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Câu 5:
\(168=2^3\cdot3\cdot7\)
\(180=2^2\cdot3\cdot5\)
UCLN(168;180)=12
BCNN(168;180)=840
Câu 4:
a: =>518-x+144=-36
=>662-x=-36
hay x=698
b: \(\Leftrightarrow3x=30\)
hay x=10
c: \(\Leftrightarrow2x-8=16:2=8\)
=>2x=16
hay x=8
Bài 1:
\(A=3+3^2+...+3^{100}\)
\(\Rightarrow3A=3^2+3^3+...+3^{101}\)
\(\Rightarrow3A-A=3^{101}-3\)
\(\Rightarrow2A=3^{101}-3\)
\(\Rightarrow A=\frac{3^{101}-3}{2}\)
b) Ta có: \(\left|x\right|=3\Rightarrow\left\{\begin{matrix}y=3\\y=-3\end{matrix}\right.\)
Thay y = 3 vào B ta có:
B = ..............
Thay y = -3 vào B ta có:
B = .................
Vậy B = ......................
Câu 3:
Ta có: \(\left|x\right|+\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\) ( mỗi số hạng \(\ge0\) )
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+x+1+x+2+x+3=6x\)
\(\Rightarrow4x+6=6x\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
Vậy x = 3
Câu 4:
Ta có: \(n^2-n-1⋮n-1\)
\(\Rightarrow n\left(n-1\right)-1⋮n-1\)
\(\Rightarrow1⋮n-1\)
\(\Rightarrow n-1\in\left\{1;-1\right\}\)
\(\Rightarrow n\in\left\{2;0\right\}\)
Vậy \(n\in\left\{2;0\right\}\)
Bài 1:
\(a,A=3,2.\frac{15}{24}-\left(80\%+\frac{2}{3}\right):3\frac{2}{3}\) \(b,B=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{4}+\frac{3}{8}-\frac{5}{12}}+\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{8}}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\)
\(=\frac{16}{5}.\frac{5}{8}-\left(\frac{4}{5}+\frac{2}{3}\right):\frac{11}{3}\) \(=\frac{\frac{6+9-10}{12}}{\frac{12+18-10}{48}}+\frac{\frac{30+24-15}{40}}{\frac{10+8-5}{40}}\)
\(=2-\frac{22}{15}.\frac{3}{11}\) \(=\frac{\frac{5}{12}}{\frac{20}{48}}+\frac{\frac{39}{40}}{\frac{13}{40}}\)
\(=2-\frac{2}{5}\) \(=\frac{5}{12}:\frac{5}{6}+\frac{39}{40}:\frac{13}{40}\)
\(=\frac{8}{5}\) \(=\frac{5}{12}.\frac{6}{5}+\frac{39}{40}.\frac{40}{13}\)
\(=\frac{1}{2}+3=3\frac{1}{2}\)
Hok tốt
Như thế này:
Từ A=.....=\(\frac{8}{5}\)
Còn từ B=....=\(3\frac{1}{2}\)
Bài 1:
a) \(\left(n+1\right)^3=2\left(n+1\right)^2\)
\(\Rightarrow\left(n+1\right)^3-2\left(n+1\right)^2=0\)
\(\Rightarrow\left(n+1\right)^2\left[\left(n+1\right)-2\right]\)
\(\Rightarrow\left(n+1\right)^2\left(n-1\right)=0\)
\(\Rightarrow\left\{\begin{matrix}\left(n+1\right)^2=0\\n-1=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}n+1=0\\n=1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}n=-1\\n=1\end{matrix}\right.\)
Vậy...
b) Ta có: \(3n-7⋮6n+5\)
\(\Rightarrow6n-14⋮6n+5\)
\(\Rightarrow\left(6n+5\right)-19⋮6n+5\)
\(\Rightarrow19⋮6n+5\)
\(\Rightarrow6n+5\in\left\{1;-1;19;-19\right\}\)
\(\left\{\begin{matrix}6n+5=1\\6n+5=-1\\6n+5=19\\6n+5=-19\end{matrix}\right.\Rightarrow\left\{\begin{matrix}n=-\frac{2}{3}\left(loại\right)\\n=-1\left(t.m\right)\\n=\frac{7}{3}\left(loại\right)\\n=-4\left(t.m\right)\end{matrix}\right.\)
Vậy \(n\in\left\{-1;-4\right\}\)
Bài 5:
\(xy+4x+3y+12=7\)
\(\Rightarrow x\left(y+4\right)+3\left(y+4\right)=7\)
\(\Rightarrow\left(x+3\right)\left(y+4\right)=7\)
\(\Rightarrow\left\{\begin{matrix}x+3=7\\y+4=1\end{matrix}\right.\) hoặc \(\left\{\begin{matrix}x+3=-7\\y+4=-1\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=4\\y=-3\end{matrix}\right.\) hoặc \(\left\{\begin{matrix}x=-10\\y=-5\end{matrix}\right.\)
Vậy...
Bài 7:
a) Ta có: \(\left(2x-6\right)^{10}+\left|3x+y\right|\ge0\)
\(\Rightarrow A=\left(2x-6\right)^{10}+\left|3x+y\right|-7\ge-7\)
\(\Rightarrow MIN_A=-7\) khi \(\left\{\begin{matrix}\left(2x-6\right)^{10}=0\\\left|3x+y\right|=0\end{matrix}\right.\)
Vậy \(MIN_A=-7\) khi \(x=3;y=-9\)
toán nớ mà lp 6 hả cha
có trả lời ko