\(1\)) \(70:\frac{4x+720}{x}=\frac{1}{2}\)

\(\Leftrightarrow\frac{4x+720}{x}=70:\frac{1}{2}\)

\(\Leftrightarrow\frac{4x+720}{x}=140\)

\(\Leftrightarrow\left(4x+720\right):x=140\)

\(\Leftrightarrow4x+720=140.x\)

\(\Leftrightarrow4x-140x=-720\)

\(\Leftrightarrow x.\left(-136\right)=-720\)

\(\Leftrightarrow x=-720:\left(-136\right)\)

\(\Leftrightarrow x=\frac{90}{17}\)

\(2\)) Mình đang nghĩ

Bài 8:
a) Đk: x#2 (*)
Với (*), A=(x - 2 + 5)/(x - 2)= 1 + 5/(x - 2)
A nguyên <=> x-2 thuộc Ư(5)={-5;-1;1;5}
=> S={-3;1;3;7}
b) Đk: x#-3
Với (*), A= (- 2x - 6 + 7)/(x + 3) = -2 + 7/(x+3)
A nguyên <=> x + 3 thuộc Ư(7)={1;-1;7;-7}
=> S = {-2;- 4;4;-10}

) Đk: x#2 (*) 
Với (*), A=(x - 2 + 5)/(x - 2)= 1 + 5/(x - 2) 
A nguyên <=> x-2 thuộc Ư(5)={-5;-1;1;5} 
=> S={-3;1;3;7} 
b) Đk: x#-3 
Với (*), A= (- 2x - 6 + 7)/(x + 3) = -2 + 7/(x+3) 
A nguyên <=> x + 3 thuộc Ư(7)={1;-1;7;-7

\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\)\(\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\)\(\frac{1}{x+2}-\frac{1}{x-7}=\frac{x}{\left(x+2\right)\left(x+7\right)}\)

\(\Leftrightarrow\)\(\frac{x+7-x-2}{\left(x+2\right)\left(x+7\right)}=\frac{x}{\left(x+2\right)\left(x+7\right)}\)

\(\Leftrightarrow x=5\)

Bài 1:

a) (2x-3). (x+1) < 0

=>2x-3 và x+1 ngược dấu

Mà 2x-3<x+1 với mọi x

\(\Rightarrow\begin{cases}2x-3< 0\\x+1>0\end{cases}\)

\(\Rightarrow\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)\(\Rightarrow-1< x< \frac{3}{2}\)

b)\(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)

\(\Rightarrow x-\frac{1}{2}\) và x+3 cùng dấu

Xét \(\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\)\(\Rightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\)

Xét \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)\(\Rightarrow\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)

=>....

Bài 2:

\(S=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{999.1001}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{999}-\frac{1}{1001}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{1001}\right)\)

\(=\frac{1}{2}\cdot\frac{998}{3003}\)

\(=\frac{499}{3003}\)

tự làm nhé. bài cô Kiều cho dễ mừ :)

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

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

`9)(3x-2)(x+5)=0`

`<=>[(x=2/3),(x=-5):}`

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

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

`<=>[(x=1/5),(x=3),(x=-3):}`

`12)x(x-3)+3(x-3)=0`

`<=>(x-3)(x+3)=0`

`<=>[(x=3),(x=-3):}`

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

`<=>(x+5)(x-1)=0`

`<=>[(x=-5),(x=1):}`

\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

\(x^2-1>x^2-4>x^2-7>x^2-10\)

\(\text{Để }\left(x^2-1\right).\left(x^2-4\right).\left(x^2-7\right).\left(x^2-10\right)< 0\)

\(\Rightarrow\hept{\begin{cases}\left(x^2-1\right)>0\\\left(x^2-4\right).\left(x^2-7\right).\left(x^2-10\right)< 0\end{cases}\text{hoặc }\hept{\begin{cases}\left(x^2-1\right).\left(x^2-4\right).\left(x^2-7\right)>0\\\left(x^2-10\right)< 0\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}x^2>1\\x^2< 4\end{cases}\text{hoặc }\hept{\begin{cases}x^2>7\\x^2< 10\end{cases}}}\)

\(\Rightarrow x^2=9\Rightarrow x=\pm3\)

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Boul đẹp trai_tán gái đổ 100% Thử tán mị xem có đổ k mà ns??:Đ