\(x^3\) + \(\dfrac{29}{28}\) = 
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3 tháng 8 2017

a,

\(\dfrac{13}{17}=1-\dfrac{4}{17}\\ \dfrac{25}{29}=1-\dfrac{4}{29}\\ \dfrac{4}{17}>\dfrac{4}{29}\Rightarrow1-\dfrac{4}{17}< 1-\dfrac{4}{29}\Leftrightarrow\dfrac{13}{17}< \dfrac{25}{29}\)

Vậy \(\dfrac{13}{17}< \dfrac{25}{29}\)

b,

\(\dfrac{59}{101}>\dfrac{56}{101}>\dfrac{56}{105}\\ \Rightarrow\dfrac{59}{101}>\dfrac{56}{105}\)

Vậy \(\dfrac{59}{101}>\dfrac{56}{105}\)

c,

\(\dfrac{14}{55}>\dfrac{14}{56}=\dfrac{1}{4}=\dfrac{20}{80}>\dfrac{20}{83}\)

Vậy \(\dfrac{14}{55}>\dfrac{20}{83}\)

3 tháng 8 2017

d,

\(\dfrac{13}{57}< \dfrac{13}{39}=\dfrac{1}{3}=\dfrac{29}{87}< \dfrac{29}{73}\)

Vậy \(\dfrac{13}{57}< \dfrac{29}{73}\)

e,

\(\dfrac{17}{21}=\dfrac{17\cdot101}{21\cdot101}=\dfrac{1717}{2121}\)

Vậy \(\dfrac{17}{21}=\dfrac{1717}{2121}\)

11 tháng 3 2017

d)

Ta có: \(\dfrac{1}{51}>\dfrac{1}{100}\)

\(\dfrac{1}{52}>\dfrac{1}{100}\)

...

\(\dfrac{1}{99}>\dfrac{1}{100}\)

\(\dfrac{1}{100}=\dfrac{1}{100}\)

\(\Rightarrow S=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{99}+\dfrac{1}{100}>\dfrac{1}{100}.50=\dfrac{1}{2}\)\(\Rightarrow S>\dfrac{1}{2}\)

11 tháng 3 2017

các con trên ???

8 tháng 3 2018

quy đồng rồi biết

8 tháng 3 2018

trả lời thì trả lời đi ko biết thì thôi đừng đăng lên

a: \(\Leftrightarrow x\cdot\dfrac{62}{7}=\dfrac{29}{9}\cdot\dfrac{56}{3}=\dfrac{1624}{27}\)

hay \(x=\dfrac{1624}{27}:\dfrac{62}{7}=\dfrac{5684}{837}\)

b: \(\Leftrightarrow\dfrac{1}{5}:x=\dfrac{12}{35}\)

nên \(x=\dfrac{1}{5}:\dfrac{12}{35}=\dfrac{1}{5}\cdot\dfrac{35}{12}=\dfrac{7}{12}\)

c: \(\Leftrightarrow\left|2x+\dfrac{1}{3}\right|=\dfrac{30-7}{42}=\dfrac{23}{42}\)

=>2x+1/3=23/42 hoặc 2x+1/3=-23/42

=>2x=3/14 hoặc 2x=-37/42

=>x=3/28 hoặc x=-37/84

3 tháng 8 2017

3. a) Ta có : 13.29 = 377
25.17 = 425
=> \(\dfrac{13}{17}< \dfrac{25}{29}\)
b) Ta có : 59.105 > 56.101
=> \(\dfrac{59}{101}>\dfrac{56}{105}\)
c) Ta có : 14.83 = 1162
20.55 = 1100
=> \(\dfrac{14}{55}>\dfrac{20}{83}\)
d) Ta có : 13.73 = 949
29.57 = 1653
=> \(\dfrac{13}{57}< \dfrac{29}{73}\)
e) Ta có : \(\dfrac{1717}{2121}=\dfrac{17}{21}\)
=> \(\dfrac{17}{21}=\dfrac{1717}{2121}\)
@Đặng Vũ Hoài Anh

3 tháng 8 2017

4. Gọi các phân số cần tìm có dạng \(\dfrac{x}{3}\)
Ta có : \(\dfrac{-1}{2}< \dfrac{x}{3}< \dfrac{1}{2}\)

=> \(\dfrac{-3}{6}< \dfrac{2x}{6}< \dfrac{3}{6}\)

=> -3 < 2x < 3
=> 2x = -2; 0; 2
=> x = -1; 0; 1 (thỏa mãn)
@Đặng Vũ Hoài Anh

2 tháng 5 2017

Bài 1:

a)

\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)

b)

\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)

c)

\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)

d)

\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)

e)

\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)

f)

\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)

g)

\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)

h)

\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)

i)

\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)

15 tháng 7 2018

\(\dfrac{29}{30}-\left(\dfrac{13}{30}+x\right)=\dfrac{7}{69}\)

\(\Leftrightarrow\dfrac{13}{23}+x=\dfrac{29}{30}-\dfrac{7}{69}\)

\(\Leftrightarrow\dfrac{13}{23}+x=\dfrac{199}{230}\)

\(\Leftrightarrow x=\dfrac{199}{230}-\dfrac{13}{23}\)

\(\Leftrightarrow x=\dfrac{69}{230}\)

4 tháng 8 2018

bài 2:tính hợp lý

1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)

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

\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)

\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)

Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)

22 tháng 4 2018

giúp mình đi mà ToT khocroi