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: \(1-cos2a=2\cdot sin^2\alpha\)

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

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

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

Lời giải:

Đặt $x^2-2x=a$ thì pt trở thành:

$(x^2-2x)^2-2(x^2-2x+1)+2=0$

$\Leftrightarrow a^2-2(a+1)+2=0$

$\Leftrightarrow a^2-2a=0$

$\Leftrightarrow a(a-2)=0$

$\Leftrightarrow a=0$ hoặc $a-2=0$

$\Leftrightarrow x^2-2x=0$ hoặc $x^2-2x-2=0$

Nếu $x^2-2x=0$

$\Leftrightarrow x(x-2)=0$

$\Leftrightarrow x=0$ hoặc $x=2$

Nếu $x^2-2x-2=0$

$\Leftrightarrow (x^2-2x+1)-3=0$

$\Leftrightarrow (x-1)^2=3$
$\Leftrightarrow x-1=\pm \sqrt{3}\Leftrightarrow x=1\pm \sqrt{3}$

ĐKXĐ: \(x\ne-1\)

\(\dfrac{1}{x+1}-\dfrac{x}{x^2-x+1}=\dfrac{3}{x^3+1}\)

=>\(\dfrac{1}{x+1}-\dfrac{x}{x^2-x+1}=\dfrac{3}{\left(x+1\right)\left(x^2-x+1\right)}\)

=>\(\dfrac{x^2-x+1-x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{3}{\left(x+1\right)\left(x^2-x+1\right)}\)

=>\(x^2-x+1-x^2-x=3\)

=>-2x+1=3

=>-2x=2

=>x=-1(loại)

vậy: \(x\in\varnothing\)

ĐKXĐ: \(x\ne-1\)

\(\dfrac{1}{x+1}-\dfrac{x}{x^2-x+1}=\dfrac{3}{x^3+1}\)

=>\(\dfrac{x^2-x+1-x\left(x+1\right)}{\left(x^2-x+1\right)\left(x+1\right)}=\dfrac{3}{\left(x+1\right)\left(x^2-x+1\right)}\)

=>\(x^2-x+1-x^2-x=3\)

=>-2x=2

=>x=-1(loại)

a) Ta có: 

\(sin54^o=\dfrac{y}{3}=>y=3\cdot sin54^o\approx2,4\left(cm\right)\\ =>x=\sqrt{3^2-y^2}=\sqrt{9-2,4^2}\approx1,8\left(cm\right)\)

b) Ta có: 

\(sin32^o=\dfrac{1,5}{y}=>y=\dfrac{1,5}{sin32^o}\approx2,8\left(cm\right)\\ =>x=\sqrt{y^2-1,5^2}=\sqrt{2,8^2-1,5^2}\approx2,4\) 

c) Ta có:

\(tan70^o=\dfrac{y}{0,8}=>y=0,8\cdot tan70^o\approx2,2\left(cm\right)\\ =>x=\sqrt{y^2+0,8^2}=\sqrt{2,2^2+0,8^2}\approx2,3\left(cm\right)\)

Cho tam giác ABC vuông tại A, ^B là góc biết số đo 

a, sinB = y/3 => y  \(\approx\)2,42 cm 

cosB = x/3 => y \(\approx\)1,76 cm 

b, sinB = 1,5/y => y = 1,5/sinB \(\approx\)2,83 cm 

tanB = 1,5/x => x = 1,5/tanB => x \(\approx\)2,4 cm 

c, tanB = y/0,8 => y = 0,8.tanB => y \(\approx\)2,19 cm 

cosB = 0,8/x => x = 0,8/cosB => x \(\approx\)2,34 cm