`1,(4x^3+3x^3):x^3+(15x^2+6x):(-3x)=0`

`<=> 4 + 3 + (-5x) + (-2)=0`

`<=> -5x+5=0`

`<=>-5x=-5`

`<=>x=1`

`2,(25x^2-10x):5x +3(x-2)=4`

`<=> 5x - 2 + 3x-6=4`

`<=> 8x -8=4`

`<=> 8x=12`

`<=>x=12/8`

`<=>x=3/2`

`3,(3x+1)^2-(2x+1/2)^2=0`

`<=> [(3x+1)-(2x+1/2)][(3x+1)+(2x+1/2)]=0`

`<=>( 3x+1-2x-1/2)(3x+1+2x+1/2)=0`

`<=>( x+1/2) (5x+3/2)=0`

`@ TH1`

`x+1/2=0`

`<=>x=0-1/2`

`<=>x=-1/2`

` @TH2`

`5x+3/2=0`

`<=> 5x=-3/2`

`<=>x=-3/2 : 5`

`<=>x=-15/2`

`4, x^2+8x+16=0`

`<=>(x+4)^2=0`

`<=>x+4=0`

`<=>x=-4`

`5, 25-10x+x^2=0`

`<=> (5-x)^2=0`

`<=>5-x=0`

`<=>x=5`

Bài 2: Tìm x

a)ĐKXĐ: \(x\ne0\)

Ta có: \(\left(4x^4+3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)

\(\Leftrightarrow\frac{-x^3\left(4x+3\right)}{x^3}+\frac{3x\left(5x+2\right)}{3x}=0\)

\(\Leftrightarrow-4x-3+5x+2=0\)

\(\Leftrightarrow x-1=0\)

hay x=1(nhận)

Vậy: x=1

b) ĐKXĐ: \(x\notin\left\{0;\frac{1}{3}\right\}\)

Ta có: \(\left(x^2-12x\right):2x-\left(3x-1\right)^2:\left(3x-1\right)=0\)

\(\Leftrightarrow\frac{x\left(x-12\right)}{2x}-\frac{\left(3x-1\right)^2}{\left(3x-1\right)}=0\)

\(\Leftrightarrow\frac{x-12}{x}-3x+1=0\)

\(\Leftrightarrow\frac{x-12}{x}=3x-1\)

\(\Leftrightarrow x-12=x\left(3x-1\right)\)

\(\Leftrightarrow3x^2-x+x-12=0\)

\(\Leftrightarrow3x^2-12=0\)

\(\Leftrightarrow3x^2=12\)

\(\Leftrightarrow x^2=4\)

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

Vậy: \(x\in\left\{2;-2\right\}\)

a) 6x(5x + 3) + 3x(1 – 10x) = 7  

⇒ 30x²+18x+3x-30x²=7

⇒21x=7

⇒x=\(\dfrac{7}{21}\)

⇒x= \(\dfrac{1}{3}\)

b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44

⇒15x-63x²-15+63x + 63x²-35x+36x-20=44

⇒79x-35=44

⇒79x=44+35

⇒79x=79

⇒x=1

d) 5x(12x + 7) – 3x(20x – 5) = - 100

⇒60x²+35x-60x²+15=-100

⇒35x+15=-100

⇒35x=-100-15

⇒35x=-115

⇒x=\(\dfrac{-115}{35}\)

⇒x=\(\dfrac{-23}{7}\)

Bài 1:

a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)

\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)

b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)

c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)

e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Bài 2:

a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

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

Đây nhé! Tích giúp mình nha

Ta có: \(\dfrac{4x^4+3x^3}{-x^3}+\dfrac{15x^2+6x}{3x}=0\)

\(\Leftrightarrow-4x-3+5x+2=0\)

\(\Leftrightarrow x-1=0\)

hay x=1

\(1.6x\left(x-10\right)-2x+20=0\)

⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)

⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)

⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)

KL....

\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)

⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)

⇔ \(x=+-1\) hoặc \(x=3\)

KL....

\(3.x^2-8x+16=2\left(x-4\right)\)

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

⇔ \(\left(x-4\right)\left(x-6\right)=0\)

⇔ \(x=4\) hoặc \(x=6\)

KL.....

\(4.x^2-16+7x\left(x+4\right)=0\)

\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)

⇔ \(x=-4hoacx=\dfrac{1}{2}\)

KL.....

\(5.x^2-13x-14=0\)

⇔ \(x^2+x-14x-14=0\)

\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)

\(\text{⇔}x=14hoacx=-1\)

KL......

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