a: \(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\x+y+2\left(x-y\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+2y+3x-3y=4\\x+y+2x-2y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x-y=4\\3x-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-y-3x+y=4-5\\3x-y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x=-1\\y=3x-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=3\cdot\dfrac{-1}{2}-5=-\dfrac{3}{2}-5=-\dfrac{13}{2}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\left(x+1\right)\left(y-1\right)=xy-1\\\left(x-3\right)\left(y+3\right)=xy-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy-x+y-1=xy-1\\xy+3x-3y-9=xy-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x+y=0\\3x-3y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-y=0\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y-x+y=0-2\\x-y=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}0y=-2\\x-y=0\end{matrix}\right.\Leftrightarrow\left(x;y\right)\in\varnothing\)

\(1.2\left(x+2\right)^2< 2x\left(x+2\right)+4\\ \Leftrightarrow2\left(x^2+4x+4\right)-2x\left(x+2\right)-4< 0\\ \Leftrightarrow2x^2+8x+4-2x^2-4x-4< 0\\ \Leftrightarrow4x< 0\\ \Leftrightarrow x< 0\\ 2.\left(x-1\right)^2+x^2< \left(x+1\right)^2+\left(x+2\right)^2\\ \Leftrightarrow x^2-2x+1+x^2< x^2+2x+1+x^2+4x+4\\ \Leftrightarrow2x^2-2x+1-2x^2-6x-5< 0\\ \Leftrightarrow-8x-4< 0\\ \Leftrightarrow8x>-4\\ \Leftrightarrow x>-\dfrac{1}{2}\\ 3.\left(x^2+1\right)\left(x-6\right)< \left(x-2\right)^3\\ \Leftrightarrow x^3-6x^2+x-6< x^3-6x^2+12x-8\\ \Leftrightarrow x-6< 12x-8\\ \Leftrightarrow12x-x>-6+8\\ \Leftrightarrow11x>2\\ \Leftrightarrow x>\dfrac{2}{11}\)

Diện tích tam giác BAC là:

\(S_{BAC}=\dfrac{1}{2}\cdot BA\cdot BC\cdot sinABC=\dfrac{1}{2}\cdot5,2\cdot3,5\cdot sin75=\dfrac{91\sqrt{6}+91\sqrt{2}}{40}\)

ABCD là hình bình hành

=>\(S_{ABCD}=2\cdot S_{BAC}=\dfrac{91\sqrt{6}+91\sqrt{2}}{20}\)

a: Ta có: ΔABC vuông tại A

mà AM là đường trung tuyến

nên MA=MB=MC

Xét ΔMAC có \(\widehat{AMB}\) là góc ngoài tại M

nên \(\widehat{AMB}=\widehat{MAC}+\widehat{MCA}=2\cdot\widehat{ACB}=2\alpha\)

=>\(sin2\alpha=sinAMB=\dfrac{AH}{AM}=AH:\dfrac{BC}{2}=\dfrac{2AH}{BC}\)

Xét ΔAHC vuông tại H có \(sinC=\dfrac{AH}{AC}\)

Xét ΔABC vuông tại A có \(cosC=\dfrac{AC}{BC}\)

\(2sin\alpha\cdot cos\alpha=2\cdot\dfrac{AH}{AC}\cdot\dfrac{AC}{BC}=\dfrac{2\cdot AH}{BC}\)

Do đó: \(sin2\alpha=2\cdot sin\alpha\cdot cos\alpha\)

b: \(cos2\alpha=cosAMH=\dfrac{HM}{AM}\)

=>\(1+cos2\alpha=1+\dfrac{HM}{AM}=\dfrac{HC}{AM}=\dfrac{2\cdot HC}{BC}=2\cdot\dfrac{AC^2}{BC^2}\)

\(2\cdot cos^2\alpha=2\cdot cos^2C=2\cdot\left(\dfrac{CA}{BC}\right)^2=2\cdot\dfrac{CA^2}{CB^2}\)

Do đó: \(1+cos2\alpha=2\cdot cos^2\alpha\)

c: \(1-cos2\alpha=1-cosAMH=1-\dfrac{HM}{AM}=\dfrac{HB}{AM}=\dfrac{2HB}{BC}=2\cdot\dfrac{AB^2}{BC^2}\)

\(2\cdot sin^2\alpha=2\cdot sin^2ACB=2\cdot\left(\dfrac{AB}{BC}\right)^2\)

Do đó: \(1-cos2a=2\cdot sin^2\alpha\)

Xét ΔAHC vuông tại H có \(tanC=\dfrac{AH}{HC}\)

=>\(\dfrac{8}{HC}=tan45=1\)

=>HC=8(cm)

Xét ΔABC vuông tại A có AH là đường cao

nên \(AH^2=HB\cdot HC\)

=>\(HB\cdot8=8^2\)

=>HB=8(cm)

BC=BH+CH=8+8=16(cm)

ΔAHC vuông tại H

=>\(HA^2+HC^2=AC^2\)

=>\(AC=\sqrt{8^2+8^2}=8\sqrt{2}\left(cm\right)\)

ΔAHB vuông tại H

=>\(HA^2+HB^2=AB^2\)

=>\(AB=\sqrt{8^2+8^2}=8\sqrt{2}\left(cm\right)\)

Xét ΔABC vuông tại A có \(sinB=\dfrac{AC}{BC}\)

=>\(\dfrac{AC}{15}=\dfrac{3}{5}\)

=>\(AC=15\cdot\dfrac{3}{5}=9\left(cm\right)\)

ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(AB=\sqrt{15^2-9^2}=12\left(cm\right)\)

Tam giác `ABC` vuông tại `A`

`=> AC =  BC . sinB = 15 . 3/5 = 9 (cm)`

Và `AB =` \(\sqrt{BC^2-AC^2}=\sqrt{15^2-9^2}=\sqrt{144}=12\) `(cm)`

Xét ΔABC vuông tại A có \(tanB=\dfrac{AC}{AB}\)

=>\(\dfrac{AC}{AB}=\sqrt{3}\)

=>\(\dfrac{AC^2}{AB^2}=3\)

=>\(AC^3=3AB^2\)

ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(4\cdot AB^2=2^2=4\)

=>\(AB^2=1\)

=>AB=1(cm)

=>\(AC=1\cdot\sqrt{3}=\sqrt{3}\left(cm\right)\)