Câu 3: 

a: Trên tia Ox, ta có: OA<OB

nên điểm A nằm giữa hai điểm O và B

=>OA+AB=OB

hay AB=2(cm)

b: Ta có: A nằm giữa hai điểm O và B

mà OA=AB

nên A là trung điểm của OB

c: Trên đoạn BO, ta có: BC<BO

nên điểm C nằm giữa hai điểm O và B

=>OC+CB=OB

hay OC=3cm

Trên tia Ox, ta có: OA<OC

nên điểm A nằm giữa hai điểm O và C

=>OA+AC=OC

hay AC=1cm

\(\dfrac{1}{2^2}>\dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\dfrac{1}{3^2}>\dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)

...

\(\dfrac{1}{100^2}>\dfrac{1}{100\cdot101}=\dfrac{1}{100}-\dfrac{1}{101}\)

Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{2}-\dfrac{1}{101}=\dfrac{99}{202}\)

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}=\dfrac{1}{99}-\dfrac{1}{100}\)

Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)

Suy ra: \(\dfrac{9}{202}< \dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{99}{100}\)

\(d,\dfrac{-25}{12}-\left(\dfrac{23}{12}+1\dfrac{1}{2}\right)=d,\left(\dfrac{-25}{12}-\dfrac{23}{12}\right)-\dfrac{3}{2}=\dfrac{-48}{12}-\dfrac{3}{2}=-4-\dfrac{3}{2}=\dfrac{-8}{2}-\dfrac{3}{2}=-\dfrac{11}{2}\\ e,\dfrac{-1}{9}.\dfrac{-3}{5}-\dfrac{5}{6}.\dfrac{3}{-5}+\dfrac{-5}{2}.\dfrac{3}{5}=\dfrac{1}{9}.\dfrac{3}{5}+\dfrac{5}{6}.\dfrac{3}{5}-\dfrac{5}{2}.\dfrac{3}{5}=\dfrac{3}{5}\left(\dfrac{1}{9}+\dfrac{5}{6}-\dfrac{5}{2}\right)=\dfrac{3}{5}.\dfrac{-14}{9}=\dfrac{-14}{15}\)

\(a)(-3/5)*x=-1/20+1/2=9/20=>x=9/20:(-3/5)=-3/4\)

Các câu kia làm tương tự nhé, chúc em học giỏi

a: =>-3/5x=-1/20+1/2=-1/20+10/20=-9/20

=>x=3/4

b: =>-1/15x-2/15=3/5

=>-1/15x=6/15+2/15=8/15

=>x=-8

c: \(\Leftrightarrow\left(2x-1\right)\left(x-3\right)\left(x+3\right)=0\)

hay \(x\in\left\{\dfrac{1}{2};3;-3\right\}\)

Bài 3:

\(a,\left(5x-2\right)+\left(-3x+1\right)=\left(-42\right)-\left(-91\right)\\ \Rightarrow5x-2+\left(-3x\right)+1=50\\ \Rightarrow2x-1=49\\ \Rightarrow2x=50\\ \Rightarrow x=25\\ b,\left(3-x\right)\left(9+3x\right)=0\\ \Rightarrow\left[{}\begin{matrix}3-x=0\\9+3x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\3x=-9\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(c,5x^2-\left(-6\right)=\left(-33\right)-\left(-44\right)\\ \Rightarrow5x^2+6=11\\ \Rightarrow5x^2=5\\ \Rightarrow x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

\(d,2\left(2x-4\right)^2-77=-45\\ \Rightarrow2\left(2x-4\right)^2=32\\ \Rightarrow\left(2x-4\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}2x-4=-4\\2x-4=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=0\\2x=8\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

bài 3

c] 5x mũ 2 -[-6] =[-33] - [-44]

cho xin một like đi

\(=\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{50}+\dfrac{1}{51}-\dfrac{102}{51\cdot52}\)

\(=\dfrac{1}{2}+\dfrac{1}{51}-\dfrac{102}{51\cdot52}\)

\(=\dfrac{1}{2}+\dfrac{52-102}{51\cdot52}=\dfrac{1}{2}+\dfrac{-50}{51\cdot52}=\dfrac{319}{663}\)

a ) 13 . 35 + 13 . 67 - 26

= 13 . ( 35 + 67 ) - 26

= 13 . 102 - 26

= 1326 - 26

= 1300

b) ( 2⁵ .12 + 4 . 2⁵ ) : 2⁷

= [ 2⁵ . ( 12 + 4 ) ] : 2⁷

=  2⁵ . 16 : 2⁷

=  2⁵ . 2⁴ : 2⁷

= 2⁹ : 2⁷

= 2² = 4

c) 2³ . [ 50 - ( 17 - 54 : 3³ )] + 16

= 8 .  [ 50 - ( 17 - 54 : 27 )] + 16

=  8 . [ 50 - ( 17 - 2 )] + 16

= 8 . [ 50 - 15 ] + 16

= 8 . 35 + 16

= 280 + 16

= 296

                                    K MK NHA