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Giải:
a) \(\dfrac{1}{2}< x< \dfrac{7}{8}\)
\(\Leftrightarrow\dfrac{12}{24}< x< \dfrac{21}{24}\)
\(\Leftrightarrow x\in\left\{\dfrac{13}{24};\dfrac{14}{24};\dfrac{15}{24};\dfrac{16}{24};\dfrac{17}{24};\dfrac{18}{24};\dfrac{19}{24};\dfrac{20}{24}\right\}\)
Mà x là số hữu tỉ có mẫu là 24
\(\Leftrightarrow x=\left\{\dfrac{13}{24};\dfrac{17}{24};\dfrac{19}{24}\right\}\)
Vậy ...
b) \(\dfrac{3}{5}< x< \dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{12}{20}< x< \dfrac{12}{15}\)
\(\Leftrightarrow x\in\left\{\dfrac{12}{19};\dfrac{12}{18};\dfrac{12}{17};\dfrac{12}{16}\right\}\)
Mà x là số hữu tỉ có tử là 12
\(\Leftrightarrow x=\left\{\dfrac{12}{19};\dfrac{12}{17}\right\}\)
Vậy ...
b: \(\left|x-\dfrac{3}{5}\right|< \dfrac{1}{3}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{3}{5}>-\dfrac{1}{3}\\x-\dfrac{3}{5}< \dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\dfrac{4}{15}< x< \dfrac{14}{15}\)
c: \(\left|x+\dfrac{11}{2}\right|>-5.5\)
mà \(\left|x+\dfrac{11}{2}\right|\ge0\forall x\)
nên \(x\in R\)
Nhiều quá, từng bài 1 nhé, bài nào làm được, tớ sẽ cố gắng.
bài 2:
a) \(x>2x\Leftrightarrow x-2x>0\Leftrightarrow-x>0\Leftrightarrow x< 0\)
Kl: x<0
b) \(a+x< a\Leftrightarrow x< 0\)
Kl: x<0
c) \(x^3>x^2\Leftrightarrow x^3-x^2>0\Leftrightarrow x^2\left(x-1\right)>0\) (*)
Mà x^2 > 0 \(\Rightarrow\) (*) \(\Leftrightarrow x-1>0\Leftrightarrow x>1\)
Kl: x>1
Câu 4:
a) \(1-2x< 7\Leftrightarrow2x>-6\Leftrightarrow x>3\)
Kl: x>3
b) \(\left(x-1\right)\left(x-2\right)>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< 2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< 1\end{matrix}\right.\)
Kl: x>2 hoặc x<1
c) \(\left(x-2\right)^2\left(x+1\right)\left(x+4\right)< 0\Leftrightarrow\left(x+1\right)\left(x+4\right)< 0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x+4< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\\x+4>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-1\\x< -4\end{matrix}\right.\\\left\{{}\begin{matrix}x< -1\\x>-4\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-1< x< -4\left(vô-lý\right)\\-4< x< -1\end{matrix}\right.\) \(\Leftrightarrow-4< x< -1\)
Kl: -4<x<-1
d) ĐK: x khác 9\(\dfrac{x^2\left(x+3\right)}{x-9}< 0\Leftrightarrow x^2\left(x+3\right)\left(x-9\right)< 0\Leftrightarrow\left(x+3\right)\left(x-9\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3>0\\x-9< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0\\x-9>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-3\\x< 9\end{matrix}\right.\\\left\{{}\begin{matrix}x< -3\\x>9\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-3< x< 9\left(N\right)\\9< x< -3\left(vô-lý\right)\end{matrix}\right.\) \(\Leftrightarrow-3< x< 9\)
Kl: -3<x<9
e) Đk: x khác 0
\(\dfrac{5}{x}< 1\Leftrightarrow\dfrac{5}{x}< \dfrac{5}{5}\Leftrightarrow x>5\left(N\right)\)
KL: x >5
f) ĐK: x khác 1
\(\dfrac{2x-5}{x-1}< 0\Leftrightarrow\left(2x-5\right)\left(x-1\right)< 0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5>0\\x-1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5< 0\\x-1>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{5}{2}\\x< 1\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x>1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{2}< x< 1\left(vô-lý\right)\\1< x< \dfrac{5}{2}\left(N\right)\end{matrix}\right.\)
Kl: 1< x< 5/2
1.
\(\left(\dfrac{-1}{8}+\dfrac{-5}{6}\right)\cdot\dfrac{6}{23}\\ =-\dfrac{23}{24}\cdot\dfrac{6}{23}\\ =-\dfrac{6}{24}=-\dfrac{1}{4}\)
2. Xem lại đề nha!
4.
\(x+0,75=-1\dfrac{1}{4}\\ x+\dfrac{3}{4}=-\dfrac{3}{4}\\ x=-\dfrac{3}{4}-\dfrac{3}{4}\\ x=-\dfrac{3}{4}+\left(-\dfrac{3}{4}\right)=-\dfrac{6}{4}=-\dfrac{3}{2}\)
5.
\(\dfrac{x}{28}=-\dfrac{4}{7}\\ \Leftrightarrow7x=-4.28\\ \Rightarrow7x=-112\\ \Rightarrow x=-112:7=-16\)
6.
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Leftrightarrow\dfrac{x}{7}=\dfrac{y}{9}\Leftrightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Vậy giá trị của tỉ số \(\dfrac{x}{y}=\dfrac{7}{9}\).
a/dễ --> tự lm
b/ \(\left(x-\dfrac{4}{7}\right)\left(1\dfrac{3}{5}+2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\1\dfrac{3}{5}+2x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\2x=\dfrac{8}{5}\Rightarrow x=\dfrac{4}{5}\end{matrix}\right.\)
Vậy...............
c/ \(\left(x-\dfrac{4}{7}\right):\left(x+\dfrac{1}{2}\right)>0\)
TH1: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{4}{7}\\x>-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{4}{7}\)
TH2: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x< -\dfrac{1}{2}\)
Vậy \(x>\dfrac{4}{7}\) hoặc \(x< -\dfrac{1}{2}\) thì thỏa mãn đề
d/ \(\left(2x-3\right):\left(x+1\dfrac{3}{4}\right)< 0\)
TH1: \(\left\{{}\begin{matrix}2x-3>0\\x+1\dfrac{3}{4}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>1,5\\x< -\dfrac{7}{4}\end{matrix}\right.\)(vô lý)
TH2: \(\left\{{}\begin{matrix}2x-3< 0\\x+1\dfrac{3}{4}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< 1,5\\x>-\dfrac{7}{4}\end{matrix}\right.\)\(\Rightarrow-\dfrac{7}{4}< x< 1,5\)
Vậy...................
Vì |5/7x-4|<2/7
suy ra 5/7x-4<2/7
5/7x<30/7
x<30/7:5/7
x<6
Chúc bạn học tốt
|5/7x-4|<2/7
=>-2/7<5/7x-4<2/7
=>26/7<5/7x<30/7
=>26<5x<30
=>26/5<x<6
6) Tìm giá trị lớn nhất : A = 0,5 - | x - 3,5 |
Vì | x - 3,5 | \(\ge\) 0
nên A= 0,5 - | x - 3,5 | \(\le\) 0,5
GTLN của A là 0,5 khi và chỉ khi x-3,5= 0
=> x= 3,5
5) Tìm x thuộc Q :(x +1)(x-2) < 0
Để (x +1)(x-2) \(\in Q\)
Thì x+1 và x-2 khác dấu
mà ta thấy x+1 > x-2 ( luôn luôn xảy ra)
=> x+1\(\ge\)0 => x= -1
x-2\(\le\) 0 => x= 2
Vậy -1 <x <2
vậy: x \(\in\) 0;1
bài 4:
gọi x. y, z, k lần lượt là số học sinh khối 6, 7, 8,9
theo đề ta có:
\(\dfrac{x}{11}=\dfrac{y}{10}=\dfrac{z}{9}=\dfrac{k}{8}\) và y-k= 22
=> \(\dfrac{x}{11}=\dfrac{y}{10}=\dfrac{z}{9}=\dfrac{k}{8}\)= \(\dfrac{y-k}{10-8}=\dfrac{22}{2}=11\)
=> x= 121
y= 110
z= 99
k= 88
Vậy khối 6, 7, 8, 9 có..............................
\(\left(x-1\right)\left(x+5\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x+5>0\Rightarrow x>-5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x+5< 0\Rightarrow x< -5\end{matrix}\right.\end{matrix}\right.\)
\(\left(x-1\right)\left(x+5\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x+5< 0\Rightarrow x< -5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x+5>0\Rightarrow x>-5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-5< x< 1\)
câu dễ tự làm
\(\Rightarrow x>-5;x< -5\)
a: 2x(x-1/7)=0
=>x(x-1/7)=0
=>x=0 hoặc x=1/7
b: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
nên \(x=\dfrac{-1}{4}:\dfrac{7}{20}=\dfrac{-20}{4\cdot7}=\dfrac{-5}{7}\)
c: \(\Leftrightarrow\dfrac{41}{9}:\dfrac{41}{18}-7< x< \left(3.2:3.2+\dfrac{45}{10}\cdot\dfrac{31}{45}\right):\left(-21.5\right)\)
\(\Leftrightarrow2-7< x< \dfrac{\left(1+3.1\right)}{-21.5}\)
\(\Leftrightarrow-5< x< \dfrac{-41}{215}\)
mà x là số nguyên
nên \(x\in\left\{-4;-3;-2;-1\right\}\)