\(-6x^8:\left(\dfrac{3}{7}x^4\right)\)

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\(-6x^8:\left(\dfrac{3}{7}x^4\right)=\left(-6:\dfrac{3}{7}\right)\cdot\left(x^8:x^4\right)=-14x^4\)

17 tháng 5 2024

   - 6\(x^8\) : (\(\dfrac{3}{7}\)\(x^4\))

= - 6\(x^8\) x \(\dfrac{7}{3x^4}\)

= - 14\(x^4\) 

25 tháng 7 2017

help me

25 tháng 7 2017

\(\left(x-1\right)\left(x+5\right)>0\)

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

\(\left(x-1\right)\left(x+5\right)< 0\)

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

\(\Rightarrow-5< x< 1\)

câu dễ tự làm

\(\Rightarrow x>-5;x< -5\)

8 tháng 11 2018

câu 1 \(A=\dfrac{3^2}{5^2}.5^2-\dfrac{9^3}{4^3}:\dfrac{3^3}{4^3}+\dfrac{1}{2}\)

\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{\left(3^2\right)^3}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}\)

\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{3^6}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}=3^2-3^3+\dfrac{1}{2}=-18+\dfrac{1}{2}=-\dfrac{35}{2}\)

\(B=\left[\dfrac{4}{11}+\dfrac{7}{22}.2\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{4^4}{8^2}\right)^{2009}\)

\(B=\left[\dfrac{4}{11}+\dfrac{7}{11}\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{\left(2^2\right)^4}{\left(2^3\right)^2}\right)^{2009}\)

\(B=1^{2010}-\left(\dfrac{1}{2^2}.\dfrac{2^8}{2^6}\right)^{2009}\)

\(B=1^{2010}-\left(\dfrac{2^8}{2^8}\right)^{2009}\)

\(B=1^{2010}-1^{2009}=1-1=0\)

8 tháng 11 2018

câu 2

a) \(2x-\dfrac{5}{4}=\dfrac{20}{15}\)

\(\Leftrightarrow2x=\dfrac{4}{3}+\dfrac{5}{4}\)

\(\Leftrightarrow2x=\dfrac{31}{12}\)

\(\Leftrightarrow x=\dfrac{31}{24}\)

b) \(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)

\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}\)

\(\Leftrightarrow x=-\dfrac{5}{6}\)

27 tháng 7 2017

\(\text{a) }\left(x-1\right)\left(x-5\right)>0\\ \text{ Để }\left(x-1\right)\left(x-5\right)>0\text{ thì }\Rightarrow x-1\text{ và }x-5\text{ cùng dấu }\\ \text{+) Xét }x-1\text{ và }x-5\text{ là số nguyên dương }\Rightarrow\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x-5>0\Rightarrow x>5\end{matrix}\right.\Rightarrow x>5\\ \text{+) Xét }x-1\text{ và }x-5\text{ là số nguyên âm }\Rightarrow\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x-5< 0\Rightarrow x< 5\end{matrix}\right.\Rightarrow x< 1\\ \text{Vậy }\left(x-1\right)\left(x-5\right)>0\text{ khi }x< 1\text{ hoặc }x>5\)

\(\text{b) }\left(x-1\right)\left(x-5\right)< 0\\ \text{ Để }\left(x-1\right)\left(x-5\right)< 0\text{ thì }\Rightarrow x-1\text{ và }x-5\text{ trái dấu }\\ \text{ Mà }x-1>x-5\\ \Rightarrow\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x-5< 0\Rightarrow x< 5\end{matrix}\right.\Rightarrow1< x< 5\\ \text{ Vậy }\left(x-1\right)\left(x-5\right)< 0\text{ khi }1< x< 5\)

27 tháng 7 2017

\(\text{c) }\dfrac{3}{4}-\dfrac{1}{4}\left|x-\dfrac{1}{7}\right|=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{1}{4}\left|x-\dfrac{1}{7}\right|=\dfrac{1}{2}\\ \Leftrightarrow\left|x-\dfrac{1}{7}\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{7}=-2\\x-\dfrac{1}{7}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{13}{7}\\x=\dfrac{15}{7}\end{matrix}\right.\\ \text{Vậy }x=-\dfrac{13}{7}\text{ hoặc }x=\dfrac{15}{7}\)

\(\text{d) }\left(x-\dfrac{1}{2}\right)^2=\dfrac{1}{16}\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=-\dfrac{1}{4}\\x-\dfrac{1}{2}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\\ \text{Vậy }x=\dfrac{1}{4}\text{ hoặc }x=\dfrac{3}{4}\)

\(\text{e) }8\left(x+1\right)-2\left(2x+5\right)=0\\ \Leftrightarrow8x+8-4x+10=0\\ \Leftrightarrow\left(8x-4x\right)+\left(8+10\right)=0\\ \Leftrightarrow4x+18=0\\ \Leftrightarrow4x=-18\\ \Leftrightarrow x=-\dfrac{9}{2}\\ \text{Vậy }x=-\dfrac{9}{2}\)

\(\text{g) }\left(6x-1\right)-\left(x+8\right)=0\\ \Leftrightarrow6x-1-x-8=0\\ \Leftrightarrow\left(6x-x\right)-\left(1+8\right)=0\\ \Leftrightarrow5x-9=0\\ \Leftrightarrow5x=9\\ \Leftrightarrow x=\dfrac{9}{5}\\ \text{Vậy }x=\dfrac{9}{5}\)

\(\text{h) }\left|7x-\dfrac{1}{4}\right|=1\\ \Leftrightarrow\left[{}\begin{matrix}7x-\dfrac{1}{4}=-1\\7x-\dfrac{1}{4}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=-\dfrac{3}{4}\\7x=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{28}\\x=\dfrac{5}{28}\end{matrix}\right.\\ \text{Vậy }x=-\dfrac{3}{28}\text{ hoặc }x=\dfrac{5}{28}\)

\(\text{q) }-2x-3=-x+7\\ \Leftrightarrow-2x-3-\left(-x+7\right)=0\\ \Leftrightarrow-2x-3+x-7=0\\ \Leftrightarrow\left(-2x+x\right)-\left(3+7\right)=0\\ \Leftrightarrow-x-10=0\\ \Leftrightarrow-x=10\\ \Leftrightarrow x=-10\\ \text{ Vậy }x=-10\)

20 tháng 8 2017

1.Tính

a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)

b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)

c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)

d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)

e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)

Bài 2

a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)

\(x=\dfrac{13}{49}\)

b.\(\left|x-1,5\right|=2\)

Xảy ra 2 trường hợp

TH1

\(x-1,5=2\)

\(x=3,5\)

TH2

\(x-1,5=-2\)

\(x=-0,5\)

Vậy \(x=3,5\) hoặc \(x=-0,5\) .

Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.

20 tháng 8 2017

Ths bn nhé

3 tháng 8 2017

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

\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)

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

b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)

\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)

\(x+\dfrac{5}{6}=\dfrac{16}{15}\)

\(x=\dfrac{16}{15}-\dfrac{5}{6}\)

\(x=\dfrac{7}{30}\)

c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)

\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)

\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)

\(\dfrac{7}{5}x=\dfrac{-43}{35}\)

\(\Rightarrow x=\dfrac{-43}{49}\)

d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)

\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)

\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}-\dfrac{3}{4}\)

\(x=\dfrac{-5}{12}\)

e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)

\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)

\(x+\dfrac{4}{5}=2,15-3,75\)

\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)

\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)

\(x=\dfrac{-12}{5}\)

f) \(\left(x-2\right)^2=1\)

\(\Rightarrow x=1\)

Sức chịu đựng có giới hạn -.-

3 tháng 8 2017

- Mình tiếp tục cho Nguyễn Phương Trâm nhé.

g, \(\left(2x-1\right)^3=-27\)

\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)

\(\Rightarrow2x-1=-3\)

\(\Rightarrow2x=-2\)

=> \(x=-1\)

- Vậy x = -1

h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)

\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)

\(\Rightarrow\left(x-1\right)^2=900 \)

\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)

=> x = 31

i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)

=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{16}\)

- Vậy x=\(\dfrac{1}{16}\)

j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)

\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)

\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{3}{4}\)

- Vạy x = \(\dfrac{3}{4}\)

k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)

=>\(4^x=4\)

=> x = 1

- Vậy x = 1

a: TH1: x>=0

=>x+x=1/3

=>x=1/6(nhận)

TH2: x<0

Pt sẽ là -x+x=1/3

=>0=1/3(loại)

b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)

c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)

\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)

\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)

\(\Leftrightarrow3x^2-63x+60=4x+72\)

=>3x^2-67x-12=0

hay \(x\in\left\{22.51;-0.18\right\}\)

27 tháng 7 2017

h) \(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x\left(1+25\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow x=2\)

haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=

1 tháng 8 2017

bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây

a: \(=\left|\dfrac{3}{2}-\dfrac{7}{3}\right|^2+\dfrac{1}{4}=\dfrac{17}{18}\)

b: \(=\left|1-2-\dfrac{1}{3}\right|+\dfrac{5}{6}=1+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{13}{6}\)

c: \(=\left|\dfrac{3}{2}-\dfrac{7}{4}\right|-\dfrac{7}{4}=-\dfrac{3}{2}\)

d: =x-5+8-x=3

27 tháng 6 2017

a, \(\dfrac{3}{4}+x=\dfrac{8}{13}\)

\(x=\dfrac{8}{13}-\dfrac{3}{4}\)

\(x=-\dfrac{7}{52}\)

b,\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{1}{4}\)

\(x=\dfrac{1}{4}-\dfrac{2}{5}\)

\(x=-\dfrac{3}{20}\)

c, \(2x\left(x-\dfrac{1}{7}\right)=0\)

\(2x-\dfrac{1}{7}=0\)

\(x-\dfrac{1}{7}=0:2\)

\(x-\dfrac{1}{7}=0\)

\(x=0-\dfrac{1}{7}\)

\(x=\dfrac{1}{7}\)

d, \(\dfrac{3}{4}+\dfrac{1}{4}\div x=\dfrac{2}{5}\)

\(\left(\dfrac{3}{4}+\dfrac{1}{4}\right):x=\dfrac{2}{5}\)

\(1:x=\dfrac{2}{5}\)

\(x=1:\dfrac{2}{5}\)

\(x=\dfrac{5}{2}\)

27 tháng 6 2017

a) \(\dfrac{3}{4}+x=\dfrac{8}{13}\)\(\Leftrightarrow\) \(x=\dfrac{8}{13}-\dfrac{3}{4}=\dfrac{-7}{52}\) vậy \(x=\dfrac{-7}{52}\)

b) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\) \(\Leftrightarrow\) \(\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\) \(\Leftrightarrow\) \(x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=\dfrac{-3}{20}\) vậy \(x=\dfrac{-3}{20}\)

c) \(2x\left(x-\dfrac{1}{7}\right)=0\) \(\Leftrightarrow\) \(2x^2-\dfrac{2}{7}x=0\)

\(\Delta\) = \(\left(\dfrac{-2}{7}\right)^2-4.2.0=\dfrac{4}{49}>0\)

\(\Rightarrow\) phương trình có 2 nghiệm phân biệt

\(x_1=\dfrac{\dfrac{2}{7}+\sqrt{\dfrac{4}{49}}}{4}=\dfrac{1}{7}\)

\(x_2=\dfrac{\dfrac{2}{7}-\sqrt{\dfrac{4}{49}}}{4}=0\)

vậy \(x=0;x=\dfrac{1}{7}\)