vẫn ghi số độ nha bạn,do mình không biết đánh số độ vào thôi

a) Ta có: \(sin\alpha=cos\left(90-\alpha\right)\Rightarrow sin42=cos48\)

\(\Rightarrow sin42-cos48=0\)

b) Ta có: \(sin\alpha=cos\left(90-\alpha\right)\Rightarrow sin61=cos29\Rightarrow sin^261=cos^229\)

\(\Rightarrow sin^261+sin^229=sin^229+cos^229=1\)

c) Ta có: \(tan\alpha=\dfrac{1}{tan\left(90-\alpha\right)}\Rightarrow tan40=\dfrac{1}{tan50}\)

\(\Rightarrow tan40.tan50=1\) mà \(tan45=1\Rightarrow tan40.tan45.tan50=1\)

a) Vì (d) song song với đường thẳng \(y=-2x+2003\Rightarrow\left\{{}\begin{matrix}a=-2\\b\ne2003\end{matrix}\right.\)

\(\Rightarrow\left(d\right):y=-2x+b\)

Vì (d) cắt trục hoành tại điểm có hoành độ = 1

\(\Rightarrow\) tọa độ điểm đó là \(\left(1;0\right)\)

\(\Rightarrow1=b\Rightarrow\left(d\right):y=-2x+1\)

b) pt hoành độ giao điểm: \(-\dfrac{1}{2}x^2=-2x+2\Rightarrow\dfrac{1}{2}x^2-2x+2=0\)

\(\Rightarrow x^2-4x+4=0\Rightarrow\left(x-2\right)^2=0\Rightarrow x=2\Rightarrow y=-\dfrac{1}{2}.2^2=-2\)

\(\Rightarrow\) tọa độ giao điểm là \(\left(2;-2\right)\)

chỉnh lại đề 1 chút: \(A=\dfrac{1+2sin\alpha.cos\alpha}{cos^2\alpha-sin^2\alpha}=\dfrac{cos^2\alpha+sin^2\alpha+2sin\alpha.cos\alpha}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}\)

\(=\dfrac{\left(cos\alpha+sin\alpha\right)^2}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}=\dfrac{cos\alpha+sin\alpha}{cos\alpha-sin\alpha}\)

\(\sqrt{\dfrac{4}{2x+3}}\) xác định khi \(\dfrac{4}{2x+3}\ge0\Rightarrow2x+3>0\Rightarrow x>-\dfrac{3}{2}\)

\(\sqrt{\dfrac{2x-1}{2-x}}\) xác định khi \(\dfrac{2x-1}{2-x}\ge0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1\ge0\\2-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1\le0\\2-x< 0\end{matrix}\right.\left(l\right)\end{matrix}\right.\Rightarrow\dfrac{1}{2}\le x< 2\)

i) \(\sqrt{8-3\sqrt{7}}+\sqrt{4-\sqrt{7}}=\sqrt{\dfrac{16-6\sqrt{7}}{2}}+\sqrt{\dfrac{8-2\sqrt{7}}{2}}\)

\(=\sqrt{\dfrac{\left(3-\sqrt{7}\right)^2}{2}}+\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}=\dfrac{\left|3-\sqrt{7}\right|}{\sqrt{2}}+\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}\)

\(=\dfrac{3-\sqrt{7}}{\sqrt{2}}+\dfrac{\sqrt{7}-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

j) \(\sqrt{5+\sqrt{21}}-\sqrt{5-\sqrt{21}}=\sqrt{\dfrac{10+2\sqrt{21}}{2}}-\sqrt{\dfrac{10-2\sqrt{21}}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}+\sqrt{3}\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}-\sqrt{3}\right)^2}{2}}=\dfrac{\left|\sqrt{7}+\sqrt{3}\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{7}-\sqrt{3}\right|}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{7}-\sqrt{3}}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

\(\dfrac{3-2\sqrt{3}}{\sqrt{3}}=\dfrac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}}=\sqrt{3}-2\)

còn \(\dfrac{2}{\sqrt{3}-1}=\dfrac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\dfrac{2\left(\sqrt{3}+1\right)}{3-1}\) là do \(\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{a}-\sqrt{b}}{a-b}\)

rồi thế dô thôi

a) Ta có: \(\angle BHE=\angle BFE=90\Rightarrow BEHF\) nội tiếp

b) Ta có: \(\angle EFB=\angle EMB=90\Rightarrow EFMB\) nội tiếp

\(\Rightarrow B,E,H,F,M\) cùng thuộc 1 đường tròn

Ta có: \(\angle OEB=\dfrac{180-\angle BOE}{2}=90-\dfrac{1}{2}\angle BOE=90-\angle BAE=\angle AEF\)

\(\Rightarrow\angle AEO=\angle BEF\)

mà \(\left\{{}\begin{matrix}\angle AEO=\angle HBC\left(HEBMnt\right)\\\angle BEF=\angle CMF\left(BEFMnt\right)\end{matrix}\right.\)

\(\Rightarrow\angle HBC=\angle CMF\) \(\Rightarrow FM\parallel BH\) mà \(BH\bot AE\Rightarrow FM\bot AE\)

c) Kẻ \(EQ'\bot AC\)

Ta có: \(\angle EFA+\angle EQ'A=90+90=180\Rightarrow AFEQ'\) nội tiếp 

\(\Rightarrow\angle AFQ'=\angle AEQ'=90-\angle Q'AE=90-\angle EBC=\angle MEB=\angle MFB\)

mà A,F,B thẳng hàng \(\Rightarrow Q',F,M\) thẳng hàng \(\Rightarrow Q\equiv Q'\)

\(2\sqrt{5}\left(2-3\sqrt{5}\right)+\left(1-2\sqrt{5}\right)^2+6\sqrt{5}=4\sqrt{5}-30+20+1-4\sqrt{5}+6\sqrt{5}\)

\(=6\sqrt{5}-9\)

\(\sqrt{4x+8}+3\sqrt{x+2}=3+\dfrac{4}{5}\sqrt{25x+50}\left(x\ge-2\right)\)

\(\Rightarrow2\sqrt{x+2}+3\sqrt{x+2}-4\sqrt{x+2}=3\Rightarrow\sqrt{x+2}=3\Rightarrow x=7\)

\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{\dfrac{4+2\sqrt{3}}{2}}+\sqrt{\dfrac{4-2\sqrt{3}}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}+\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}+\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)