Tìm x biết:
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-(\frac{2}{3}-\frac{1}{4})\) \()\) Với x\(\in Z\)
K
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10 tháng 8 2019
Câu 1,
x+y=-1/3 ; y+z=5/4 ; x+z= 4/3
=> 2(x+y+z)=9/4
=> x+y+z=9/8
Ta lại có: x+y=-1/3
=> z=9/8 -(-1/3)=35/24
Ta lại có: z+y=5/4
=> y=-5/24
=> x=.....
Câu 2:
\(-4\le x\le-\frac{11}{18}\)
YS
12 tháng 5 2017
Bài giải:
a, \(11.xx-66=4.x+11\)
\(11x^2-66=4.x+11\)
\(11x^2-66-4.x-11=0\)
\(11x^2-77-4x=0\)
\(11x^2-4x-77=0\)
\(x=\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4.11.\left(-77\right)}}{2.11}\)
\(x=\frac{4+\sqrt{16}+3388}{22}\)
\(x=\frac{4+\sqrt{3404}}{22}\)
\(x=\frac{4+2\sqrt{851}}{22}\)
\(x=\frac{2-\sqrt{851}}{11}\)
\(\Rightarrow\)Có hai trường hợp: \(x_1=\frac{2-\sqrt{851}}{11};x_2=\frac{2+\sqrt{851}}{11}\)
Tớ bận rồi, cậu coi câu trên đã nhé ! Tớ xin lỗi, khi nào tớ sẽ làm tiếp =))
TB
25 tháng 6 2017
\(\frac{2}{3}\) .\(\frac{3}{4}\)\(\le\)\(\frac{x}{18}\) \(\le\)\(\frac{7}{3}\).\(\frac{1}{3}\)
\(\frac{1}{2}\le\frac{x}{18}\le\frac{7}{9}\)
\(\frac{9}{18}\le\frac{x}{18}\le\frac{14}{18}\)
\(\Rightarrow x\in\){9:10;11;12;13;14}
ML
25 tháng 6 2017
\(\frac{2}{3}.\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\)
\(\frac{2}{3}.\left(\frac{5}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}.\frac{1}{3}\)
\(\frac{2}{3}.\frac{11}{12}\le\frac{x}{18}\le\frac{7}{9}\)
\(\frac{11}{18}\le\frac{x}{18}\le\frac{7}{9}\)
\(\frac{11}{18}\le\frac{x}{18}\le\frac{14}{18}\)
Vậy \(x\in\left\{11;12;13\right\}\)
S
18 tháng 5 2017
Bài 3:
a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
3A = \(1-\frac{1}{2^6}\)
=> 3A < 1
=> A < \(\frac{1}{3}\)(đpcm)
b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)
4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
4B = \(3-\frac{1}{3^{99}}\)
=> 4B < 3
=> B < \(\frac{3}{4}\) (2)
Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)
HL
17 tháng 8 2017
a)Ta có: 1/2-(1/3+1/4)= -1/12
1/48-(1/16-1/6)=1/8
suy ra: -1/12<x<1/8
<=> -2/24<x<3/24
=>x thuộc:(-1/24 ;0 ;1/24 ;2/24 ;3/24)
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(\Rightarrow\frac{-1}{12}\le\frac{x}{12}< 1-\frac{5}{12}\)
\(\Rightarrow\frac{-1}{12}\le\frac{x}{12}< \frac{7}{12}\)
\(\Rightarrow-1\le x< 7\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5;6\right\}\)
= 9/12 - 10/12 =< x/12 < 1 -( 8/12 - 3/12)
= -1/12 =< x/12 < 7/12
=> x thuộc -1,0,1,2,3,4,5,6