Sửa đề: +229+230-231-232

=(1+2-3-4)+(5+6-7-8)+...+(225+226-227-228)+(229+230-231-232)

=(-2)+(-2)+...+(-2)

=(-2)*116=-232

a: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+1+3x^2=-33\)

\(\Leftrightarrow39x=-34\)

hay \(x=-\dfrac{34}{39}\)

b: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-2\right)\left(x+2\right)=1\)

\(\Leftrightarrow x^3-27-x^3+4x=1\)

\(\Leftrightarrow4x=28\)

hay x=7

c: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)

\(\Leftrightarrow x^3+8-x^3+9x=26\)

\(\Leftrightarrow x=2\)

Cách 1 : A=100+98+96+...+2-97-95-...-1

A= 100 + (98-97) + (96-95) + ... +(2-1)

Từ 1 đến 98 có 98 số => có 98 : 2 cặp mà hiệu = 1

A = 100 + 49 x 1 = 149

B = 1+2-3-4+5+6-7-8+9+10-11-12+...-299-300+301+302

B = 1 + 2 + (302 - 300) + (301 - 299) + ... + (10 - 8) + (9-7) + (6-4) + (5-3) 

Từ 3 đến 302 có 300 số => có 300 : 2 cặp hiệu = 2

B = 1 + 2 + 150 x 2 = 303

Cách  2 :

A = 100 + (98-97) + (96-95) + ……. + (2-1)

Ta thấy:  97; 95; ….; 1  có (97 – 1) : 2 + 1 = 49 (số hạng)

A = 100 + (1+1+1+….+1)     (có 49 số 1).

A = 100 + 49 = 149

a, A = 100+(98-97)+(86-95)+....+(2-1) = 100+1+1+...+1 (49 số 1) = 149

b, B = 1+(2-3-4+5)+(6-7-8+9)+....(297-298-299+330)+331-332

= 1+0+0+....+0+331-332 = 0 

Nếu đúng thì k mk nha

\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)

Mai lam

`@` `\text {Ans}`

`\downarrow`

`a)`

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

`=> 5*x^3 = 0+5`

`=> 5*x^3 = 5`

`=> x^3 = 5 \div 5`

`=> x^3 = 1`

`=> x^3 = 1^3`

`=> x=1`

Vậy, `x=1.`

`b)`

\(( x+1)^2 = 16\)

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

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

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

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

Vậy, `x \in {3; -5}`

`c)`

\(( x+1)^3 = 27\)

`=> (x+1)^3 = 3^3`

`=> x+1=3`

`=> x=3-1`

`=> x=2`

Vậy, `x=2.`

`d)`

\(( x-1)^3 = 343\)

`=> (x-1)^3 = 7^3`

`=> x-1=7`

`=> x=7+1`

`=> x=8`

Vậy, `x=8.`

`e)`

\((2x - 1^3) = 125\) hay đề là `(2x-1)^3 = 125` vậy ạ?

Mình làm cả 2 TH nhé!

`(2x-1^3)=125`

`=> 2x-1=125`

`=> 2x=125+1`

`=> 2x=126`

`=> x=126 \div 2`

`=> x=63`

TH2:

`(2x-1)^3 = 125`

`=> (2x-1)^3 = 5^3`

`=> 2x-1=5`

`=> 2x=5+1`

`=> 2x=6`

`=> x=6 \div 2`

`=> x=3`

Vậy, `x=3.`

(a) \(5x^3-5=0\Leftrightarrow5x^3=5\Leftrightarrow x^3=1\Leftrightarrow x=1\)

(b) \(\left(x+1\right)^2=16\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

(c) \(\left(x+1\right)^3=27\Leftrightarrow x+1=3\Leftrightarrow x=2\)

(d) \(\left(x-1\right)^3=343\Leftrightarrow x-1=7\Leftrightarrow x=8\)

(e) \(\left(2x-1\right)^3=125\Leftrightarrow2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)

a) x = 2 

b) x = 2     

c) x = 2

d) x = 1.