\(\left(x+3\right)\left(1-x\right)>0.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3>0.\\1-x>0.\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0.\\1-x< 0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-3.\\x< 1.\end{matrix}\right.\\\left\{{}\begin{matrix}x< -3.\\x>1.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow-3< x< 1.\)

\(\left(x^2-1\right)\left(x^2-4\right)< 0.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-1< 0.\\x^2-4>0.\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-1>0.\\x^2-4< 0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2< 1.\\x^2>4.\end{matrix}\right.\\\left\{{}\begin{matrix}x^2>1.\\x^2< 4.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1.\\x>-1.\end{matrix}\right.\\\left[{}\begin{matrix}x>2.\\x< -2.\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1.\\x< -1.\end{matrix}\right.\\\left[{}\begin{matrix}x< 2.\\x>-2.\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-1< x< 1.\\\left[{}\begin{matrix}x>2.\\x< -2.\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1.\\x< -1.\end{matrix}\right.\\-2< x< 2.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>2.\\x< -2.\\-2< x< -1.\\1< x< 2.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x< -2.\\x>2.\end{matrix}\right.\)

\(\sqrt{1-\sqrt{x^4-x^2}}=x-1\)

\(\sqrt{1-\left|x^2\right|-\left|x\right|}=x-1\)

\(\sqrt{1-x^2-x}=x-1\)

\(x\sqrt{1-x}=x-1\)

\(\sqrt{1-x}=\frac{x-1}{x}\)

\(1-x=\left(\frac{x-1}{x}\right)^2\)

\(1-x=\frac{x^2-1}{x^2}\)

\(1-x=-1\)

\(x=2\)

vay \(x=2\)

\(x=2\)

ĐK: \(\orbr{\begin{cases}x\ge1\\x\le-1\end{cases}}\) và \(1-\sqrt{x^2\left(x^2-1\right)}\ge0\)

Kết hợp điều kiện có nghiệm \(x-1\ge0\) ta có \(1\le x\le\sqrt{\frac{1+\sqrt{5}}{2}}\)

\(\sqrt{1-\sqrt{x^4-x^2}}=x-1\Leftrightarrow1-x\sqrt{x^2-1}=x^2-2x+1\)

\(\Rightarrow x\left(2-x\right)=x\sqrt{x^2-1}\Rightarrow2-x=\sqrt{x^2-1}\) (Vì \(x\ne0\) )

\(\Rightarrow x^2-4x+4=x^2-1\Rightarrow x=\frac{5}{4}\left(tm\right)\)

( x + 1 )² = 9

\(\Rightarrow\orbr{\begin{cases}\left(x+1\right)^2=3^2\\\left(x+1\right)^2=\left(-3\right)^2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x+1=3\\x+1=-3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)

(x-1)^2=9  
<=>(x-1)^2 - 3^2 = 0 
<=>(x-1 +3) ( x-1 -3) = 0 
<=> (x+2) .(x-4) = 0 
<=> x= -2 hay x = 4 

Vậy.............

 (x+1)^2=3^2   

=> x+1=3 =>x=2

a)

`2/3+5/2-3/4`

`=10/4-3/4+2/3`

`=7/4+2/3`

`=21/12+8/12`

`=29/12`

b)

`2/5xx1/2:1/3`

`=2/10xx3/1`

`=6/10=3/5`

c)

`2/9:2/9xx1/3`

`=2/9xx9/2xx1/3`

`=1xx1/3`

`=1/3`

a, \(\dfrac{2}{3}\) + \(\dfrac{5}{2}\) - \(\dfrac{3}{4}\)

= \(\dfrac{8}{12}\) + \(\dfrac{30}{12}\) - \(\dfrac{9}{12}\)

= \(\dfrac{38-9}{12}\)

= \(\dfrac{29}{12}\)

b, \(\dfrac{2}{5}\) x \(\dfrac{1}{2}\) : \(\dfrac{1}{3}\)

= \(\dfrac{1}{5}\) x \(\dfrac{3}{1}\)

= \(\dfrac{3}{5}\)

c, \(\dfrac{2}{9}\) : \(\dfrac{2}{9}\) x \(\dfrac{1}{3}\)

= 1 x \(\dfrac{1}{3}\)

= \(\dfrac{1}{3}\)

chưa ai trả lời à!!!

A=100+(98-97)-(96-95)-(2-1)

A=100+1-1-1

A=97

A = 100 + 98 + 96 + ... + 2 - 97 - 95 - ... - 1

A = 100 + (98 - 97) + (96 - 95) + ... + (2 - 1)

A = 100 + 1 + 1 + ... + 1 (49 số 1)

A = 100 + 49 = 149

          Q = 14 . 29 + 14 . 71 + ( 1 + 2 + 3 + ... + 99)(199199 . 198 - 198198 . 199)

= 14 . ( 29 + 71 ) + ( 1 + 2 + ... + 99)( 199 . 1001 . 198 - 198 . 1001 . 199 )

= 14 . 100 + ( 1  + 2 + ... + 99) . 0

= 1400 + 0

= 1400

\(Q=14.29+14.71+\left(1+2+3+4+....+99\right).\left(199199.198-198198.199\right)\)

\(=14.\left(29+71\right)+\left(1+2+3+4+..+99\right).\left(199.101.198-198.1001.199\right)\)

\(=14.100+\left(1+2+3+4+...+99\right).0\)

\(=1400+0\)

\(=1400\)

cảm ơn hai nhìu nha