a) Ta có: \(A=2\cdot\cot37^0\cdot\cot53^0+\sin^228^0+\sin^262^0-\dfrac{3\cdot\tan54^0}{\cot36^0}\)

\(=2\cdot\tan53^0\cdot\cot53^0+\sin^228^0+\cos^228^0-\dfrac{3\cdot\tan54^0}{\tan54^0}\)

\(=2+1-3\)

=0

a) \(A=2sin30^o-2cos60^o+tan45^o\)

\(=2\left(sin30^o-có60^o\right)+1\)

\(=2\left(sin30^o-sin30^o\right)+1=1\)

b) \(B=3sin^225^o+3sin^265^o-tan35^o+cot55^o-\frac{cot32^o}{tan58^o}\)

\(=3\left(sin^225^o+cos^225^o\right)-\left(tan35^o-cot55^o\right)-\frac{cot32^o}{cot32^o}\)

\(=3-\left(tan35^o-tan35^o\right)-1\)

\(=2\)

c) \(C=tan67^o-cos23^o+cos^216^p+cos^274^o-\frac{4cot37^o}{2tan53^o}\)

= \(tan67^o-tan67^o+sin^274^o+cos^274^o-\frac{4cot37^o}{2cot37^o}\)

\(=1-2=-1\)

d) \(D=2cot37^ocot53^o+sin^228^o-\frac{3tan54^o}{cot36^o}+sin^262^o\)

\(=2cot37^otan37^o+sin^228^o+cos^228^o-\frac{3tan54^o}{tan54^o}\)

\(=2+1-3=0\)

Mấy bài kiểu này bạn chỉ cần áp dụng tính chất tỉ số lượng giác của hai góc phụ nhau và các hệ thức trong bài tập số 14 (SGK - Tr.77) là sẽ ra thôi

Chúc bạn học tốt nhé!

\(=2\cdot sin53^0\cdot cos53^0+1-3=sin106^0-2\)

Chọn C

c

c

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT=\dfrac{a}{na+mb}+\dfrac{b}{nb+ma}\)

\(=\dfrac{a^2}{na^2+mab}+\dfrac{b^2}{nb^2+mab}\)

\(\ge\dfrac{\left(a+b\right)^2}{na^2+nb^2+2mab}\). Cần chứng minh BĐT

\(\dfrac{\left(a+b\right)^2}{na^2+nb^2+2mab}\ge\dfrac{2}{m+n}\)

Điều này đúng vì tương đương với \(\left(a-b\right)^2\left(m-n\right)\ge0\forall a,b,m,n>0;m>n\)

A, B, M thẳng hàng khi \(\overrightarrow{AM}=k\overrightarrow{AB}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=k\\2=k.7\end{matrix}\right.\Rightarrow x=\dfrac{23}{7}\Rightarrow M\left(\dfrac{23}{7};0\right)\Rightarrow D\)

\(ab\cdot\sqrt{\dfrac{a}{3b}}-a^2\sqrt{\dfrac{3b}{a}}\)

\(=a\sqrt{ab}-a^2\cdot\dfrac{\sqrt{3b}}{\sqrt{a}}\)

\(=a\sqrt{ab}-a\sqrt{a}\cdot\sqrt{3b}\)

\(=a\sqrt{ab}\left(1-\sqrt{3}\right)\)

\(\Leftrightarrow m=\dfrac{a\sqrt{ab}\left(1-\sqrt{3}\right)}{\sqrt{3ab}}=\dfrac{a\left(\sqrt{3}-3\right)}{3}\)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

\(\Leftrightarrow yz+zx+xy=0\)

\(\Leftrightarrow\left[{}\begin{matrix}yz=-zx-xy\\zx=-xy-yz\\xy=-yz-zx\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{1}{x^2+2yz}=\dfrac{1}{x^2-xz-xy+yz}=\dfrac{1}{\left(x-y\right)\left(x-z\right)}\)

CMTT\(\Rightarrow\dfrac{1}{y^2+2zx}=\dfrac{1}{\left(y-z\right)\left(y-x\right)}\)

\(\dfrac{1}{z^2+2xy}=\dfrac{1}{\left(z-x\right)\left(z-y\right)}\)

\(\Rightarrow A=\dfrac{1}{\left(x-y\right)\left(x-z\right)}+\dfrac{1}{\left(y-z\right)\left(y-x\right)}+\dfrac{1}{\left(z-x\right)\left(z-y\right)}\)

\(A=\dfrac{y-z}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\dfrac{z-x}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\dfrac{x-y}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

\(A=\dfrac{y-z+z-x+x-y}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=0\left(đpcm\right)\)

a)

\(\left\{{}\begin{matrix}\left(2m-1\right)^2-4\left(m^2-m\right)\ge0\left(1\right)\\\dfrac{1}{m^2-m}>0\left(2\right)\\\dfrac{2m-1}{m^2-m}>0\left(3\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow m^2-m>0\Rightarrow\left[{}\begin{matrix}m< 0\\m>1\end{matrix}\right.\) (I)

Kết hợp \(\left(2\right)\Rightarrow\left(3\right)\Leftrightarrow2m-1>0\Rightarrow m>\dfrac{1}{2}\)(II)

\(\left(1\right)\Leftrightarrow4m^2-4m+1-4m^2+4m=1\ge0\forall m\) (III)

Từ (I) (II) (III) \(\Rightarrow m>1\)

Kết luận nghiệm BPT m>1

b)

\(\left\{{}\begin{matrix}\left(m-2\right)^2-\left(m+3\right)\left(m-1\right)\ge0\left(1\right)\\\dfrac{m-2}{m+3}< 0\left(2\right)\\\dfrac{m-1}{m+3}>0\left(3\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow m^2-4m+4-m^2-2m+3=-6m+7\ge0\Rightarrow m\le\dfrac{7}{6}\)(I)

\(\left(2\right)\Leftrightarrow-3< m< 2\) (2)

\(\left(3\right)\Leftrightarrow\left[{}\begin{matrix}m< -3\\m>1\end{matrix}\right.\)(3)

Nghiệm Hệ BPT là: \(1< m\le\dfrac{7}{6}\)