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

a: =>2x-3-3x+1=34

=>-x-2=34

=>-x=36

hay x=-36

b: =>x+1=0 hoặc 5-5x=0

=>x=-1 hoặc x=1

c: \(\Leftrightarrow\left(2x+1\right)^3=45+19=64\)

=>2x+1=4

=>2x=3

hay x=3/2

d: \(\Leftrightarrow4x^2=16\)

=>x=2 hoặc x=-2

a) \(\left(2x-3\right)+\left(-3x+1\right)=34\)

\(2x-3-3x+1=34\)

\(-x=36\)

\(x=-36\)

b) \(\left(x+1\right)\left(5-5x\right)=0\)

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

c) \(\left(2x+1\right)^3=49-\left(-19\right)=64=4^3\)

\(2x+1=4\)

\(x=\dfrac{3}{2}\)

d) \(4x^2-3=13\)

\(4x^2=16\)

\(x^2=4\)

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

a) \(\left(2x-3\right)\left(-3x+1\right)=34\)

\(\left[{}\begin{matrix}2x-3=34\\-3x+1=34\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{37}{2}\\x=-11\end{matrix}\right.\)

b) \(\left(x+1\right)\left(5-5x\right)=0\)

\(\left[{}\begin{matrix}x+1=0\\5-5x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-1\end{matrix}\right.\)

c) \(\left(2x+1\right)^3=45-\left(-19\right)\)

\(\left(2x+1\right)^3=64\)

\(\left(2x+1\right)^3=2^6\)

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

a: \(=\left(6+\dfrac{1}{3}+2+\dfrac{2}{3}\right)+\left(-3-\dfrac{2}{5}-1-\dfrac{3}{5}\right)+4=9-5+4=8\)

b: \(=\dfrac{5}{6}-\dfrac{1}{9}+\dfrac{4}{5}-\dfrac{4}{6}-\dfrac{7}{9}+\dfrac{3}{5}+\dfrac{3}{5}-\dfrac{1}{9}-\dfrac{1}{6}\)

=-1+2=1

Vì AB+BC>AC; AC+BC>AB; AB+AC>BC

nên ba điểm A,B,C không thẳng hàng

Vì: AB+BC>AC

 AC+BC>AB

AB+AC>BC

=> A,B,C không thẳng hàng

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

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

...

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

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

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

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

a, \(-\dfrac{3}{5}x=-\dfrac{1}{20}+\dfrac{1}{2}=\dfrac{9}{20}\Leftrightarrow x=-\dfrac{3}{4}\)

b, \(-\dfrac{2}{5}x+\dfrac{1}{3}x-\dfrac{28}{105}=\dfrac{3}{5}\Leftrightarrow-\dfrac{1}{15}x=\dfrac{13}{15}\Leftrightarrow x=-13\)

c, \(\left(\dfrac{1}{2}x-\dfrac{1}{4}\right)\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=\dfrac{3}{4};x=3;x=-3\)

a: \(\Leftrightarrow2n-2+3⋮n-1\)

\(\Leftrightarrow n-1\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{2;0;4;-2\right\}\)

b \(\Leftrightarrow12n-9⋮3n+1\)

\(\Leftrightarrow12n+4-13⋮3n+1\)

\(\Leftrightarrow3n+1\in\left\{1;-1;13;-13\right\}\)

hay \(n\in\left\{0;-\dfrac{2}{3};4;-\dfrac{14}{3}\right\}\)

c: \(\Leftrightarrow n\left(n-1\right)+1⋮n-1\)

\(\Leftrightarrow n-1\in\left\{1;-1\right\}\)

hay \(n\in\left\{2;0\right\}\)

Nhìn cứ giống toán lớp 8,9 hơn lớp 6 đúng hok

a, \(2\left(n-1\right)+3⋮n-1\Rightarrow3⋮n-1\Leftrightarrow n-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

b, \(4n-3⋮3n+1\Leftrightarrow12n-9⋮3n+1\)

\(\Leftrightarrow4\left(3n+1\right)-13⋮3n+1\Rightarrow-13⋮3n+1\Rightarrow3n+1\inƯ\left(-13\right)=\left\{\pm1;\pm13\right\}\)

c, \(n^2-n+1=n\left(n-1\right)+1⋮n-1\Rightarrow n-1\inƯ\left(1\right)=\left\{\pm1\right\}\)

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

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

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