`@` `\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)Câu a sai đề

b)x=4

k mik nha

Học tốt

viết lại đề câu a nha

1) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

\(\Leftrightarrow\left(2x-5\right).-2=0\)

\(\Leftrightarrow-4x+10=0\)

\(\Leftrightarrow-4x=-10\)

\(\Leftrightarrow x=\frac{5}{2}.\)

Vậy \(S=\left\{\frac{5}{2}\right\}\)

2)\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right).\left(x^2-3x+9+x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)

\(\Leftrightarrow\left(x+3\right).x.\left(x-2\right)=0\)

\(\Leftrightarrow x+3=0\)hoặc \(x=0\)hoặc \(x-2=0\)

\(\Leftrightarrow x=-3\)hoặc \(x=0\)hoặc \(x=2\)

Vậy \(S=\left\{-3;0;2\right\}\)

a) \(\left(x+2\right)^2-9=0\)

\(=>\left(x+2\right)^2-3^2=0\\ =>\left(x+2-3\right).\left(x+2+3\right)=0\)

\(=>\left(x-1\right).\left(x+5\right)=0\)

\(=>\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}=>\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

Vậy x= 1 hoặc x= -5

b) \(x^2-2x+1=25\)

\(=>x^2-2.x.x+1^2=25\)

\(=>\left(x-1\right)^2-25=0\\ =>\left(x-1\right)^2-5^2=0\)

\(=>\left(x-1-5\right).\left(x-1+5\right)=0\)

\(=>\left(x-6\right).\left(x+4\right)=0=>\orbr{\begin{cases}x-6=0\\x+4=0\end{cases}}\)

\(=>\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)

Vậy x= 6 hoặc x= -4

c) \(4x\left(x-1\right)-\left(2x+5\right)\left(2x-5\right)=1\)

\(=>4x\left(x-1\right)-\left[\left(2x\right)^2-5^2\right]=1\)

\(=>4x\left(x-1\right)-4x^2+25-1=0\)

\(=>4x\left(x-1\right)-4x^2+24=0\)

\(=>4x\left(x-1\right)-\left(4x^2-24\right)=0\\ =>4x\left(x-1\right)-4\left(x^2-6\right)=0\)

..................... tắc ròi -.-"

d) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+3\right)=15\)

\(=>x^3+27-x^3-3x=15\)

\(=>27-3x-15=0=>12-3x=0=>3\left(4-x\right)=0\)

Vì \(3>0=>4-x=0=>x=4\)

Vậy x= 4

e) \(3\left(x+2\right)^2+\left(2x+1\right)^2-7\left(x+3\right)\left(x-3\right)=28\)

\(=>3\left(x^2+2.x.2+2^2\right)+4x^2+4x+1-7\left(x^2-9\right)=28\)

\(=>3\left(x^2+4x+4\right)+4x^2+4x+1-7x^2+63=28\)

\(=>3x^2+12x+12+4x^2+4x+1-7x^2+63=28\)

\(=>16x+75=28=>16x=-47=>x=\frac{-47}{16}\)

Cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt :>'-'

Cảm ơn cậu nhiều nhé!