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\(\cdot\dfrac{10}{x}< \dfrac{x}{11}\Rightarrow x^2< 110\Rightarrow-\sqrt{110}< x< \sqrt{110}\\ \cdot\dfrac{x}{11}< \dfrac{12}{x}\Rightarrow x^2< 132\Rightarrow-\sqrt{132}< x< \sqrt{132}\)\(\Rightarrow-\sqrt{132}< x< \sqrt{132}\)
\(\dfrac{10}{x}< \dfrac{x}{11}< \dfrac{12}{x}\)
<=> \(\dfrac{110}{11x}< \dfrac{x^2}{11x}< \dfrac{132}{11x}\)
<=> 110 < x2 < 132
<=> x2 = 121
<=> x = 11
@Khánh Linh
a: \(\dfrac{7}{11}< x-\dfrac{1}{7}< \dfrac{10}{13}\)
\(\Leftrightarrow\dfrac{7}{11}+\dfrac{1}{7}< x< \dfrac{10}{13}+\dfrac{1}{7}\)
hay 60/77<x<83/91
b: \(\dfrac{7}{9}< \dfrac{13}{11}-x< \dfrac{15}{16}\)
\(\Leftrightarrow\dfrac{-7}{9}>x-\dfrac{13}{11}>-\dfrac{15}{16}\)
\(\Leftrightarrow-\dfrac{7}{9}+\dfrac{13}{11}>x>\dfrac{-15}{16}+\dfrac{13}{11}\)
\(\Leftrightarrow\dfrac{40}{99}>x>\dfrac{43}{176}\)
Do \(\dfrac{6}{5}< x-\dfrac{3}{2}< \dfrac{12}{5}\)
\(\Leftrightarrow x-\dfrac{3}{2}\in\left\{\dfrac{7}{5};\dfrac{8}{5};\dfrac{9}{5};\dfrac{10}{5};\dfrac{11}{5}\right\}\)
\(\Rightarrow\)Ta có bảng:
| \(x-\dfrac{3}{2}\) | \(x\) | \(KL\) |
| \(\dfrac{7}{5}\) | \(\dfrac{29}{10}\) | \(TM\) |
| \(\dfrac{8}{5}\) | \(\dfrac{31}{10}\) | \(TM\) |
| \(\dfrac{9}{5}\) | \(\dfrac{33}{10}\) | \(TM\) |
| \(\dfrac{10}{5}\) | ... | ... |
| \(\dfrac{11}{5}\) | ... | ... |
Mấy ô còn lại bạn tự tính rồi KL nhé, mình mỏi tay quá!
a; \(\dfrac{6}{x}\) < \(\dfrac{x}{7}\) < \(\dfrac{8}{x}\)
vì \(x\) \(\in\) N* ta có: 6.7 < \(x.x\) < 7.8
42 < \(x^2\) < 56
\(x^2\) = 49
\(x\) = \(\pm\) 7
Vì \(x\) \(\in\) N*; \(x\) = 7
b; \(\dfrac{x}{11}\) < \(\dfrac{12}{x}\) < \(\dfrac{x}{9}\)
9.12< \(x^2\) < 11.12
108 < \(x^2\) < 132
\(x^2\) = 121
\(\left[{}\begin{matrix}x=-11\\x=11\end{matrix}\right.\)
Vì \(x\in\) N*
\(x\) = 11
a) ta co:
1/18<x/12<y/9<1/4
=>2/36<x.3/36<y.4/36<9/36
=>x.3thuộc{3;6};y.4thuộc{4;8}
=>x thuộc{1;2};y thuộc{1:2}
b) ta co
7/8<x/40<9/10
=>70/80<x.2/40<72/80
=>x.2 =71
=>x=71/2
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}>\dfrac{-2}{7}\\\dfrac{x}{3}< \dfrac{11}{4}\end{matrix}\right.\) \(\begin{matrix}\left(1\right)\\\left(2\right)\end{matrix}\)
(1) <=> \(\dfrac{x}{3}+\dfrac{2}{7}>0\Leftrightarrow\dfrac{7x+6}{21}>0\Leftrightarrow7x+6>0;x>-\dfrac{6}{7}\) \(x\in z\Rightarrow x\ge0\)(a)
(2)<=>\(\dfrac{x}{3}-\dfrac{11}{4}< 0;4x-33< 0;x< \dfrac{33}{4};x\in z;x\le8\) (b)
từ (a) và (b) => x ={0;1;2;3;4;5'6;7;8}
\(\dfrac{1}{6}.x+\dfrac{1}{10}.x-\dfrac{4}{15}.x+1=0\)
=> \(\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{4}{15}\right).x+1\)= 0
=> \(0.x+1=0\)
=> 0 . x = 0 - 1
=> 0 . x = -1
=> x = 0 : -1
=> x = 0
~ Chúc bạn học giỏi ! ~
\(\dfrac{1}{6}x+\dfrac{1}{10}x-\dfrac{4}{15}x+1=0\)
\(\Leftrightarrow\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{4}{15}\right)x+1=0\)
\(\Leftrightarrow0x+1=0\)\(\)
Vì 0x luôn = 0 => 0x + 1 > 0
<=> x \(\in\varnothing\)
a: \(\Leftrightarrow70+18< x< 120+126+70\)
=>88<x<316
hay \(x\in\left\{89;90;...;315\right\}\)
b: \(\Leftrightarrow-\dfrac{9}{3}< x< \dfrac{8}{5}+\dfrac{9}{5}=\dfrac{17}{5}\)
=>-3<x<3,4
hay \(x\in\left\{-2;-1;0;1;2;3\right\}\)
x=11
Tức là :10/11 < 11/11 hay 1 < 12/11