a: \(\left(A\cap B\right)\cap C=(4;10]\cap\left(5;+\infty\right)=(5;10]\)

c: A\B=[3;4]

B\C=(4;5]

C\A=[3;5]

d: (A\B) giao C=[3;4] giao (5;+\(\infty\))=[4;5)

\(\overrightarrow{u}=\left(\dfrac{1}{2};-5\right);\overrightarrow{v}=\left(k;-4\right)\)

để vecto u vuông góc với vecto v thì 1/2*k+(-4)*(-5)=0

=>k*1/2=-20

=>k=-40

Chọn B

1)

a) \(5n-8⋮4-n\)

\(\Rightarrow-20+5n+12⋮4-n\)

\(\Rightarrow-5\left(4-n\right)+12⋮4-n\)

\(\Rightarrow12⋮4-n\)

\(\Rightarrow4-n\in\left\{-1;1;-2;2;-3;3;-4;4;-6;6;-12;12\right\}\)

+) \(4-n=-1\Rightarrow n=5\)

+) \(4-n=1\Rightarrow n=3\)

+) \(4-n=-2\Rightarrow n=6\)

+) \(4-n=2\Rightarrow n=2\)

+) \(4-n=-3\Rightarrow n=7\)

+) \(4-n=3\Rightarrow n=1\)

+) \(4-n=-4\Rightarrow n=8\)

+) \(4-n=4\Rightarrow n=0\)

+) \(4-n=-6\Rightarrow n=10\)

+) \(4-n=6\Rightarrow n=-2\)

+) \(4-n=-12\Rightarrow n=16\)

+) \(4-n=12\Rightarrow n=-8\)

Vậy \(n\in\left\{5;3;6;2;7;1;8;0;10;-2;16;-8\right\}\)

b) Ta có:\(n^2+3n+6⋮n+3\)

\(\Rightarrow n\left(n+3\right)+6⋮n+3\)

\(\Rightarrow6⋮n+3\)

\(\Rightarrow n+3\in\left\{-1;1;-2;2;-3;3;-6;6\right\}\)

+) \(n+3=-1\Rightarrow n=-4\)

+) \(n+3=1\Rightarrow n=-2\)

+) \(n+3=-2\Rightarrow n=-5\)

+) \(n+3=2\Rightarrow n=-1\)

+) \(n+3=-3\Rightarrow n=-6\)

+) \(n+3=3\Rightarrow n=0\)

+) \(n+3=-6\Rightarrow n=-9\)

+) \(n+3=6\Rightarrow n=3\)

Vậy \(n\in\left\{-4;-2;-5;-1;-6;0;-9;3\right\}\)

\(\sin\alpha+\cos\alpha=\sqrt{2}\) (1)

=> \(\left(\sin a+\cos a\right)^2=2\)

=> \(\sin^2\alpha+\cos^2\alpha+2\cdot\sin\alpha\cdot\cos\alpha=2\)

\(\Rightarrow1+2\cdot\sin\alpha\cdot\cos\alpha=2\Rightarrow2\cdot\sin\alpha\cdot\cos\alpha=1\)

\(\Rightarrow\sin\alpha\cdot\cos\alpha=\dfrac{1}{2}\)

Có: \(\left(\sin\alpha-\cos\alpha\right)^2=\left(\sin\alpha+\cos\alpha\right)^2-4\cdot\sin\alpha\cdot\cos a=2-2=0\)

=> \(\sin\alpha-\cos\alpha=0\) (2)

Từ (1),(2) => \(2\sin\alpha=\sqrt{2}\Rightarrow\sin a=\dfrac{\sqrt{2}}{2}\Rightarrow\alpha=45\)(ktm)

Vậy không có a nào t/m điều kiện

Câu 1:

Áp dụng BĐT Cauchy:

\(1+x^3+y^3\geq 3\sqrt[3]{x^3y^3}=3xy\)

\(\Rightarrow \frac{\sqrt{1+x^3+y^3}}{xy}\geq \frac{\sqrt{3xy}}{xy}=\sqrt{\frac{3}{xy}}\)

Hoàn toàn tương tự:

\(\frac{\sqrt{1+y^3+z^3}}{yz}\geq \sqrt{\frac{3}{yz}}; \frac{\sqrt{1+z^3+x^3}}{xz}\geq \sqrt{\frac{3}{xz}}\)

Cộng theo vế các BĐT thu được:

\(\text{VT}\geq \sqrt{\frac{3}{xy}}+\sqrt{\frac{3}{yz}}+\sqrt{\frac{3}{xz}}\geq 3\sqrt[6]{\frac{27}{x^2y^2z^2}}=3\sqrt[6]{27}=3\sqrt{3}\) (Cauchy)

Ta có đpcm

Dấu bằng xảy ra khi $x=y=z=1$

Câu 4:

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{2}{x}+\frac{3}{y}\right)(x+y)\geq (\sqrt{2}+\sqrt{3})^2\)

\(\Leftrightarrow 1.(x+y)\geq (\sqrt{2}+\sqrt{3})^2\Rightarrow x+y\geq 5+2\sqrt{6}\)

Vậy \(A_{\min}=5+2\sqrt{6}\)

Dấu bằng xảy ra khi \(x=2+\sqrt{6}; y=3+\sqrt{6}\)

------------------------------

Áp dụng BĐT Cauchy:

\(\frac{ab}{a^2+b^2}+\frac{a^2+b^2}{4ab}\geq 2\sqrt{\frac{ab}{a^2+b^2}.\frac{a^2+b^2}{4ab}}=1\)

\(a^2+b^2\geq 2ab\Rightarrow \frac{3(a^2+b^2)}{4ab}\geq \frac{6ab}{4ab}=\frac{3}{2}\)

Cộng theo vế hai BĐT trên:

\(\Rightarrow B\geq 1+\frac{3}{2}=\frac{5}{2}\) hay \(B_{\min}=\frac{5}{2}\). Dấu bằng xảy ra khi $a=b$

2, a,

\(f\left(-2\right)=5-2\times\left(-2\right)=9\)

\(f\left(-1\right)=5-2\times\left(-1\right)=7\)

\(f\left(0\right)=5-2\times0=5\)

\(f\left(3\right)=5-2\times3=-1\)

b, \(y=5\Leftrightarrow5-2x=5\Leftrightarrow x=0\)

\(y=3\Leftrightarrow5-2x=3\Leftrightarrow x=1\)

\(y=-1\Leftrightarrow5-2x=-1\Leftrightarrow x=3\)

toan lop7 nha may ban