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1.-12(x-5)+7(3-x)=5.
#-12x+60+21-7x=5
#-12x-7x=5-60-21
#-19x=-76
#x=-76:(-19)
#x=4(TMĐK:x€Z)
Vậy x=4
#là dấu suy ra nhé! Máy mình không có dấu suy ra!
\(=\lim\dfrac{1+\dfrac{1}{n}+\dfrac{5}{n^2}}{2+\dfrac{1}{n^2}}=\dfrac{1}{2}\)
1:
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2-1-9n^2}{\sqrt{n^2-1}-3n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{-8n^2-1}{\sqrt{n^2-1}-3n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(-8-\dfrac{1}{n^2}\right)}{n\left(\sqrt{1-\dfrac{1}{n^2}}-3\right)}=\lim\limits_{n\rightarrow\infty}-\dfrac{8}{1-3}\cdot n=\lim\limits_{n\rightarrow\infty}4n=+\infty\)
2:
\(\lim\limits_{n\rightarrow\infty}\sqrt{4n^2+5}+n\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{4n^2+5-n^2}{\sqrt{4n^2+5}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5}{\sqrt{4n^2+5}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(3+\dfrac{5}{n^2}\right)}{n\left(\sqrt{4+\dfrac{5}{n^2}}-1\right)}\)
\(=\lim\limits_{n\rightarrow\infty}n\cdot\left(\dfrac{3}{\sqrt{4}-1}\right)=+\infty\)
a;Chia n cả tử và mẫu
b;Chia cho n4 mà tử dần đến 0 mẫu dần đến 1 nên lim =0
a) \(\lim \frac{{2{n^2} + 6n + 1}}{{8{n^2} + 5}} = \lim \frac{{{n^2}\left( {2 + \frac{6}{n} + \frac{1}{{{n^2}}}} \right)}}{{{n^2}\left( {8 + \frac{5}{{{n^2}}}} \right)}} = \lim \frac{{2 + \frac{6}{n} + \frac{1}{n}}}{{8 + \frac{5}{n}}} = \frac{2}{8} = \frac{1}{4}\)
b) \(\lim \frac{{4{n^2} - 3n + 1}}{{ - 3{n^3} + 6{n^2} - 2}} = \lim \frac{{{n^3}\left( {\frac{4}{n} - \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}} \right)}}{{{n^3}\left( { - 3 + \frac{6}{n} - \frac{2}{{{n^3}}}} \right)}} = \lim \frac{{\frac{4}{n} - \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}}}{{ - 3 + \frac{6}{n} - \frac{2}{{{n^3}}}}} = \frac{{0 - 0 + 0}}{{ - 3 + 0 - 0}} = 0\).
c) \(\lim \frac{{\sqrt {4{n^2} - n + 3} }}{{8n - 5}} = \lim \frac{{n\sqrt {4 - \frac{1}{n} + \frac{3}{{{n^2}}}} }}{{n\left( {8 - \frac{5}{n}} \right)}} = \frac{{\sqrt {4 - 0 + 0} }}{{8 - 0}} = \frac{2}{8} = \frac{1}{4}\).
d) \(\lim \left( {4 - \frac{{{2^{{\rm{n}} + 1}}}}{{{3^{\rm{n}}}}}} \right) = \lim \left( {4 - 2 \cdot {{\left( {\frac{2}{3}} \right)}^{\rm{n}}}} \right) = 4 - 2.0 = 4\).
e) \(\lim \frac{{{{4.5}^{\rm{n}}} + {2^{{\rm{n}} + 2}}}}{{{{6.5}^{\rm{n}}}}} = \lim \frac{{{{4.5}^{\rm{n}}} + {2^2}{{.2}^{\rm{n}}}}}{{{{6.5}^{\rm{n}}}}} = \lim \frac{{{5^n}.\left[ {4 + 4.{{\left( {\frac{2}{5}} \right)}^{\rm{n}}}} \right]}}{{{{6.5}^n}}} = \lim \frac{{4 + 4.{{\left( {\frac{2}{5}} \right)}^{\rm{n}}}}}{6} = \frac{{4 + 4.0}}{6} = \frac{2}{3}\).
g) \(\lim \frac{{2 + \frac{4}{{{n^3}}}}}{{{6^{\rm{n}}}}} = \lim \left( {2 + \frac{4}{{{{\rm{n}}^3}}}} \right).\lim {\left( {\frac{1}{6}} \right)^{\rm{n}}} = \left( {2 + 0} \right).0 = 0\).
a. Chắc đề là: \(\lim\dfrac{2-5^{n-2}}{3^n+2.5^n}=\lim\dfrac{2\left(\dfrac{1}{5}\right)^{n-2}-1}{9\left(\dfrac{3}{5}\right)^{n-2}+50}=-\dfrac{1}{50}\)
b. \(=\lim\dfrac{2\left(\dfrac{1}{5}\right)^n-25}{\left(\dfrac{3}{5}\right)^n-2}=\dfrac{25}{2}\)
2.
Đặt \(f\left(x\right)=x^4+x^3-3x^2+x+1\)
Hàm f(x) liên tục trên R
\(f\left(0\right)=1>0\) ; \(f\left(-1\right)=-3< 0\)
\(\Rightarrow f\left(0\right).f\left(-1\right)< 0\Rightarrow f\left(x\right)=0\) luôn có ít nhất 1 nghiệm thuộc khoảng \(\left(-1;0\right)\)
Hay pt đã cho luôn có ít nhất 1 nghiệm âm lớn hơn -1
3.
Ta có: M là trung điểm AD, N là trung điểm SD
\(\Rightarrow\) MN là đường trung bình tam giác SAD
\(\Rightarrow MN||SA\Rightarrow\left(MN,SC\right)=\left(SA,SC\right)\)
Ta có: \(AC=\sqrt{AB^2+BC^2}=a\sqrt{2}\)
\(SA=SC=a\)
\(\Rightarrow SA^2+SC^2=AC^2\Rightarrow\Delta SAC\) vuông tại S hay \(SA\perp SC\)
\(\Rightarrow\) Góc giữa MN và SC bằng 90 độ
\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)
\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)
\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)
\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)
\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)
Bậc tử lớn hơn bậc mẫu, giới hạn bằng \(+\infty\)
Cụ thể thì:
\(=lim\frac{\sqrt{x+\frac{1}{x^3}-\frac{11}{x^4}}}{2+\frac{1}{x}+\frac{1}{x^2}}=\frac{+\infty}{2}=+\infty\)
Ta có lim 1 − 2 n 5 n + 5 n 2 = lim n 2 1 n 2 − 2 n n 2 5 n + 5 = lim 1 n 2 − 2 n 5 n + 5 = 0 5 = 0 .
Chọn đáp án D