Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
5/17 + -4/9 + -20/31 + -11/31 < x < -3/7 + 7/15 + 4/-7 + 8/15 + 2/3
-175/153<x\(\le\)2/3
-175/153<x\(\le\)102/153
suy ra -175<x\(\le\)153
vậy x bằng (-174/153,-173/153,-172/153,.........,101/153,102/153
Bài 2:
a; 17 - 11 - (-39)
= 17 - 11 + 39
= 6 + 39
= 45
b; 125 - 4[ 3 -7 .(-2)]
= 125 - 4.[3 + 14]
= 125 - 4.17
= 125 - 68
= 57
bài1:a
-3 + 12
= 12 - 3
= 9
b)(-24) : 8 = -3
c)-9 - 13
= -9 + (-13)
=-(9 + 13)
= -22
a) 3x + 27 = 9
3x = 9 - 27
3x = -18
x = -18 : 3
x = - 6
Vậy x=-6
b) 2x + 12 = 3(x - 7 )
2x + 12 = 3x - 21
12 + 21 = 3x - 2x
33 = x
Vậy x=33
c) 2x2 - 1 = 49
2x2=49+1
2x2=50
x2=50:2
x2=25
x2=52
=> x= + 5
Vậy x=+5
d) |x + 9 | . 2 =10
|x + 9 | = 10 : 2
|x + 9 | = 5
\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5-9=-4\\x=-5-9=-14\end{cases}}\)
Vậy \(x\in\left\{-4;-14\right\}\)
\(a,x\in\left\{-4;-3;-2;-1;0;1;2\right\}\)
\(b,x\in\left\{-8;-7;-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6;7;8;9;10;11\right\}\)
\(c,x\in\left\{-5;5\right\}\)
\(d;|x|=|-7|\)
<=>\(|x|=7\)
=>\(x\in\left\{-7;7\right\}\)
\(e,|x|=-|6|\)
<=>\(|x|=-6\)
=>\(x\in\varnothing\)
\(g,|-37|-|x|=|-5|\)
<=>\(37-|x|=5\)
<=>\(|x|=32\)
=>\(x\in\left\{-32;32\right\}\)
\(h,3+|x|=9\)
<=>\(|x|=6\)
=>\(x\in\left\{-6;6\right\}\)
\(i,3< |x|< 7\)
=>\(|x|\in\left\{4;5;6\right\}\)
=>\(x\in\left\{-6;-5;-4;4;5;6\right\}\)
Câu 1:
A = \(\dfrac{-7}{12}+\dfrac{11}{8}-\dfrac{5}{9}=\dfrac{-42}{72}+\dfrac{99}{72}-\dfrac{40}{72}=\dfrac{-42+99-40}{72}=\dfrac{17}{72}\)
\(B=\dfrac{1}{7}-\dfrac{8}{7}:8-3:\dfrac{3}{4}.2^2=\dfrac{1}{7}-\dfrac{8}{7}.\dfrac{1}{8}-3.\dfrac{4}{3}.4=\dfrac{1}{7}-\dfrac{1}{7}-16=0-16=-16\)
\(C=1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}=\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}.\dfrac{5}{11}=\dfrac{3}{7}-\dfrac{2}{3}=\dfrac{9-14}{21}=-\dfrac{5}{21}\)
Vậy A=\(\dfrac{17}{72};B=-16;C=\dfrac{-5}{21}\)
Câu 2:
a. \(\dfrac{-11x}{12}+\dfrac{3}{4}=-\dfrac{1}{6}\)
\(\Rightarrow\dfrac{-11x}{12}=-\dfrac{1}{6}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{-11x}{12}=\dfrac{-2}{12}-\dfrac{9}{12}\)
\(\Rightarrow\dfrac{-11x}{12}=\dfrac{-11}{12}\)
\(\Rightarrow-11x=\dfrac{-11.12}{12}\)
\(\Rightarrow-11x=-11\Rightarrow x=1\)
Vậy x=1
b. \(3-(\dfrac{1}{6}-x).\dfrac{2}{3}=\dfrac{2}{3}\Rightarrow3-\left(\dfrac{1}{6}-x\right)=1\)
\(\Rightarrow-(\dfrac{1}{6}-x)=1-3\Rightarrow\dfrac{1}{6}+x=-2\)
\(\Rightarrow x=2-\dfrac{1}{6}\Rightarrow x=\dfrac{11}{6}\)
Vậy x = \(\dfrac{11}{6}\)
Câu 4:
Ta có: \(\dfrac{1}{2.3}=\dfrac{1}{6}\)
\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3}{6}-\dfrac{2}{6}=\dfrac{1}{6}\)
\(\Rightarrow\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\frac{12}{27}+\frac{2}{3}-\frac{2}{9}\le x\le\frac{11}{7}+\frac{2}{5}-\frac{7}{5}+\frac{3}{7}\)
\(\Leftrightarrow\frac{8}{9}\le x\le1\)
Vậy \(\frac{8}{9}\le x\le1\)
à nhầm ko đọc kĩ đề nên làm lại
\(\left(\frac{12}{27}+\frac{2}{3}\right)-\frac{2}{9}\le x\le\left(\frac{11}{7}+\frac{2}{5}\right)-\frac{7}{5}+\frac{3}{7}\)
\(\Leftrightarrow\frac{8}{9}\le x\le1\)
Mà \(x\in Z\)
\(\Rightarrow x=1\)
Vậy x=1