900= 5750

199x = -14150

x= -14150/199

\(\left(x^2+1\right)^2-\left(2x+100\right)^2=0\)

\(\Leftrightarrow\left(x^2+1-2x-100\right)\left(x^2+1-2x+100\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-2x-99=0\\x^2-2x+101=0\left(loại\right)\end{cases}}\)

\(\Leftrightarrow\left(x-1\right)^2=100\)

\(\Leftrightarrow\orbr{\begin{cases}x=11\\x=-9\end{cases}}\) ( thỏa mãn )

(x + 1) + (x + 2) + (x + 3) + ... + (x + 100) = 0

(x + x + x + ... + x) + (1 + 2 + 3 + ... + 100) = 0

100x + (100 + 1) . 100 : 2 = 0

100x + 5050 = 0

100x = -5050

x = -5050 : 100

x = -50,5

x = -50,5

Bài 1:tìm x thuộc Z

a)x.(x-1)=0

\(\Leftrightarrow\left[\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy: \(x=0;1\)

b)(x-3).(x+4)=0

\(\Leftrightarrow\left[\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

Vậy: \(x=3;-4\)

c)(2x-4).(x+2)=0

\(\Leftrightarrow2\left(x-2\right).\left(x+2\right)=0\)

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

Vậy: \(x=2;-2\)

d)(x+1)^2.(x-2)^2=0

\(\Leftrightarrow\left[\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Vậy: \(x=-1;2\)

e) x(x+1).(x+2)^2.(x+3)^3=0

\(\Leftrightarrow\left[\begin{matrix}x=0\\x+1=0\\x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\x=-1\\x=-2\\x=-3\end{matrix}\right.\)

Vậy: \(x=0;-1;-2;-3\)

f)(x-9)^5.(x-5)^8=0

\(\Leftrightarrow\left[\begin{matrix}x-9=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=9\\x=5\end{matrix}\right.\)

Vậy: \(x=9;5\)

g)x(x+100)^10.(x+2000)^20.(x+300)^300=0

\(\Leftrightarrow\left[\begin{matrix}x=0\\x+100=0\\x+200=0\\x+300=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\x=-100\\x=-200\\x=-300\end{matrix}\right.\)

Vậy: \(x=0;-100;-200;-300\)

h)(x-2)^2=0

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy: \(x=2\)