c)4.(x+1)-3.(x+1)=14
e)l x+4 l . l x+5 l > 0
f)1+2+3+....+x=1275 help me
o l m . v n
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a) x (x - 2) = 0
=> x = 0 hoặc x - 2 = 0 <=> x = 2
Vậy pt có nghiệm là S = {0; 2}
b) x2 + 2012x = 0
<=> x (x + 2012) = 0
=> x = 0 hoặc x + 2012 = 0 <=> x = -2012
Vậy pt có nghiệm là S = {0; -2012}
c) 4 (x + 1) - 3(x + 1) = 14
<=> (4 - 3) (x + 1) = 14
<=> 1 (x + 1) = 14
<=> x = 14 - 1
<=> x = 13
Vậy pt có nghiệm là S = {13}
d) x2 - 9 < 0
<=> x2 < 9
<=> x < 3
Vậy pt có nghiệm là S = {x thuộc R/ x < 3}
#Học tốt
\(x+1+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\Rightarrow x+1+x+2+x+3+...+x+100=5750\)
\(\Rightarrow100x+1+2+3+...+100=5750\)
\(\Rightarrow100x+\left[\left(\dfrac{100-1}{1}+1\right):2\right]\left(100+1\right)=5750\)
\(\Rightarrow100x+5050=5750\)
\(\Rightarrow100x=700\Rightarrow x=7\)
\(25-\left(30+x\right)=x-\left(123-67\right)\)
\(\Rightarrow25-30+x=x-123+67\)
\(\Rightarrow-5+x=x-56\)
\(\Rightarrow x\in\varnothing\)
\(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\Rightarrow x=5\\\left(x-5\right)^2-1=0\Rightarrow\left(x-5\right)^2=1\Rightarrow x=6;4\end{matrix}\right.\)
\(\left(x^2+1\right)\left(x-3\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+1>0\Rightarrow x^2>-1\\x-3< 0\Rightarrow x< 3\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+1< 0\Rightarrow x^2< -1\\x-3>0\Rightarrow x>3\end{matrix}\right.\end{matrix}\right.\)
Vậy...
Câu 2:
a: Để A là số nguyên thì \(x-1\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{2;0;4;-2\right\}\)
b: Để -4/2x-1 là số nguyên thì \(2x-1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{1;0;\dfrac{3}{2};-\dfrac{1}{2};\dfrac{5}{2};-\dfrac{3}{2}\right\}\)
c: Để 3x+7/x-1là số nguyên thì \(3x-3+10⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)
hay \(x\in\left\{2;0;3;-1;6;-4;11;-9\right\}\)
d: Để 4x-1/x-3 là số nguyên thì \(4x-12+11⋮x-3\)
\(\Leftrightarrow x-3\in\left\{1;-1;11;-11\right\}\)
hay \(x\in\left\{4;2;14;-8\right\}\)
a/ \(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)+\dfrac{4}{5}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{5}{2}-x=1+\dfrac{4}{5}\)
\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{9}{5}+\dfrac{5}{2}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{43}{10}\)
\(\Rightarrow x=\dfrac{43}{5}\)
b/ \(\dfrac{1}{6}\left(2x-3\right)=\dfrac{1}{2}\left(-x+\dfrac{1}{4}\right)-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{1}{3}x-\dfrac{1}{2}=-\dfrac{1}{2}x+\dfrac{1}{8}-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{1}{3}x+\dfrac{1}{2}x=\dfrac{1}{8}-\dfrac{2}{3}+\dfrac{1}{2}\)
\(\Rightarrow\dfrac{5}{6}x=-\dfrac{1}{24}\Rightarrow x=-\dfrac{1}{20}\)
c/ làm như b
d/ \(\left(x-1\right)^4=\left(x-1\right)^6\)
\(\Rightarrow\left[{}\begin{matrix}x-1=-1\\x-1=0\\x-1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;2\right\}\)
a/ \(\left|x-\frac{1}{3}\right|=2\frac{1}{5}\)
\(\left|x-\frac{1}{3}\right|=\frac{11}{5}\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{11}{5}\\x-\frac{1}{3}=\frac{-11}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{11}{5}+\frac{1}{3}\\x=\frac{-11}{5}+\frac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{33}{15}+\frac{5}{15}\\x=\frac{-33}{15}+\frac{5}{15}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{38}{15}\\x=\frac{28}{15}\end{matrix}\right.\)
Vậy:
c/
\(\left|x+\frac{1}{4}\right|-\frac{3}{4}=5\%\)
\(\left|x+\frac{1}{4}\right|-\frac{3}{4}=\frac{1}{20}\)
\(\left|x+\frac{1}{4}\right|=\frac{1}{20}+\frac{15}{20}\)
\(\left|x+\frac{1}{4}\right|=\frac{4}{5}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{4}=\frac{4}{5}\\x+\frac{1}{4}=\frac{-4}{5}\end{matrix}\right.\)
Bạn tự tính tiếp nhé!
2)
a) \(B=2\left|x\right|-3\left|y\right|\text{ tại }x=\dfrac{1}{2};\text{ }y=-3\)
Thay \(x=\dfrac{1}{2}\) và \(y=-3\) vào biểu thức B ta được:
\(2\left|\dfrac{1}{2}\right|-3\left|-3\right|=2.\dfrac{1}{2}-3.3=1-9=-8\)
Vậy giá trị cuar biểu thức \(B=2\left|x\right|-3\left|y\right|\text{ tại }x=\dfrac{1}{2};\text{ }y=-3\) là -8