`a)[2x-3]/4-[4x-5]/3=[5-x]/6`

`<=>3(2x-3)-4(4x-5)=2(5-x)`

`<=>6x-9-16x+20=10-2x`

`<=>8x=1`

`<=>x=1/8`

(`b->c` mk đã làm r).

1.

a, => 21-x+3 < 0

=> 24-x < 0

=> x < 24

b, => 7+x > 0

=> x > -7

c, => x-1 < 0 ; x+2 > 0 ( vì x-1 < x+2 )

=> x < 1 ; x > -2

=> -2 < x < 1

Tk mk nha

Bài 3: 

a: x(x-1)=0

=>x=0 hoặc x-1=0

=>x=0 hoặc x=1

b: (x-3)(x+4)=0

=>x-3=0 hoặc x+4=0

=>x=3 hoặc x=-4

c: (2x-4)(x+2)=0

=>2x-4=0 hoặc x+2=0

=>x=2 hoặc x=-2

d: (x+1)²(x-2)²=0

=>x+1=0 hoặc x-2=0

=>x=-1 hoặc x=2

Bài 2: - Xét dấu :

P1 : (-).(+).(-).(-) -> Kết quả cuối cùng là số âm.

P2 : (-).(-).(-).(-).(+) -> Kết quả cuối cùng là số dương.

===> P1 < P2.

Bài 3 :

a) \(x\cdot\left(x-1\right)=0\)

\(\Rightarrow\left[\begin{matrix}x=0\\x-1=0\end{matrix}\right.\rightarrow\left[\begin{matrix}x=0\\x=1\end{matrix}\right.\)

b) \(\left(x-3\right)\cdot\left(x+4\right)=0\)

\(\left[\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\rightarrow\left[\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

c) \(\left(2x-4\right)\cdot\left(x+2\right)=0\rightarrow\left[\begin{matrix}2x-4=0\\x+2=0\end{matrix}\right.\rightarrow\left[\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

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

e) \(x\cdot\left(x+1\right)\cdot\left(x+2\right)^2\cdot\left(x+3\right)^3=0\)

\(\Rightarrow\left[\begin{matrix}x=0\\x+1=0\\x+2=0\\x+3=0\end{matrix}\right.\rightarrow\left[\begin{matrix}x=0\\x=-1\\x=-2\\x=-3\end{matrix}\right.\)

f) \(\left(x-9^5\right)\cdot\left(x-5\right)^8=0\)

\(\Rightarrow\left[\begin{matrix}x-9=0\\x-5=0\end{matrix}\right.\rightarrow\left[\begin{matrix}x=9\\x=5\end{matrix}\right.\)

g) \(x\cdot\left(x+100\right)^{10}\cdot\left(x+2000\right)^{20}\cdot\left(x+300\right)^{3000}=0\)

\(\Rightarrow\left[\begin{matrix}x=0\\x+100=0\\x+2000=0\\x+300=0\end{matrix}\right.\rightarrow\left[\begin{matrix}x=0\\x=-100\\x=-2000\\x=-300\end{matrix}\right.\)

h) \(\left(x-2\right)^2=0\rightarrow x=2\)

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

1

b;

B=1+ (7-5) + (11-9) + ...+(101-99)

B=1+2+2+..+2

B=1+25.2=51

2.

a.

ĐK : x+2 >=0 => x>=-2

\(\left|x+2\right|-x=2\\ \Rightarrow\left|x+2\right|=2+x\\ \Rightarrow\left[{}\begin{matrix}x+2=x+2\\x+2=-x-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}0x=0\\2x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}0x=0\\x=-2\end{matrix}\right.\)

Vậy x=-2

2.

d;

\(\left|x-5\right|\)=14-x

\(\Leftrightarrow\left[{}\begin{matrix}x-5=14-x\\x-5=x-14\end{matrix}\right.\)

em giải 2 cái này ra để tìm x

\(\text{a) x. (x + 2)= 0}\)

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

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

vậy_____

\(d.\left(x-5\right)\left(x^2+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x^2+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x\in\varnothing\end{cases}}\)

Mình làm mẫu câu a còn các câu khác tương tự nha

a, x.(x+2) = 0

=> x=0 hoặc x+2=0

=> x=0 hoặc x=-2

Vậy x thuộc {-2;0}

Tk mk nha