a: Ta có: \(AE=EB=\dfrac{AB}{2}\)

\(BF=FC=\dfrac{BC}{2}\)

\(DK=KC=\dfrac{DC}{2}\)

mà AB=BC=CD

nên AE=EB=BF=FC=DK=KC

Xét tứ giác AECK có

AE//CK

AE=CK

Do đó: AECK là hình bình hành

b: Xét ΔDCF vuông tại C và ΔCBE vuông tại B có

DC=CB

CF=BE

Do đó: ΔDCF=ΔCBE

=>\(\widehat{DFC}=\widehat{CEB}\)

mà \(\widehat{CEB}+\widehat{BCE}=90^0\)

nên \(\widehat{BCE}+\widehat{DFC}=90^0\)

=>CE\(\perp\)DF

a) \(\dfrac{12}{5}>\dfrac{10}{5}=2=\dfrac{4}{2}>\dfrac{3}{2}\) (Số 2 làm trung gian)

Hay \(\dfrac{12}{5}>\dfrac{3}{2}\)

b) Ta có: 

`2023 < 2024 =>` \(\dfrac{2023}{2024}< 1\)

`2026 > 2025 =>` \(\dfrac{2026}{2025}>1\)

=> \(\dfrac{2023}{2024}< 1< \dfrac{2026}{2025}\) (1 làm trung gian)

Hay \(\dfrac{2023}{2024}< \dfrac{2026}{2025}\)

a: \(\dfrac{12}{5}=2,4;\dfrac{3}{2}=1,5\)

mà 2,4>1,5

nên \(\dfrac{12}{5}>\dfrac{3}{2}\)

b: \(\dfrac{2023}{2024}< \dfrac{2024}{2024}=1;\dfrac{2026}{2025}>\dfrac{2025}{2025}=1\)

Do đó: \(\dfrac{2023}{2024}< \dfrac{2026}{2025}\)

Bài 4:

a: 4x-y=1

=>y=4x-1

Vậy: Nghiệm tổng quát là \(\left\{{}\begin{matrix}x\in R\\y=4x-1\end{matrix}\right.\)

b: x+3y=-2

=>x=-3y-2

Vậy: Nghiệm tổng quát là \(\left\{{}\begin{matrix}y\in R\\x=-3y-2\end{matrix}\right.\)

Bài 5:

x+2y-3=0

=>2y=-x+3

=>\(y=\dfrac{-x+3}{2}\)

Vậy: Nghiệm tổng quát là \(\left\{{}\begin{matrix}x\in R\\y=\dfrac{-x+3}{2}\end{matrix}\right.\)

Biểu diễn tập nghiệm: 

a: 2<3

=>\(2+8,5\cdot6< 3+8,5\cdot6\)

b: 2<3

=>\(\sqrt{2}< \sqrt{3}\)

=>\(-\sqrt{2}>-\sqrt{3}\)

=>\(30-\sqrt{2}>30-\sqrt{3}\)

c:

Vì 3>2

nên \(3\sqrt{3}>3\sqrt{2}\)

=>\(-3\sqrt{3}< -3\sqrt{2}\)

mà 35<36

nên \(35-3\sqrt{3}< 36-3\sqrt{2}\)

a: 2a+3>2b+3

=>2a>2b

=>a>b

b: -3a-1>=-3b-1

=>\(-3a>=-3b\)

=>3a<=3b

=>a<=b

c: 5-2a<5-a-b

=>5-2a+a<5-b

=>5-a<5-b

=>a-5>b-5

=>a>b

a: \(x^2+y^2>=2xy\)

=>\(x^2-2xy+y^2>=0\)

=>\(\left(x-y\right)^2>=0\)(luôn đúng)

b: \(x^2+4xy>=-4y^2\)

=>\(x^2+4xy+4y^2>=0\)

=>\(\left(x+2y\right)^2>=0\)(luôn đúng)

c: \(2\left(x^2+y^2\right)>=\left(x+y\right)^2\)

=>\(2x^2+2y^2-x^2-2xy-y^2>=0\)

=>\(x^2-2xy+y^2>=0\)

=>\(\left(x-y\right)^2>=0\)(luôn đúng)

a: \(\dfrac{x}{y}+\dfrac{y}{x}>=2\cdot\sqrt{\dfrac{x}{y}\cdot\dfrac{y}{x}}=2\)

b: \(\dfrac{1}{x}+\dfrac{1}{y}>=\dfrac{4}{x+y}\)

=>\(\dfrac{x+y}{xy}>=\dfrac{4}{x+y}\)

=>\(\left(x+y\right)^2>=4xy\)

=>\(x^2+2xy+y^2-4xy>=0\)

=>\(x^2-2xy+y^2>=0\)

=>\(\left(x-y\right)^2>=0\)(luôn đúng)

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

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