a) \(x=0,05\)

b) \(x=1,125\)

c) \(x=0,96\)

d) \(x=0,025\)

Bạn tự làm đi dễ mà . Cố mag vận động đầu óc đừng copy làm bài nữa khó lắm mới hỏi thôi

Chuẩn đấy

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

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

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

- Mình có thể giúp bài 1 :)

 5^x +  5^ ( x + 2 ) = 650

5^x +  5^x . 5² = 650

 5^x .( 1 + 25 ) = 650

 5^x . 26 = 650 

5^x = 650 : 26

 5^x = 25

 5^x = 5² 

=> x = 2 

Vậy x = 2 

The Reflection Of Light~ Ủa đề  là 5^x-5^x+2= 650 sao ông lại đổi thành 5^x-5^(x+2) zậy

Bài 1L

a) \(\left(x-7\right)\left(x+3\right)< 0\)

TH1:

\(\hept{\begin{cases}x-7>0\\x+3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>7\\x< -3\end{cases}}}\)( loại )

TH2:

\(\hept{\begin{cases}x-7< 0\\x+3>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 7\\x>-3\end{cases}\Leftrightarrow}-3< x< 7}\)( chọn )

Vậy \(-3< x< 7\)

Bài 2:

a) \(\left(5x+8\right)-\left(2x-15\right)+21=2x-5\)

\(\Leftrightarrow5x+8-2x+15+21=2x-5\)

\(\Leftrightarrow5x-2x-2x=-5-21-8-15\)

\(\Leftrightarrow x=-49\)

Vậy ...

Bài 1:

b) \(\left(2x+6\right)\left(x-5\right)\le0\)

TH1:

\(\hept{\begin{cases}2x+6\le0\\x-5\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le-3\\x\ge5\end{cases}}}\)( loại )

TH2:

\(\hept{\begin{cases}2x+6\ge0\\x-5\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge-3\\x\le5\end{cases}\Leftrightarrow}-3\le x\le5}\)

Vậy \(-3\le x\le5\)

`2//(5x-8)-3(4x-5)=4(3x-4)`

`<=>5x-8-12x+15=12x-16`

`<=>-19x=-23`

`<=>x=23/19`     Vậy `x=23/19`

`3//2(x^3-1)-2x^2(x+2x^4)+(4x^5+4)x=6`

`<=>2x^3-2-2x^3-4x^6+4x^6+4x=6`

`<=>4x=8`

`<=>x=2`     Vậy `x=2`

`A`

1) \(\dfrac{3x}{4x-8}\)

\(ĐKXĐ:4x-8\ne0\Leftrightarrow x\ne2\)

2) \(\dfrac{2x}{x^2-9}\)

\(ĐKXĐ:x^2-9\ne0\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

3) \(\dfrac{6}{x^3+1}=\dfrac{6}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(ĐKXĐ:\)\(x+1\ne0\Leftrightarrow x\ne-1\)

(do \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))

4) \(\dfrac{6x^2}{x^2-2x+1}=\dfrac{6x^2}{\left(x-1\right)^2}\)

\(ĐKXĐ:x-1\ne0\Leftrightarrow x\ne1\)

5) \(\dfrac{x-2}{x^2+3}\)

Do \(x^2+3>0\forall x\in R\)

Vậy biểu thức trên xác định với mọi x

6) \(\dfrac{2x}{x^2+3x+2}=\dfrac{2x}{\left(x+1\right)\left(x+2\right)}\)

\(ĐKXĐ:\)\(\left\{{}\begin{matrix}x+1\ne0\\x+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne-2\end{matrix}\right.\)

\(2x+\left(1+2+3+...+100\right)=15150\)

\(2x+\left[\left(1+100\right)+\left(2+99\right)+...+\left(50+51\right)\right]=15150\)

\(2x+\left[101+101+...+101\right]=15150\)CÓ 50 SỐ 101

\(2x+\left[101\times50\right]=15150\)

\(2x=15150:5050\)

\(2x=3\)

\(x=3:2\)

\(x=1.5\)

a, 2x + (1+2+3+4+...+100) = 15150 

=> 2x + \(\frac{\left(1+100\right).\left[\left(100-1\right)+1\right]}{2}\)= 15150 

=> 2x + \(\frac{101.100}{2}\)= 15150 

=> 2x + 5050 = 15150 

=> 2x             = 15150 - 5050 

=> 2x             = 10100

=> x              =  10100 : 2 

=> x              = 5050 

Vậy x = 5050 

b, .(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)+(x+8)=36 

=> (x + x + x + x +x + x +x +x ) + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 36 

=> 8x + 36 = 36 

=> 8x         = 0 

=>  x          = 0 

Vậy x = 0 

c, 0+0+4+6+8+...+2x=110 

Sửa đề :0 + 2 + 4 + 6 + 8 + ... + 2x = 110 = 2 + 4 + 6 + 8 + ... + 2x = 110 

SSH  : \(\frac{\left(2\text{x}-2\right)}{2}+1=x-1+1=x\)

Tổng : \(\frac{\left(2\text{x}+2\right).x}{2}=110\Leftrightarrow\frac{2.\left(x+1\right).x}{2}=110\)

                                                    \(\Leftrightarrow\left(x+1\right)x=110\)

                                                     \(\Leftrightarrow\left(10+1\right).10=110\)

 => x = 10 

Vậy x = 10 

#)Giải :

a) 2x + ( 1 + 2 + 3 + 4 + ... + 100 ) = 15150

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

=> 2x + 5050 = 15150

=> 2x = 10100

=> x = 5050

b) ( x + 1 ) + ( x + 2 ) + ... + ( x + 8 ) + ( x + 9 ) = 36

=> ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 9 ) = 36

=> ( x + x + x + ... + x ) + 45 = 36

=> ( x + x + x + ... + x ) = -9

=> x = -1

c) Đề kiểu j v ? xem lại đi bạn ???

a) Ta có: \(\dfrac{1}{4}-\left|x+\dfrac{1}{2}\right|=\dfrac{1}{8}\)

\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{1}{8}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{8}\\x+\dfrac{1}{2}=-\dfrac{1}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{8}\\x=\dfrac{-5}{8}\end{matrix}\right.\)

b) Ta có: \(\left|2x-5\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=x+3\left(x\ge\dfrac{5}{2}\right)\\2x-5=-x-3\left(x< \dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=3+5\\2x+x=-3+5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=\dfrac{2}{3}\left(nhận\right)\end{matrix}\right.\)

a)125 : x = 2² - (-1)

125 : x = 4 + 1

125 : x = 5

x = 125 : 5

x = 25

-------------------------------------------------

b) 2x - 8 = -4

2x = (-4) + 8

2x = 4

x = 4 : 2 

x = 2

-----------------------------------------------------------

c) Xem lại đề.

\(125:x=2^2-\left(-1\right)\)

\(=>125:x=4+1\)

\(=>125:x=5\)

\(=>x=125:5\)

\(=>x=25\)

_____

\(2x-8=-4\)

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

\(=>2x=4\)

\(=>x=4:2\)

\(=>x=2\)

_______

\(6^{2x+5}=216\)

\(=>6^{2x+5}=6^3\)

\(=>2x+5=3\)

\(=>2x=3-5\)

\(=>2x=-2\)

\(=>x=\left(-2\right):2\)

\(=>x=-1\)

\(#NqHahh\)

125:x=2² - (-1)

125:x=4+1

125:x=5

X=125:5

X=25

2x -8 = -4

2x=-4+8

2x=4

X=4:2

X=2

6²(x+5)  = 216

36(x+5) = 216

X+5 = 216:36

X+5=6

X=6-5

X=1

Theo đề ra, ta có: \(\frac{x-1}{y+2}=\frac{3}{5}\)

\(\Rightarrow\frac{x-1}{3}=\frac{y+2}{5}=\frac{x-1+y+2}{8}=\frac{23-1+2}{8}=\frac{24}{8}=3\)

\(\frac{x-1}{3}=3\Rightarrow x=3.3+1=10\)

\(\frac{y+2}{5}=3\Rightarrow y=5.3-2=13\)