a ) 

\(5x\left(x-3\right)+7\left(x-3\right)=0\)

\(\Rightarrow\left(5x+7\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x+7=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-7\\x=3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{7}{5}\\x=3\end{cases}}}\)

Vậy ...

b ) 

\(x^{2017}=x^{2018}\)

\(\Rightarrow x^{2017}-x^{2018}=0\)

\(\Rightarrow x^{2017}\left(1-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^{2017}=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

Vậy ...

c ) 

\(2x^2=x\)

\(\Rightarrow2x^2:x=1\)

\(\Rightarrow2x=1\)

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

Vậy ...

e ) 

\(x^5=x^4\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)( làm tương tự như phần b )  

~ 

d ) 

\(x^3+x^2=0\)

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

\(\Rightarrow\orbr{\begin{cases}x^2=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

Vậy ...

a)

<=> 10x - 35 + 16x - 10 = 5 

<=> 10x + 16x = 5 + 35 + 10

<=> 26x = 50

<=> x = 50/26 = 25/13

a: \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)

\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)

\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)

=>16x-7=13x+2

=>3x=9

hay x=3

b: \(\dfrac{x+1}{2016}+\dfrac{x}{2017}=\dfrac{x+2}{2015}+\dfrac{x+3}{2014}\)

\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x}{2017}+1\right)=\left(\dfrac{x+2}{2015}+1\right)+\left(\dfrac{x+3}{2014}+1\right)\)

=>x+2017=0

hay x=-2017

e: \(\left(2x-3\right)^2=144\)

=>2x-3=12 hoặc 2x-3=-12

=>2x=15 hoặc 2x=-9

=>x=15/2 hoặc x=-9/2

a,  x - 3 : 2 = 5 14 : 5 12

=>  x - 3 : 2 = 5 2

=>  x - 3 : 2 = 25

=> x – 3 = 25

=> x = 53

b,  30 : x - 7 = 15 19 : 15 18

=>  30 : x - 7 = 15

=> x – 7 = 2

=> x = 9

c,  x 70 = x

=>  x 70 - x = 0

=>  x ( x 69 - 1 ) = 0

=> 

d,  2 x + 1 3 = 9 . 81

=>  2 x + 1 3 = 9 3

=> 2x + 1 = 9

=> x = 4

e,  5 x + 5 x + 2 = 650

=>  5 x 1 + 5 2 = 650

=>  5 x . 26 = 650

=>  5 x = 25

=> x = 2

f,  4 x - 1 2 = 25 . 9

=> 4 x - 1 2 = 5 2 . 3 2

=>  4 x - 1 2 = 15 2

=> 4x – 1 = 15

=> x = 4

Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1 

=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1 

=> (15x - 8x) + (25 + 12) = 11x + 37 

=> 7x + 37 = 11x + 37 

=> 11x - 7x = 0 

=>  x = 0 

a, \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)

\(\Leftrightarrow x^2-4x+4-\left(x^2+6x+9\right)-4x-4=5\)

\(\Leftrightarrow x^2-4x+4-x^2-6x-9-4x-4=5\)

\(\Leftrightarrow-14x-9=5\)

\(\Leftrightarrow-14x=14\)

\(\Leftrightarrow x=-1\)

Vậy....

b, \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(\Leftrightarrow\left(2x\right)^2-3^2-\left(x^2-2x+1\right)-3x^2+15x=-44\)

\(\Leftrightarrow4x^2-9-x^2+2x-1-3x^2+15x=-44\)

\(\Leftrightarrow-10+17x=-44\)

\(\Leftrightarrow17x=-34\)

\(\Leftrightarrow x=-2\)

Vậy....

c, \(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)

\(\Leftrightarrow\left(5x\right)^2+10x+1-\left[\left(5x\right)^2-3^2\right]=30\)

\(\Leftrightarrow\left(5x\right)^2+10x+1-\left(5x\right)^2+9=30\)

\(\Leftrightarrow10x+10=30\)

\(\Leftrightarrow10x=20\)

\(\Leftrightarrow x=2\)

Vậy....

d, \(\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-2\right)^2=7\)

\(\Leftrightarrow x^2+6x+9+x^2-4-2\left(x^2-4x+4\right)=7\)

\(\Leftrightarrow2x^2+6x+5-2x^2+8x-8=7\)

\(\Leftrightarrow14x-3=7\)

\(\Leftrightarrow14x=10\)

\(\Leftrightarrow x=\frac{10}{14}=\frac{5}{7}\)

Vậy...

câu hỏi đây

bn Nhật Hạ lm đúng r đấy

conkia

bai dai dong qua

a) (x-2)^3-x(x+1)(x-1)+6x(x-3)=0

\(x^3-6x^2+12x-8-x\left(x^2-1\right)+6x\left(x-3\right)=0\)

\(x^3-6x^2+12x-8-x^3+x+6x^2-18x=0\)

\(-5x-8=0\)

\(x=-\frac{8}{5}\)

Mai mik làm mấy bài kia sau

2/

b) ( cái bài này chịu)

c) (x+1)^3-(x-1)^3-6(x-1)^2=-10

(x+1-x+1)\(\left[\left(x+1\right)^2+\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]\)\(-6\left(x^2-2x+1\right)=-10\)

\(2\left(x^2+2x+1+x^2-1+x^2-2x+1\right)-6x^2+12x-6=-10\)

\(2\left(3x^2+1\right)-6x^2+12x-6=0\)

\(6x^2+2-6x^2+12x-6=-10\)

\(12x=-10+4\)

\(12x=-6=>x=-\frac{1}{2}\)

d) (5x-1)^2-(5x-4)(5x+4)=7

\(25x^2-10x+1-25x^2+16=7\)

-10x = 7 - 17

-10x = -10

x= 1

Câu còn lại bn làm tương tự

3/

a) 

Ta có: 

(a+b+c)^2=3(ab+bc+ca)

a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 3ab + 3bc + 3ac

a^2 + b^2 + c^2 + 2ab + 2ac + 2bc - 3ab - 3bc - 3ac = 0

a^2 + b^2 + c^2  - ac - bc - ab = 0

2a^2 + 2b^2 + 2c^2  - 2ac - 2bc - 2ab = 0

(a²-2ab+b²⁾+(a²-2ac+c²) + (b²-2bc +c²) = 0

(a-b)^2 + (a-c)^2 + (b-c)^2 =0

=> a=b=c