tìm x biết
a)11(x-6)=4x+11
b)4\(\dfrac{1}{3}\)(\(\dfrac{1}{6}\)-\(\dfrac{1}{2}\))\(\le\)x\(\le\)\(\dfrac{2}{3}\)(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{3}{4}\))với x\(\in\)z
tim x \(\in\) z biet
a . \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{3}{4}\right)\le x\le\dfrac{1}{24}\cdot\left(\dfrac{1}{3}-\dfrac{1}{3}\right)\)
b. \(-4\dfrac{1}{3}\left(\dfrac{1}{2}+\dfrac{1}{6}\right)\le x\le\dfrac{-2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}\cdot\dfrac{3}{4}\right)\)
a. \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{3}{4}\right)\le x\le\dfrac{1}{24}.\left(\dfrac{1}{3}-\dfrac{1}{3}\right)\)
\(\dfrac{1}{2}-\dfrac{13}{12}\le x\le\dfrac{1}{24}.0\) ( lười viết nên điền kết quả luôn )
\(\dfrac{-7}{12}\le x\le0\)
\(0,5833...\le x\le0\)
Vì \(x\in Z\)\(\Rightarrow x\in\left\{0\right\}\)
Vậy...
b. \(-4\dfrac{1}{3}\left(\dfrac{1}{2}+\dfrac{1}{6}\right)\le x\le\dfrac{-2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}.\dfrac{3}{4}\right)\)
\(\dfrac{-26}{9}\le x\le\dfrac{1}{36}\)
\(-2,8888...\le x\le0,277...\)
Vì \(x\in Z\Rightarrow x\in\left\{-2;-1;0\right\}\)
Vậy ...
BT3: Tìm x, biết:
7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\)
8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)
7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\)
\(\Rightarrow\dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\)
\(\Rightarrow-2\le x\le2\)
\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)
8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)
\(\Rightarrow\dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{6}{12}-\dfrac{2}{12}\right)\)
\(\Rightarrow\dfrac{2}{3}\cdot\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}\cdot\dfrac{4}{12}\)
\(\Rightarrow\dfrac{22}{36}\le\dfrac{x}{18}\le\dfrac{28}{36}\)
\(\Rightarrow\dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\)
\(\Rightarrow x\in\left\{11;12;13;14\right\}\)
8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{3}{6}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}.\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}.\dfrac{2}{6}\\ \dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\\ \Rightarrow11\le x\le14\\ \Rightarrow x\in\left\{11;12;13;14\right\}\)
7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\\ \dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\\ \Rightarrow-2\le x\le2\\ \Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)
Tìm số nguyên x biết: a) \(-4\dfrac{3}{5}.2\dfrac{4}{23}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)
b)\(-4\dfrac{1}{3}.\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\le x\le-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< =x< =\dfrac{-13}{5}:\dfrac{21}{15}\)
=>-10<=x<=-13/7
hay \(x\in\left\{-10;-9;...;-2\right\}\)
b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =-\dfrac{2}{3}\cdot\dfrac{-11}{12}\)
=>-13/9<=x<=11/18
hay \(x\in\left\{-1;0\right\}\)
1. Tìm x \(\in\) Z: \(4\dfrac{1}{2}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
\(\dfrac{9}{2}\left(\dfrac{2}{12}-\dfrac{6}{12}\right)\le x\le\dfrac{2}{3}\left(\dfrac{4}{12}-\dfrac{6}{12}-\dfrac{9}{12}\right)\)
\(\Leftrightarrow\dfrac{9}{2}.\dfrac{-4}{12}\le x\le\dfrac{2}{3}.\dfrac{-11}{12}\)\(\Leftrightarrow\dfrac{-3}{2}\le x\le\dfrac{-11}{18}\)
\(\Rightarrow x=-1\)
\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}< x\le\dfrac{13}{4}+\dfrac{14}{8}\)
\(\dfrac{1}{4}+\dfrac{5}{12}+\dfrac{-1}{13}< x< \dfrac{7}{5}+\dfrac{2}{10}+\dfrac{1}{2}\) \(\dfrac{79}{15}+\dfrac{7}{5}+\dfrac{-8}{3}\le x\le\dfrac{10}{3}+\dfrac{15}{4}+\dfrac{23}{12}\)
\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)
\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)
\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)
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) \(\dfrac{79}{15}+\dfrac{7}{5}+\dfrac{-8}{3}\le x\le\dfrac{10}{3}+\dfrac{15}{4}+\dfrac{23}{12}\)
\(\dfrac{79}{15}+\dfrac{21}{15}+\dfrac{-40}{15}\le x\le\dfrac{40}{12}+\dfrac{45}{12}+\dfrac{23}{12}\)
\(4\le x\le9\)
\(\Rightarrow\) \(x\in\) {\(4;5;6;7;8;9\)}
Giải bất phương trình sau
a)\(\dfrac{2-x}{3}\)\(-x-2\le\dfrac{x-17}{2}\)
b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\)
a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\) \(\Leftrightarrow\) \(6\left(\dfrac{2-x}{3}-x-2\right)\le6\left(\dfrac{x-17}{2}\right)\) \(\Leftrightarrow\) 4-2x-6x-12\(\le\)3x-51 \(\Leftrightarrow\) -2x-6x-3x\(\le\)-51-4+12 \(\Leftrightarrow\) -11x\(\le\)-43 \(\Rightarrow\) x\(\ge\)43/11.
b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\) \(\Leftrightarrow\) \(12\left(\dfrac{2x+1}{3}+\dfrac{4-x}{4}\right)\le12\left(\dfrac{3x+1}{6}+\dfrac{4-x}{12}\right)\) \(\Leftrightarrow\) 8x+4+12-3x\(\le\)6x+2+4-x \(\Leftrightarrow\) 8x-3x-6x+x\(\le\)2+4-4-12 \(\Leftrightarrow\) 0x\(\le\)-10 (vô lí).
a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\)
\(\Leftrightarrow2\left(2-x\right)-6\left(x+2\right)\le3\left(x-17\right)\)
\(\Leftrightarrow4-2x-6x-12\le3x-51\)
\(\Leftrightarrow-11x\le-43\)
\(\Leftrightarrow x\ge\dfrac{43}{11}\)
Vậy S = {\(x\) | \(x\ge\dfrac{43}{11}\) }
b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\)
\(\Leftrightarrow4\left(2x+1\right)-3\left(x-4\right)\le2\left(3x+1\right)-\left(x-4\right)\)
\(\Leftrightarrow8x+4-3x+12\le6x+2-x+4\)
\(\Leftrightarrow0x\le-10\) (vô lý)
Vậy \(S=\varnothing\)
Tìm x biết
a)\(\dfrac{11}{12}\).x + \(\dfrac{3}{4}\)= -\(\dfrac{1}{6}\)
b)3-\(\left(\dfrac{1}{6}-x\right)\).\(\dfrac{2}{3}\)=\(\dfrac{2}{3}\)
\(a,\dfrac{11}{12}x+\dfrac{3}{4}=-\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{11}{12}x=-\dfrac{1}{6}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{11}{12}x=-\dfrac{11}{12}\)
\(\Leftrightarrow x=-\dfrac{11}{12}:\dfrac{11}{12}\)
\(\Leftrightarrow x=-\dfrac{11}{12}.\dfrac{12}{11}\)
\(\Leftrightarrow x=-1\)
\(b,3-\left(\dfrac{1}{6}-x\right).\dfrac{2}{3}=\dfrac{2}{3}\)
\(\Leftrightarrow3-\dfrac{2}{3}.\left(\dfrac{1}{6}-x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow3-\dfrac{1}{9}+\dfrac{2}{3}x=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{2}{3}-3+\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{2}{3}x=-\dfrac{20}{9}\)
\(\Leftrightarrow x=-\dfrac{20}{9}:\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{10}{3}\)
Giải đầy đủ pls
Bài 3
\(\dfrac{55}{23}+\dfrac{-22}{23}\le x\le\dfrac{1}{5}-\dfrac{-1}{6}+\dfrac{79}{30}\) có bao nhiêu số nguyên X thỏa mãn
A 1 B 2 C 3 D 4
Bài 4
Nếu \(\dfrac{-11}{12}< \dfrac{5}{x}< \dfrac{-11}{15}\) Thì x là bao nhiêu
A 5 B 6 C -5 D -6
Bài 5
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
A 1 B 2 C \(\dfrac{99}{100}\) D \(\dfrac{1}{100}\)
Bài 3
\(\dfrac{55}{23}+\dfrac{-22}{23}\le x\le\dfrac{1}{5}-\dfrac{-1}{6}+\dfrac{79}{30}\)
\(=\dfrac{33}{23}\)\(\le x\le\dfrac{90}{30}\)
\(=\dfrac{33}{23}\le x\le3\)
Mà \(x\in Z\) \(\Rightarrow\)\(x=2\)
Có 1 giá trị thỏa mãn
Chọn A
Bài 4
\(\dfrac{-11}{12}< \dfrac{5}{x}< \dfrac{-11}{15}\)
Chọn D
Bài 5
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(M=1-\dfrac{1}{100}\)
\(M=\dfrac{100}{100}-\dfrac{1}{100}\)
\(M=\dfrac{99}{100}\)
CHọn C
Tìm số nguyên x biết:
a) \(-4\dfrac{3}{5}\) x \(2\dfrac{4}{3}\) \(\le\) x \(\le\) \(-2\dfrac{3}{5}\) : \(1\dfrac{6}{15}\)
b) \(-4\dfrac{1}{3}\) (\(\dfrac{1}{2}-\dfrac{1}{6}\) ) \(\le\) x \(\le\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{7}\right)\)
c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}\dfrac{x+6}{30}\)
c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}\)
\(\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1\)
\(\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0\)
\(\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0\)
\(\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\ne0\right)\)
\(\Rightarrow x=-36\)
Vậy ...
a/ Ta có: \(-4\dfrac{3}{5}.2\dfrac{4}{3}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)
\(\Rightarrow\dfrac{-23}{5}.\dfrac{10}{3}\le x\le\dfrac{-13}{5}:\dfrac{21}{15}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{5}.\dfrac{15}{21}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{7}\)
\(\Rightarrow-15,\left(3\right)\le x\le-1,\left(857142\right)\)
Vì x \(\in\) Z nên x \(\in\left\{-1;-2;-3;...;-15\right\}\)
Chúc bạn học tốt!!!
mk quên mất, còn câu b/ nữa, câu c thì bạn Nháy giải rồi nên mk khỏi giải nha.
b/\(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\le x\le\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{7}\right)\)
\(\Rightarrow\)\(\dfrac{-13}{3}\left(\dfrac{3}{6}-\dfrac{1}{6}\right)\le x\le\dfrac{2}{3}\left(\dfrac{14}{42}-\dfrac{21}{42}-\dfrac{18}{42}\right)\)
\(\Rightarrow\dfrac{-13}{3}.\dfrac{1}{3}\le x\le\dfrac{2}{3}.\dfrac{-25}{42}\)
\(\Rightarrow\dfrac{-13}{9}\le x\le\dfrac{-25}{63}\)
\(\Rightarrow-1,\left(4\right)\le x\le0,\left(396825\right)\)
Vì x\(\in Z\) nên x \(\in\left\{-1;0\right\}\)
Mk chỉ nghĩ z thôi chứ có j thì góp ý thêm nha.