Tìm x biết: \(\dfrac{13}{14}+\dfrac{13}{35}+\dfrac{13}{65}+\dfrac{13}{104}+...+\dfrac{13}{1274}-x+2\dfrac{1}{5}=0\)
Thực hiện các phép tính:
\(a.\dfrac{-5}{9}-\dfrac{-5}{12} \) \(b.\dfrac{-5}{12}:\dfrac{15}{4}\) c. \(\dfrac{1}{13}\) x \(\dfrac{8}{13}\) + \(\dfrac{5}{13}\) x \(\dfrac{1}{13}\) - \(\dfrac{14}{13}\)
a) \(\dfrac{-5}{9}-\dfrac{-5}{12}=\dfrac{-5}{9}+\dfrac{5}{12}=\dfrac{-20}{36}+\dfrac{15}{36}=-\dfrac{5}{36}\)
b) \(\dfrac{-5}{12}:\dfrac{15}{4}=\dfrac{-5}{12}\times\dfrac{4}{15}=\dfrac{-1}{9}\)
c) \(\dfrac{1}{13}\cdot\dfrac{8}{13}+\dfrac{5}{13}\cdot\dfrac{1}{13}-\dfrac{14}{13}=\dfrac{1}{13}\cdot\left(\dfrac{8}{13}+\dfrac{5}{13}\right)-\dfrac{14}{13}=\dfrac{1}{13}\cdot1-\dfrac{14}{13}=\dfrac{1}{13}-\dfrac{14}{13}=-1\)
1.Thực hiện phép tính
a) \(\dfrac{4}{5}\) + \(\dfrac{6}{5}\) : \(\dfrac{7}{15}\) - \(\dfrac{3}{7}\)
b) \(\dfrac{9}{13}\) . \(\dfrac{5}{3}\) + \(\dfrac{-6}{13}\) .\(\dfrac{9}{13}\) - \(\dfrac{9}{13}\) . \(\dfrac{20}{13}\)
2.Tìm x,biết:
a) \(\dfrac{2}{5}\) . x - \(\dfrac{1}{3}\) = 1\(\dfrac{2}{3}\)
b) \(\dfrac{4}{9}\) - \(\dfrac{5}{9}\) : x = \(\dfrac{7}{3}\)
c) \(\dfrac{4}{5}\) . x + \(\dfrac{-7}{3}\) . x = \(\dfrac{\left(-2\right)^3}{5}\)
\(\dfrac{11}{8}+\dfrac{13}{6}=\dfrac{85}{2x}\)
\(\dfrac{3x}{182}=\dfrac{-6}{14}.\dfrac{35}{91}\)
Tìm x , biết
Thực hiện phép tính, tính nhanh nếu có thể
a, \(\dfrac{7}{13}\) + \(\dfrac{12}{13}\) + \(\dfrac{6}{-13}\)
b, ( \(\dfrac{4}{5}\) + \(\dfrac{1}{2}\) ) x ( \(\dfrac{6}{13}\) - 2 )
c, 75% : \(\dfrac{1}{2}\) - \(2\dfrac{1}{2}\) x ( \(-1\dfrac{1}{3}\) )
d, 0,375 x \(1\dfrac{3}{5}\) + 60% x \(\dfrac{2}{7}\) + \(\dfrac{3}{5}\) x \(\dfrac{5}{7}\)
e, \(\dfrac{-5}{6}\) x \(\dfrac{7}{13}\) + \(\dfrac{19}{13}\) : \(\dfrac{6}{-5}\) + \(\dfrac{2}{5}\)
f, 1,25 x \(\dfrac{7}{9}+\dfrac{5}{4}x\dfrac{15}{9}-1\dfrac{1}{4}:\dfrac{19}{3}\)
a: \(=\dfrac{7+12-6}{13}=1\)
b: \(=\dfrac{13}{10}\cdot\dfrac{6-26}{13}=\dfrac{-20}{10}=-2\)
c: \(=\dfrac{3}{4}\cdot2-\dfrac{5}{2}\cdot\dfrac{-4}{3}=\dfrac{3}{2}+\dfrac{20}{6}=\dfrac{3}{2}+\dfrac{10}{3}=\dfrac{29}{6}\)
d: \(=\dfrac{3}{8}\cdot\dfrac{8}{5}+\dfrac{3}{5}\cdot\dfrac{2}{7}+\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{5}+\dfrac{3}{5}=\dfrac{6}{5}\)
Bài 4. Trên tia Ox lấy điểm A và B sao cho OA = 8cm, OB = 12cm.
a, Điểm A có nằm giữa hai điểm O và B không? Vì sao? Tính AB.
b, Gọi M, N lần lượt là trung điểm của OA, OB. Điểm M có nằm giữa hai điểm O và N không? Vì sao? Tính MN.
c, Điểm N có là trung điểm của đoạn thẳng AM không? Vì sao?
cho mìn hỏi câu b nhoa
\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}< x\le\dfrac{13}{4}+\dfrac{14}{8}\)
\(\dfrac{1}{4}+\dfrac{5}{12}+\dfrac{-1}{13}< x< \dfrac{7}{5}+\dfrac{2}{10}+\dfrac{1}{2}\) \(\dfrac{79}{15}+\dfrac{7}{5}+\dfrac{-8}{3}\le x\le\dfrac{10}{3}+\dfrac{15}{4}+\dfrac{23}{12}\)
\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)
\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)
\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)
Vậy \(x\in\left\{22;23;24;...\right\}\)
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) \(\dfrac{79}{15}+\dfrac{7}{5}+\dfrac{-8}{3}\le x\le\dfrac{10}{3}+\dfrac{15}{4}+\dfrac{23}{12}\)
\(\dfrac{79}{15}+\dfrac{21}{15}+\dfrac{-40}{15}\le x\le\dfrac{40}{12}+\dfrac{45}{12}+\dfrac{23}{12}\)
\(4\le x\le9\)
\(\Rightarrow\) \(x\in\) {\(4;5;6;7;8;9\)}
\(\dfrac{-19}{23}\) x \(\dfrac{13}{14}\) + \(\dfrac{13}{14}\)\(\) x \(\dfrac{-15}{23}\) - \(\dfrac{13}{14}\) x \(\dfrac{1}{23}\)
Mngười giải giúp em chi tiết nha! Tks
\(\dfrac{-19}{23}\cdot\dfrac{13}{14}+\dfrac{13}{14}\cdot\dfrac{-15}{23}-\dfrac{13}{14}\cdot\dfrac{1}{23}\\ =\dfrac{13}{14}\cdot\left(\dfrac{-19}{23}+\dfrac{-15}{23}-\dfrac{1}{23}\right)\\ =\dfrac{13}{14}\cdot\dfrac{-35}{23}=\dfrac{-65}{46}\)
BT3: Tìm x, biết
13) \(\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+\dfrac{1}{4}:\dfrac{1}{8}\)
14) \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
1) \(\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+\dfrac{1}{4}:\dfrac{1}{8}\)
\(\Leftrightarrow\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+2\)
\(\Leftrightarrow\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{1}{4}\)
\(\Leftrightarrow-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-3\dfrac{3}{5}\right)\)
\(\Leftrightarrow-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-\dfrac{18}{5}\right)\)
\(\Leftrightarrow-1\dfrac{1}{5}+x=-\dfrac{9}{10}\)
\(\Leftrightarrow x=\left(-\dfrac{9}{10}\right)-\left(-1\dfrac{1}{5}\right)\)
\(\Leftrightarrow x=\dfrac{3}{10}\)
Câu 1: Tìm x, biết:
a)\(x^2-\dfrac{16}{25}=0\) b)\(\dfrac{2}{5}-\left|\dfrac{1}{2}-x\right|=6\)
C2.Tính giá của biểu thức:
a)\(A=1\dfrac{5}{13}-0,25-\left(2\dfrac{5}{9}+\dfrac{18}{13}-\dfrac{1}{4}\right)\)
b)\(\dfrac{\dfrac{3}{5}.7^2-3.5^6+\dfrac{3}{5}.3^9}{\dfrac{3}{4}.7^2-\dfrac{3}{4}.5^7+\dfrac{3}{4}.3^9}\)
a)
x^2-16/25=0
x^2-4^2/5^2=0
=>x-4/5=0
x=0+4/5
x=0/5
Bài 1:
a,\(|x-3|+|2-x|=0\)
b,\(\left(2-\dfrac{3}{4}x\right).\left(x+1\right)=0\)
bài 2:
a,A=\(\dfrac{\dfrac{-6}{7}+\dfrac{6}{13}-\dfrac{6}{29}}{\dfrac{9}{7}-\dfrac{9}{13}+\dfrac{9}{29}}\)
b,B=\(\dfrac{\dfrac{2}{15}-\dfrac{2}{21}+\dfrac{2}{39}}{0,25-\dfrac{5}{28}+\dfrac{5}{52}}\)
c,C=\(\dfrac{50-\dfrac{4}{15}+\dfrac{2}{15}-\dfrac{2}{17}}{100-\dfrac{8}{13}+\dfrac{4}{15}-\dfrac{4}{17}}:\dfrac{1+\dfrac{2}{21}-\dfrac{5}{121}}{\dfrac{65}{121}-\dfrac{26}{71}-13}\)
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: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)
Bài 2:
a: \(=\dfrac{-6\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{29}\right)}{9\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{29}\right)}=\dfrac{-6}{9}=\dfrac{-2}{3}\)
b: \(=\dfrac{\dfrac{2}{15}-\dfrac{2}{21}+\dfrac{2}{39}}{\dfrac{10}{40}-\dfrac{10}{56}+\dfrac{10}{104}}\)
\(=\dfrac{\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{13}\right)}{\dfrac{10}{8}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{2}{3}:\dfrac{5}{4}=\dfrac{2}{3}\cdot\dfrac{4}{5}=\dfrac{8}{15}\)
c: \(=\dfrac{2\left(25-\dfrac{2}{13}+\dfrac{1}{15}-\dfrac{1}{17}\right)}{4\left(25-\dfrac{2}{13}+\dfrac{1}{15}-\dfrac{1}{17}\right)}:\dfrac{1+\dfrac{2}{21}-\dfrac{5}{121}}{13\left(\dfrac{5}{121}-\dfrac{2}{21}-1\right)}\)
=2/4:(-1)/13=2/4x(-13)=-13/2