`@` `\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?

*vn:vô nghiệm.

a. \(\left(x^2-2\right)\left(x^2+x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\pm\sqrt{2}\)

-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).

b. \(16x^2-8x+5=0\)

\(\Leftrightarrow16x^2-8x+1+4=0\)

\(\Leftrightarrow\left(4x-1\right)^2+4=0\) (vô lí)

-Vậy S=∅.

c. \(2x^3-x^2-8x+4=0\)

\(\Leftrightarrow x^2\left(2x-1\right)-4\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)

-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).

d. \(3x^3+6x^2-75x-150=0\)

\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)

-Vậy \(S=\left\{-2;\pm5\right\}\)

\(A=6x^2-2x-6x^2-6x+3+8x\)

\(A=\left(6x^2-6x^2\right)+\left(-2x-6x+8x\right)+3\)

\(A=3\)

A = 2x.(3x – 1) – 6x.(x + 1) – (3 – 8x)

A = 2x.3x – 2x.1 – 6x.x – 6x.1 – 3 + 8x

A =  – 2x –  – 6x – 3 + 8x

A = -3
Sai đệ không chịu trách nhiệm đâu nha sư huynh :)))

Ta có: 2x(3x − 1) − 6x(x + 1) + (3 + 8x) 

= 6x^2 - 2x - 6x^2 - 6x + 3 + 8x

= (6x^2 - 6x^2) + (-2x - 6x + 8x) + 3

= 0 + 0 + 3 = 0

* \(2x\left(12x-5\right)-8x\left(3x-1\right)=30\Leftrightarrow24x^2-10x-24x^2+8x=30\) \(\Leftrightarrow-10x+8x=30\Leftrightarrow-2x=30\Leftrightarrow x=\dfrac{30}{-2}=-15\) vậy \(x=-15\)

* \(3x\left(3-2x\right)+6x\left(x-1\right)=15\Leftrightarrow9x-6x^2+6x^2-6x=15\)

\(\Leftrightarrow9x-6x=15\Leftrightarrow3x=15\Leftrightarrow x=\dfrac{15}{3}=5\) vậy \(x=5\)

\(a,2x\left(12x-5\right)-8x\left(3x-1\right)=30\)

\(\Leftrightarrow24x^2-10x-24x^2+8x=30\)

\(\Leftrightarrow-2x=30\)

\(\Leftrightarrow x=-15\)

\(b,3x\left(3-2x\right)+6x\left(x-1\right)=15\)

\(\Leftrightarrow9x-6x^2+6x^2-6x=15\)

\(\Leftrightarrow3x=15\)

\(\Leftrightarrow x=-5\)

a) \(2x\left(12x-5\right)-8x\left(3x-1\right)=30\)

\(\Leftrightarrow24x^2-10x-24x^2+8x=30\)

\(\Leftrightarrow-10x+8x=30\Leftrightarrow-2x=30\Leftrightarrow x=\dfrac{30}{-2}=-15\)

vậy \(x=-15\)

b) đề có sai o bn

a) đề = 24x^2-10x-24x^2+8x=30

= -10x+8x =30

\(\Rightarrow\)x=-15

b)đề \(\Leftrightarrow\)9x-6x^2=6x^2-6x=15

\(\Leftrightarrow\)9x-6x^2-6x^2+6x=15

\(\Leftrightarrow\)15x=15

\(\Rightarrow\)x=1

a,\(x\in\left\{5;1,5;\dfrac{-4}{3}\right\}\)

một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?

tìm nghiệm nha mọi người quên chưa nói

a) \(x^2+2x+3=0\)

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

\(\Rightarrow x^2+2x=-3\)

\(\Rightarrow x^2+2x+1=-3+1\)

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

Điều này là vô lý vì bình phương của 1 số luôn lớn hơn hoặc bằng 0 mà -2 < 0.

Vậy đa thức vô nghiệm.

b) \(x^2+6x+5=0\)

\(x^2+x+5x+5=0\)

\(x\left(x+1\right)+5\left(x+1\right)=0\)

\(\left(x+5\right)\left(x+1\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=-5\\x=-1\end{cases}}\)