4 câu đầu hìn như sai đề :v

`m)(3/2-2/(-5)):x-1/2=3/2`

`<=>(3/2+2/5):x=3/2+1/2=2`

`<=>19/10:x=2`

`<=>x=19/10:2=19/20`

`n)(3/2-5/11-3/13)(2x-2)=(-3/4+5/22+3/26)`

`<=>(3/2-5/11-3/13)(2x-2)+3/4-5/22-3/26=0`

`<=>(3/2-5/11-3/13)(2x-2)+1/2(3/2-5/11-3/13)=0`

`<=>(3/2-5/11-3/13)(2x-2+1/2)=0`

Mà `3/2-5/11-3/13>0`

`<=>2x-2+1/2=0`

`<=>2x-3/2=0`

`<=>2x=3/2<=>x=3/4`

h, \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\left(x\ne0\right)\)

\(\Leftrightarrow\dfrac{x^2}{2}-1=\dfrac{x}{12}\)

\(\Leftrightarrow x^2-\dfrac{x}{6}-2=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{12}+\dfrac{1}{144}-\dfrac{289}{144}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{12}\right)^2=\dfrac{289}{144}\)

\(\Leftrightarrow x=\dfrac{1}{12}\pm\dfrac{\sqrt{289}}{12}\)

Vậy ...

i, \(\Leftrightarrow x^2-\dfrac{2.x.7}{12}+\dfrac{49}{144}-\dfrac{1}{144}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{2}\right)^2=\dfrac{1}{144}\)

\(\Leftrightarrow x=\dfrac{7}{2}\pm\dfrac{1}{12}\)

Vậy ...

h) Ta có: \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\)

\(\Leftrightarrow\dfrac{x^2-2}{2x}=\dfrac{1}{12}\)

\(\Leftrightarrow12x^2-24-2x=0\)

\(\Delta=\left(-2\right)^2-4\cdot12\cdot\left(-24\right)=1156\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{2-34}{24}=\dfrac{-8}{3}\\x_2=\dfrac{2+34}{24}=\dfrac{36}{24}=\dfrac{3}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{8}{3};\dfrac{3}{2}\right\}\)

m) Ta có: \(\left(\dfrac{3}{2}-\dfrac{2}{-5}\right):x-\dfrac{1}{2}=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{19}{10}:x=2\)

hay \(x=\dfrac{19}{20}\)

Vậy: \(S=\left\{\dfrac{19}{20}\right\}\)

\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)

\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)

\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)

Vậy \(x\in\left\{22;23;24;...\right\}\)

\(\dfrac{????????}{????????????}\)

a) \(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)

   \(\dfrac{11}{55}+\dfrac{10}{55}< \dfrac{x}{55}< \dfrac{22}{55}+\dfrac{1}{55}\)

   \(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{23}{55}\)

\(\Rightarrow\) \(x=22\)

b) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}< x\le\dfrac{13}{4}+\dfrac{14}{8}\)

  \(\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{1}{6}< x\le\dfrac{26}{8}+\dfrac{14}{8}\)

  \(1< x\le5\)

  \(\Rightarrow\) \(x\in\) {\(2;3;4;5\)}

c) \(\dfrac{1}{3}+\dfrac{5}{12}+\dfrac{-1}{13}< x< \dfrac{7}{5}+\dfrac{2}{10}+\dfrac{1}{2}\)

 Ko biết làm

d) \(\dfrac{79}{15}+\dfrac{7}{5}+\dfrac{-8}{3}\le x\le\dfrac{10}{3}+\dfrac{15}{4}+\dfrac{23}{12}\)

   \(\dfrac{79}{15}+\dfrac{21}{15}+\dfrac{-40}{15}\le x\le\dfrac{40}{12}+\dfrac{45}{12}+\dfrac{23}{12}\)

   \(4\le x\le9\)

   \(\Rightarrow\) \(x\in\) {\(4;5;6;7;8;9\)}

\(\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}-\dfrac{1}{3}\)

\(=\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)-\dfrac{1}{3}\)

\(=-1+1-\dfrac{1}{3}\)

\(=0-\dfrac{1}{3}\)

\(=\dfrac{-1}{3}\)

------------------------------------------

\(14.\dfrac{3}{2}+\dfrac{6}{5}:\left(-\dfrac{2}{5}\right)\)

\(=14.\dfrac{3}{2}+\dfrac{6}{5}.\dfrac{-5}{2}\)

\(=21+\dfrac{6}{5}.\dfrac{-5}{2}\)

\(=21+\left(-3\right)\)

\(=18\)

------------------------------------------------

\(\sqrt{\dfrac{1}{4}+\dfrac{2}{3}-\left(\dfrac{1}{3}\right)^2}\)

\(=\sqrt{\dfrac{1}{4}+\dfrac{2}{3}-\dfrac{1}{9}}\)

\(=\sqrt{\dfrac{3}{12}+\dfrac{8}{12}-\dfrac{1}{9}}\)

\(=\sqrt{\dfrac{11}{12}-\dfrac{1}{9}}\)

\(=\sqrt{\dfrac{99}{108}-\dfrac{12}{108}}\)

\(=\sqrt{\dfrac{29}{36}}\)

\(=\dfrac{\sqrt{29}}{6}\)

\(\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}-\dfrac{1}{3}\)
\(=\dfrac{5}{4}+\dfrac{5}{13}-\dfrac{1}{4}-\dfrac{18}{13}-\dfrac{1}{3}\)
\(=\left(\dfrac{5}{4}-\dfrac{1}{4}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)-\dfrac{1}{3}\)
\(=1+\left(-1\right)-\dfrac{1}{3}=0-\dfrac{1}{3}=-\dfrac{1}{3}\)

\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

có dc tính nhanh ko:)?

a) =  2/4 + 1/24 = 13/24

b) = 1/2 x ( 3/5 + 12/5 ) = 1/2 x 3 = 3/2

2/4 + 1/24 = 13/24

3/10 + 6/5 = 3/2

`h)x/2-1/x=1/12(x ne 0)`

`<=>6x^2-12=x`

`<=>6x^2-x-12=0`

`<=>6x^2-9x+8x-12=0`

`<=>3x(2x-3)+4(2x-3)=0`

`<=>(2x-3)(3x+4)=0`

`<=>` \(\left[ \begin{array}{l}x=\dfrac32\\x=-\dfrac43\end{array} \right.\) 

`i)x^2-7/6x+1/3=0`

`<=>6x^2-7x+2=0`

`<=>6x^2-3x-4x+2=0`

`<=>3x(2x-1)-2(2x-1)=0`

`<=>(2x-1)(3x-2)=0`

`<=>` \(\left[ \begin{array}{l}x=\dfrac12\\x=\dfrac23\end{array} \right.\) 

Câu cuối không có dấu "=" nên không tìm được x :v

- Hai câu h, i bấm nốt đáp án để đẹp nha ;-; câu k thiếu đề :v

h) Ta có: \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\)

\(\Leftrightarrow\dfrac{x^2-2}{2x}=\dfrac{1}{12}\)

\(\Leftrightarrow12\left(x^2-2\right)-2x=0\)

\(\Leftrightarrow12x^2-2x-24=0\)

\(\Delta=\left(-2\right)^2-4\cdot12\cdot\left(-24\right)=1156\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{2+34}{12}=\dfrac{36}{12}=3\\x_2=\dfrac{2-34}{12}=\dfrac{-32}{12}=-\dfrac{8}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{3;-\dfrac{8}{3}\right\}\)

i) Ta có: \(x^2-\dfrac{7}{6}x+\dfrac{1}{3}=0\)

\(\Leftrightarrow6x^2-7x+2=0\)

\(\Leftrightarrow6x^2-3x-4x+2=0\)

\(\Leftrightarrow3x\left(2x-1\right)-2\left(2x-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)

bài 1 :

\(a,\dfrac{2}{7}+\dfrac{1}{3}=\dfrac{6}{21}+\dfrac{7}{21}=\dfrac{13}{21}\)

\(b,\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{9}{15}-\dfrac{5}{15}=\dfrac{4}{15}\)

\(c,\dfrac{13}{4}:5=\dfrac{13}{4}:\dfrac{5}{1}=\dfrac{13}{4}x\dfrac{1}{5}=\dfrac{13}{20}\)

\(d,\dfrac{6}{23}x\dfrac{1}{18}=\dfrac{1}{69}\)

bài 2 :

\(a,x+\dfrac{1}{3}=\dfrac{5}{12}\)

   \(x=\dfrac{5}{12}-\dfrac{1}{3}\)

  \(x=\dfrac{1}{12}\)

\(b,x:\dfrac{7}{4}=\dfrac{2}{5}\)

   \(x=\dfrac{2}{5}x\dfrac{7}{4}\)

   \(x=\dfrac{7}{10}\)

bài 3 :

đổi : 2 dm 1cm = 21cm

chiều cao hình bình hành là;

       21 x\(\dfrac{3}{7}=\)9(cm)

diện tích hình bình hành là;

       21 x 9 =189 (cm²)

                 đáp số : 189 cm²

bài 4 :

\(\dfrac{2}{3}x\dfrac{2}{10}+\dfrac{2}{3}x\dfrac{5}{10}x\dfrac{3}{3}\)

\(\dfrac{2}{3}x\left(\dfrac{2}{10}+\dfrac{5}{10}\right)x\dfrac{2}{3}\)

=\(\dfrac{2}{3}x1x\dfrac{2}{3}\)

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

=\(\dfrac{4}{9}\)

Bài 1)

a) \(\dfrac{6}{21}+\dfrac{7}{21}=\dfrac{13}{21}\)

b) \(\dfrac{9}{15}-\dfrac{5}{15}=\dfrac{4}{15}\)

c) \(\dfrac{13}{4}x\dfrac{1}{5}=\dfrac{13}{20}\)

d) \(\dfrac{6}{414}=\dfrac{1}{69}\)

Bài 2)

a) \(x=\dfrac{5}{12}-\dfrac{1}{3}\)

\(x=\dfrac{1}{12}\)

b) \(x=\dfrac{2}{5}x\dfrac{7}{4}\)

\(x=\dfrac{7}{10}\)

Bài 3)

2dm 1cm = 21 cm

Chiều cao tấm bìa la

\(21x\dfrac{3}{7}=9\left(cm\right)\)

Diện tích tấm bìa là

\(21x9=189\left(cm2\right)\)

Bài 4)

\(\dfrac{2}{3}x\dfrac{2}{10}+\dfrac{2}{3}x\dfrac{5}{10}x\dfrac{2}{3}=\dfrac{2}{3}x\left(\dfrac{2}{10}+\dfrac{5}{10}\right)x\dfrac{2}{3}=\dfrac{2}{3}x\dfrac{7}{10}x\dfrac{2}{3}=\dfrac{14}{45}\)

tách đi bạn

a) (2x - 3)(6 - 2x) = 0

=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)

c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)

d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)

e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)

f)\(\dfrac{x-12}{4}=\dfrac{1}{2}=\dfrac{x-12}{4}=\dfrac{2}{4}\)

⇒\(x-12=2\)

   \(x=2+12\)

  x = 14

g)2\(\dfrac{1}{4}.\left(x-7\dfrac{1}{3}\right)=1,5\)

\(\dfrac{9}{4}.\left(x-\dfrac{22}{3}\right)=1,5\) 

      \(\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}:\dfrac{9}{4}\)

       \(x-\dfrac{22}{3}=\dfrac{2}{3}\)

       \(x=\dfrac{2}{3}+\dfrac{22}{3}\) 

      \(x=8\)

3/7x2=6/7

5/12x1=5/12

3x4/7=12/7

0x5/9=0