nhé

a)(2x-1)⁶=(2x-1)⁸

=> (2x-1)⁸-(2x-1)⁶=0

=> (2x-1)⁶.((2x-1)²-1)=0  

+)th1(2x-1)⁶=0

+)th2((2x-1)²-1)=0

a) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\Rightarrow\left(2x-1\right)\in\left\{\pm1;0\right\}\)

TH1 : \(2x-1=0\)                       TH2 : \(2x-1=-1\)                      TH3 : \(2x-1=1\)

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

                      \(x=\frac{1}{2}\)                                          \(x=0\)                                                  \(x=1\)

Vậy \(x\in\left\{\frac{1}{2};0;1\right\}\)

b) Tương tự

2: Tìm x

a) Ta có: x+25=40

nên x=40-25=15

Vậy: x=15

b) Ta có: 198-(x+4)=120

\(\Leftrightarrow x+4=198-120=78\)

hay x=78-4=74

Vậy: x=74

c) Ta có: \(\left(2x-7\right)\cdot3=125\)

\(\Leftrightarrow2x-7=\dfrac{125}{3}\)

\(\Leftrightarrow2x=\dfrac{125}{3}+7=\dfrac{125}{3}+\dfrac{21}{3}=\dfrac{146}{3}\)

\(\Leftrightarrow x=\dfrac{146}{3}:2=\dfrac{146}{6}=\dfrac{73}{3}\)

Vậy: \(x=\dfrac{73}{3}\)

d) Ta có: \(x+16⋮x+1\)

\(\Leftrightarrow x+1+15⋮x+1\)

mà \(x+1⋮x+1\)

nên \(15⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(15\right)\)

\(\Leftrightarrow x+1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

hay \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)

Vậy: \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)

\(a,x+25=40\\ \Rightarrow x=40-25\\ \Rightarrow x=15\\ b,198-\left(x+4\right)=120\\ \Rightarrow-\left(x+4\right)=120-198\\ \Rightarrow-\left(x+4\right)=-78\\ \Rightarrow x+4=78\\ \Rightarrow x=78-4\\ \Rightarrow x=74\\ c,\left(2x-7\right).3=125\\ \Rightarrow2x-7=\dfrac{125}{3}\\ \Rightarrow2x=\dfrac{125}{3}+7\\ \Rightarrow2x=\dfrac{146}{3}\\ \Rightarrow x=\dfrac{146}{3}:2\Rightarrow x=\dfrac{73}{3}\\ d,\left(x+16\right)⋮\left(x+1\right)\\ \Rightarrow\left[\left(x+1\right)+15\right]⋮\left(x+1\right)\\ mà:\left(x+1\right)⋮\left(x+1\right)\\ \Rightarrow15⋮\left(x+1\right)\\ \Rightarrow\left(x+1\right)\inƯ\left(15\right)\\ \Rightarrow\left(x+1\right)\in\left\{-15;-1;1;15\right\}\\ \Rightarrow x\in\left\{-16;-2;0;14\right\}\)

Tự kết luận nhé bạn

Bài toán 4: Viết các số sau dưới dạng tổng các luỹ thừa của 10. 
213 = 2 . 100 + 1 . 10 +3 = 2. 10^2 + 1.10 + 3 . 10^0

421=4.100 + 2.10 + 1 = 4.10^2 + 2.10 + 1. 10^0

2009; = 2. 1000 + 9 = 2. 10^3 + 9 . 10^0

abc = a . 100 + b . 10 + c = a.10^2 + b.10 + c.10^0

abcde = a.10000 + b . 1000 + c . 100  + d . 10 + e = a . 10^4 + b. 10^3 + c.10^2 + d .10 + e . 10 ^0 

thanks bn

`@` `\text {Ans}`

`\downarrow`

`a)`

`3x(4x-1) - 2x(6x-3) = 30`

`=> 12x^2 - 3x - 12x^2 + 6x = 30`

`=> 3x = 30`

`=> x = 30 \div 3`

`=> x=10`

Vậy, `x=10`

`b)`

`2x(3-2x) + 2x(2x-1) = 15`

`=> 6x- 4x^2 + 4x^2 - 2x = 15`

`=> 4x = 15`

`=> x = 15/4`

Vậy, `x=15/4`

`c)`

`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`

`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`

`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`

`=> 40x^2 -17x - 1 = 1`

`d)`

`(x+2)(x+2)-(x-3)(x+1)=9`

`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`

`=> 6x + 7 =9`

`=> 6x = 2`

`=> x=2/6 =1/3`

Vậy, `x=1/3`

`e)`

`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`

`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`

`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`

`=> 12x +8 = 0`

`=> 12x = -8`

`=> x= -8/12 = -2/3`

Vậy, `x=-2/3`

`g)`

`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`

`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`

`=> -3x + 4 =14`

`=> -3x = 10`

`=> x= - 10/3`

Vậy, `x=-10/3`

Hello các bạn còn đó ko?

a)ĐKXĐ: \(x\notin\left\{0;-1\right\}\)

Ta có: \(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)

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

Suy ra: \(x^2-1+x-2x+1=0\)

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

\(\Leftrightarrow x\left(x-1\right)=0\)

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

Vậy: S={1}

b) ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

Ta có: \(\dfrac{5}{x-3}-\dfrac{2x-3}{x+3}=\dfrac{2x\left(1-x\right)}{x^2-9}\)

\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(2x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)

Suy ra: \(5x+15-2x^2+6x+3x-9-2x+2x^2=0\)

\(\Leftrightarrow12x+6=0\)

\(\Leftrightarrow12x=-6\)

hay \(x=-\dfrac{1}{2}\)(thỏa ĐK)

Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)

e tách ra nhé.☘

;))

\(\text{a)}\Rightarrow x-1-x-1-x+2=5\)

\(\Rightarrow-x=5\)

\(\Rightarrow x=-5\)

     \(\text{Vậy x=-5}\)

\(\text{b)}\left(2x-1\right)^2-\left(2x+3\right)^2=7\)

\(\Rightarrow\left(4x^2-4x+1\right)-\left(4x^2+12x+9\right)=7\)

\(\Rightarrow4x^2-4x+1-4x^2-12x-9=7\)

\(\Rightarrow-16x-8=7\)

\(\Rightarrow-16x=15\)

\(\Rightarrow x=\frac{-15}{16}\)

      \(\text{Vậy }x=\frac{-15}{16}\)

\(\text{c)}\Rightarrow16x^2-9-\left(16x^2-8x+1\right)=8\)

\(\Rightarrow-9+8x-1=8\)

\(\Rightarrow8x=18\)

\(\Rightarrow x=\frac{18}{8}=\frac{9}{4}\)

      \(\text{Vậy }x=\frac{9}{4}\)

\(\text{Phần d số rất lẻ, có thể bạn chép sai đề nên mình ko chữa nha~}\)

dmm

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

\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=2x^3+2\left(4x^2-4x+1\right)-9\)

\(\Leftrightarrow2x^3-3x^2+15x-7=2x^3+8x^2-8x-7\)

\(\Leftrightarrow-11x^2+23x=0\)

\(\Leftrightarrow x\left(-11x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{11}\end{matrix}\right.\)

a)

$(2x+1)^2-(2x+1)(2x-1)=(2x+1)[(2x+1)-(2x-1)]$

$=2(2x+1)$

b)

$(4x+3)(x-1)-2x(2x+1)=4x^2-x-3-4x^2-2x=-3x-3=-3(x+1)$

c)

$(2x+3)^2-(4x+1)(x+5)=(4x^2+12x+9)-(4x^2+21x+5)$

$=-9x+4$

d)

$(x+2)^3-(x-1)(x^2+x+1)=(x^3+6x^2+12x+8)-(x^3-1)$

$=6x^2+12x+9$

e)

$(x+2)(x^2-2x+1)-(x+3)(x-3)=(x^3-3x+2)-(x^2-9)$

$=x^3-x^2-3x+11$

f)

$(x+3)(x^2-3x+9)-(x^2+2x+4)(x-2)$

$=x^3+3^3-(x^3-2^3)=3^3+2^3=35$