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a) \(1\frac{2}{7} = 1 + \frac{2}{7} = \frac{9}{2}\)
\(\begin{array}{l}x:1\frac{2}{7} = - 3,5\\x:\frac{9}{7} = - \frac{7}{2}\\x = - \frac{7}{2}.\frac{9}{7}\\x = - \frac{9}{2}\end{array}\)
b) \(0,4.x - \frac{1}{5}.x = \frac{3}{4}\)
\(\begin{array}{l}\frac{2}{5}.x - \frac{1}{5}.x = \frac{3}{4}\\\left( {\frac{2}{5} - \frac{1}{5}} \right).x = \frac{3}{4}\\\frac{1}{5}.x = \frac{3}{4}\\x = \frac{3}{4}:\frac{1}{5}\\x = \frac{3}{4}.5\\x = \frac{{15}}{4}\end{array}\)
a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
Gọi d là ƯCLN của 12n+1 và 30n+2
=> 12n+1 chia hết cho d. 30n+2 chia hết cho d
=> (12n+1) - (30n+2) chia hết cho d
=.> 5(12n+1) - 2(30n+2) chia hết cho d
=> 1 chia hết cho d
Ta có d C Ư(1) = [-1;1]
Vây phân số \(\frac{12n+1}{30n+2}\)là phân số tối giản
\(2\frac{3}{4}x=3\frac{1}{7}:0,01\)
=> \(\frac{11}{4}x=\frac{22}{7}:\frac{1}{100}\)
=> \(x=\frac{\frac{22}{7}\cdot100}{\frac{11}{4}}=\frac{\frac{2200}{7}}{\frac{11}{4}}=\frac{2200}{7}\cdot\frac{4}{11}=\frac{800}{7}\)
\(2\frac{1}{3}:\frac{1}{3}=\frac{7}{9}:x\)
=> \(\frac{7}{3}:\frac{1}{3}=\frac{7}{9}:x\)
=> \(\frac{7}{3}\cdot\frac{3}{1}=\frac{7}{9}:x\)
=> \(\frac{7}{9}:x=7\)
=> \(x=\frac{7}{9}:7=\frac{7}{9}\cdot\frac{1}{7}=\frac{1}{9}\)
\(3,2x+\left(-1,2\right)x+2,7=-4,9\)
=> \(\left[3,2+\left(-1,2\right)\right]x=-7,6\)
=> 2.x = -7,6
=> x = -3,8
\(x:\frac{9}{14}=\frac{7}{3}:x\)
=> \(x\cdot\frac{14}{9}=\frac{7}{3}\cdot\frac{1}{x}\)
=> \(\frac{14x}{9}=\frac{7}{3x}\)
nên sửa lại đề này
\(\frac{37-x}{x+13}=\frac{3}{7}\)
=> 7(37 - x) = 3(x + 13)
=> 259 - 7x = 3x + 39
=> 259 - 7x - 3x - 39 = 0
=> 220 - 10x = 0
=> 220 = 10x
=> x = 22
\(\frac{x}{15}=\frac{-60}{x}\)=> x2 = -900 => x không thỏa mãn
a) \(2\frac{3}{4}.x=3\frac{1}{7}:0,01\)
\(=>2\frac{3}{4}.x=2200=>\frac{11}{4}x=2200=>x=800\)
b) \(2\frac{1}{3}:\frac{1}{3}=\frac{7}{9}:x\)
\(=>7=\frac{7}{9}:x\)
hay \(\frac{7}{9}:x=7=>x=\frac{1}{9}\)
c) \(3,2x+\left(-1,2\right)x+2,7=-4,9\)
\(=>2x+2,7=-4,9\)
\(=>2x=-7,6=>x=-3,8\)
d) \(x:\frac{9}{14}=\frac{7}{3}:x\)
\(=>x.x=\frac{7}{3}.\frac{9}{14}\)
\(=>x^2=\frac{3}{2}\)
....................( ko hiểu)
e) \(\frac{37-x}{x+13}=\frac{3}{7}\)
\(=>7.\left(37-x\right)=3.\left(x+13\right)\)
\(=>259-7x=3x+39\)
\(-7x-3x=39-269\)
\(=>-10x=-220=>x=22\)
f) \(\frac{x}{15}=\frac{-60}{x}\)
\(=>x.x=-60.15=>x^2=-900=>x=-30\)
cậu có thể tham khảo bài alfm trên đây ạ, chúc học tốt:>
b: \(\Leftrightarrow x+8\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-7;-9;-3;-13\right\}\)
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
\(a,\left(8-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\\ b,2x\left(x+81\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)
a)\(\left(8-x\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\)
b)\(2x\left(x+81\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)