1. Tính
a) \(\dfrac{2}{3}\)-4.(\(\dfrac{1}{2}\)+\(\dfrac{3}{4}\))
b) \(\dfrac{-2}{3}\).\(\dfrac{3}{11}\)+\(\dfrac{-16}{9}\).\(\dfrac{3}{11}\)
c) \(\dfrac{11}{125}\)-\(\dfrac{17}{18}\)-\(\dfrac{5}{7}\)+\(\dfrac{4}{9}\)+\(\dfrac{17}{14}\)
Giúp mình với
a, \(\dfrac{2}{3}-4\left(\dfrac{1}{2}+\dfrac{3}{4}\right)=\dfrac{2}{3}-4.\dfrac{5}{4}=\dfrac{2}{3}-5=-\dfrac{13}{3}\)
b, \(-\dfrac{2}{3}.\dfrac{3}{11}+\dfrac{-16}{9}.\dfrac{3}{11}=\dfrac{3}{11}.\left(-\dfrac{2}{3}+\dfrac{-16}{9}\right)\)
\(=\dfrac{3}{11}.\dfrac{-22}{9}=-\dfrac{2}{3}\)
Câu c lấy máy tính tính nha!
c,
c) \(\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{10}{14}+\dfrac{8}{18}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}+\left(-\dfrac{17}{18}+\dfrac{8}{18}\right)+\left(-\dfrac{10}{14}+\dfrac{17}{14}\right)\)
\(=\dfrac{11}{125}+\left(-\dfrac{9}{18}\right)+\dfrac{7}{14}\)
\(=\dfrac{11}{125}+\left(-\dfrac{1}{2}\right)+\dfrac{1}{2}\)
\(=\dfrac{11}{125}+0=\dfrac{11}{125}\)
Tính
B= (3 1 phần 5: 3,2+4,5 nhân 1 32 phần 45 ): -21 1 phần 2
Lưu ý : 3 cách 1 phần 5 có nghĩa là hỗn số và -21 1 phần 2 cũng là hỗn số
\(B=\left(3\dfrac{1}{5}:3,2+4,5.1\dfrac{32}{45}\right):-21\dfrac{1}{2}\)
\(B=\left(3,2:3,2+7,7\right):-21,5\)
\(B=\left(1+7,7\right):-21,5=8,7:-21,5=\dfrac{-87}{215}\)
Chúc bạn học tốt!!!
2x.(x-\(\dfrac{1}{7}\))=0
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
*\(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
Vậy phương trình đã cho có \(S=\left\{0;\dfrac{1}{7}\right\}\)
* \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\left(x\ne0\right)\)
\(\Leftrightarrow\dfrac{3}{4}+\dfrac{1}{4x}=\dfrac{2}{5}\)
\(\Leftrightarrow15x+1=8x\)
\(\Leftrightarrow7x=-1\)
\(\Leftrightarrow x=\dfrac{-1}{7}\left(tmđk\right)\)
Vậy phương trình đã cho có \(S=\left\{\dfrac{-1}{7}\right\}\)
a, \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
Vậy \(x\in\left\{0;\dfrac{1}{7}\right\}\)
b, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}\Rightarrow x=\dfrac{-5}{7}\)
Vậy \(x=\dfrac{-5}{7}\)
Chúc bạn học tốt!!!
a) \(2x.\left(x-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
b) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}\)
\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)
\(x=\dfrac{-5}{7}\)
Chúc bạn học tốt 
1. Tính
a) \(\dfrac{2}{3}\)+\(\dfrac{3}{4}\).(\(\dfrac{-4}{9}\))
b) (\(\dfrac{3}{4}\)-0,2).(0,4-\(\dfrac{4}{5}\))
Giúp mình với
\(a,\dfrac{2}{3}+\dfrac{3}{4}.\left(-\dfrac{4}{9}\right)\)
\(=\dfrac{2}{3}+\left(-\dfrac{1}{3}\right)\)
\(=\dfrac{1}{3}\)
\(b,\left(\dfrac{3}{4}-0,2\right)\left(0,4-\dfrac{4}{5}\right)\)
\(=\left(\dfrac{3}{4}-\dfrac{1}{5}\right)\left(\dfrac{2}{5}-\dfrac{4}{5}\right)\)
\(=\dfrac{11}{20}.\dfrac{-2}{5}\)
\(=-\dfrac{11}{50}\)
Tìm các giá trị của x để các biểu thức sau có giá trị dương:
a) A=\(x^2\) +4x b)B=(x-3)(x+7) c)C=\(\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{3}-x\right)\)
a, \(A=x^2+4x\)
Để A đạt giá trị dương thì \(x^2+4x>0\Rightarrow x.\left(x+4\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x+4>0\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x< 0\\x+4< 0\end{matrix}\right.\)
+, Xét \(\left\{{}\begin{matrix}x>0\\x+4>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\x>-4\end{matrix}\right.\)
\(\Rightarrow x>0\)
+, Xét \(\left\{{}\begin{matrix}x< 0\\x+4< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 0\\x< -4\end{matrix}\right.\)
\(\Rightarrow x< -4\)
Vậy...............
Các câu còn lại làm tương tự! Chúc bạn học tốt!!!
( -2 phần 3 + 3 phần 7) : 4 phần 5 + ( -1 phần 3 + 4 phần 7) : 4 phần 5
5 phần 9 : (1 phần 11 - 5 phần 22) + 5 phần 9 : ( 1 phần 15 - 2 phần 3
a,\(\left(\dfrac{-2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\left(-\dfrac{2}{3}+\dfrac{3}{7}\right).\dfrac{5}{4}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right).\dfrac{5}{4}\)
\(=\left(\dfrac{-2}{3}+\dfrac{3}{7}+\dfrac{-1}{3}+\dfrac{4}{7}\right).\dfrac{5}{4}\)
\(=\left(-1+1\right).\dfrac{5}{4}=0\)
Chúc bạn học tốt!!!
Câu b làm tương tự!
b) \(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\)
\(=\dfrac{5}{9}:\left(-\dfrac{3}{22}\right)+\dfrac{5}{9}:\left(-\dfrac{3}{5}\right)\)
\(=-\dfrac{5}{9}\cdot\dfrac{22}{3}-\dfrac{5}{9}\cdot\dfrac{5}{3}\)
\(=-\dfrac{110}{27}-\dfrac{25}{27}\)
\(=-5\)
Tìm x ϵ Q biết:
(x-3)(x+2/3)>0
\(\left(x-3\right)\left(x+\dfrac{2}{3}\right)>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>3\\x>-\dfrac{2}{3}\end{matrix}\right.\) Hoặc \(\left\{{}\begin{matrix}x-3< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 3\\x< -\dfrac{2}{3}\end{matrix}\right.\)
=> \(-\dfrac{2}{3}< x< 3\)
p/s : Không biết đúng hay sai bởi vì đang áp dụng cách mới :v
\(\left(x-3\right)\left(x+\dfrac{2}{3}\right)>0\)
TH1:
\(\Leftrightarrow\left\{{}\begin{matrix}x-3>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>3\\x>-\dfrac{2}{3}\end{matrix}\right.\) \(\Rightarrow x>3\)
TH2: \(\left\{{}\begin{matrix}x-3< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 3\\x< -\dfrac{2}{3}\end{matrix}\right.\) \(\Rightarrow x< -\dfrac{2}{3}\)
Vậy để \(\left(x-3\right)\left(x+\dfrac{2}{3}\right)>0\)
thì \(x>3\) hoặc \(x< -\dfrac{2}{3}\)
