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2) $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$
$=>\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1$
$=>\dfrac{x+4}{2000}+\dfrac{2000}{2000}+\dfrac{x+3}{2001}+\dfrac{2001}{2001}=\dfrac{x+2}{2002}+\dfrac{2002}{2002}+\dfrac{x+1}{2003}+\dfrac{2003}{2003}$
$=>\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}$
$=>\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0$
$=>(x+2004)(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}=0$
$=>x+2004=0$
$=>x=-2004$
3) Ta có : $A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}$
$=>A=\dfrac{1}{2}+\dfrac{1}{12}+...+\dfrac{1}{99.100}>\dfrac{1}{2}+\dfrac{1}{12}=\dfrac{7}{12}$
$=>A>\dfrac{7}{12}(1)$
Ta lại có : $A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1-\dfrac{1}{2}+\dfrac{1}{3})-(\dfrac{1}{4}-\dfrac{1}{5})-...-(\dfrac{1}{98}-\dfrac{1}{99})-\dfrac{1}{100}<(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$
$=>A<\dfrac{5}{6}(2)$
Từ (1)(2) => đpcm.
a: \(=\left(\dfrac{5^4}{5^2\cdot7^2}\right)^{15}:\left(\dfrac{5^6}{2\cdot7\cdot31}\right)^7\)
\(=\dfrac{5^{30}}{7^{30}}:\dfrac{5^{42}}{2^7\cdot7^7\cdot31^7}\)
\(=\dfrac{5^{30}}{7^{30}}\cdot\dfrac{2^7\cdot7^7\cdot31^7}{5^{42}}=\dfrac{2^7\cdot31^7}{7^{23}\cdot5^{12}}\)
b: \(=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\)
\(=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-7\cdot4\right)}{5^{29}\cdot2^8\cdot7^{48}}=5\cdot\left(-27\right)=-135\)
\(a)\dfrac{-5}{21}-\dfrac{1}{3}+3\dfrac{1}{2}.\left(\dfrac{-2}{3}\right)^3\)
\(=\dfrac{-5}{21}+\dfrac{-7}{21}+\dfrac{7}{2}.\dfrac{-8}{27}\)
\(=-\dfrac{4}{7}+\dfrac{-28}{27}\)
\(=\dfrac{-108}{189}+\dfrac{-196}{189}\)
\(=-\dfrac{304}{189}\)
\(b)-2\dfrac{1}{3}+\left(\dfrac{3}{8}-\dfrac{3}{4}\right)^3:\dfrac{5}{9}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\left(\dfrac{3}{8}-\dfrac{6}{8}\right)^3.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\left(-\dfrac{3}{8}\right)^3.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\dfrac{-27}{512}.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\dfrac{-243}{2560}-\dfrac{1}{2}\)
\(=\dfrac{-17920}{7680}+\dfrac{-729}{7680}+\dfrac{-3840}{7680}\)
\(=\dfrac{-22489}{7680}\)
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..............................
Giải:
a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)
\(\Leftrightarrow x=\dfrac{-63}{10}\)
Vậy ...
b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-4}{11}\)
Vậy ...
Các câu sau làm tương tự câu b)
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
\(\dfrac{x}{2^2}+\dfrac{x}{2^3}+\dfrac{x}{2^4}=\dfrac{x}{3^2}+\dfrac{x}{3^3}+\dfrac{x}{3^4}\)
\(\Leftrightarrow\dfrac{x}{2^2}+\dfrac{x}{2^3}+\dfrac{x}{2^4}-\dfrac{x}{3^2}-\dfrac{x}{3^3}-\dfrac{x}{3^4}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-\dfrac{1}{3^4}\right)=0\)
\(\Leftrightarrow x=0\)
Vậy x = 0
a)
ta có \(\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{1}{13}.\dfrac{1}{14}=\dfrac{3}{7}.9.\dfrac{1}{2}.\dfrac{1}{13}-\dfrac{1}{13}.\dfrac{1}{14}\)\(=\dfrac{1}{13}.\left(\dfrac{3}{7}.\dfrac{9}{2}-\dfrac{1}{14}\right)=\dfrac{1}{13}.\dfrac{26}{14}=\dfrac{1.26}{13.14}\)\(=\dfrac{1.13.2}{13.7.2}=\dfrac{1}{7}\)
b)\(x-\left(\dfrac{5}{2}+2x\right)=x-\dfrac{5}{2}-2x=-x-\dfrac{5}{2}=\dfrac{7}{4}\)
\(\Rightarrow-x=\dfrac{7}{4}+\dfrac{5}{2}=\dfrac{17}{4}\)
\(\Rightarrow x=-\dfrac{17}{4}\)(vì -x là số đối của x)
\(\dfrac{3}{2}x-\dfrac{7}{3}=-\dfrac{1}{4}\)
\(\dfrac{3}{2}x\) \(=\dfrac{1}{4}+\dfrac{7}{3}\)
\(\dfrac{3}{2}x\) \(=\dfrac{3+28}{12}=\dfrac{31}{12}\)
\(x\) \(=\dfrac{31}{12}.\dfrac{2}{3}=\dfrac{31.1}{6.3}=\dfrac{31}{18}\)
mơn nha :>