a: Ta có: BC⊥BA tại B

nên BC là tiếp tuyến của (A;AB)

b: Xét (A) có 

CB là tiếp tuyến

CD là tiếp tuyến

Do đó: CB=CD
hay C nằm trên đường trung trực của BD(1)

Ta có: AB=AD

nên A nằm trên đường trung trực của BD(2)

Từ (1) và (2) suy ra AC là đường trung trực của BD

hay AC\(\perp\)BD

Câu 1.

\(R_N=R_1+R_2=2+3=5\Omega\)

\(I_1=I_2=I=\dfrac{\xi}{r+R_N}=\dfrac{9}{1+5}=1,5A\)

\(U_N=I\cdot R_N=1,5\cdot5=7,5V\)

\(U_1=I_1\cdot R_1=1,5\cdot2=3V;U_2=7,5-3=4,5V\)

\(P_1=I_1^2\cdot R_1=1,5^2\cdot2=4,5W\)

\(P_2=1,5^2\cdot3=6,75W\)

\(1.\left(x+4\right)^2-\left(x-1\right)\left(x+1\right)=16\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x=-1\Leftrightarrow x=-\frac{1}{8}\)

\(2.\left(x-1\right)^2+\left(x+3\right)^2+2\left(x-1\right)\left(x+3\right)=4\Leftrightarrow\left(x-1+x+3\right)^2=4\)

\(\Leftrightarrow\left(2x+2\right)^2=4\Leftrightarrow\orbr{\begin{cases}2x+2=2\\2x+2=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

3.\(\left(x-1\right)^2-x\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left[\left(x-1\right)-x\right]=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)

\(4.\left(3x-1\right)^2+\left(5x-2\right)^2-2\left(3x-1\right)\left(5x-2\right)=9\Leftrightarrow\left(3x-1-5x+2\right)^2=9\)

\(\Leftrightarrow\left(2x-1\right)^2=9\Leftrightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

5.\(\left(x-1\right)\left(x^2+x+1\right)-x\left(x-2\right)\left(x+2\right)=5\Leftrightarrow x^3-1-\left(x^3-4x\right)=5\)

\(\Leftrightarrow4x=6\Leftrightarrow x=\frac{3}{2}\)

6.\(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(x-2\right)\left(x+2\right)=2\)

\(\Leftrightarrow x^3-3x^2+3x-1-\left(x^3+27\right)+x^2-4=2\)

\(\Leftrightarrow-2x^2+3x-34=0\text{ vô nghiệm}\)

Câu 5:

\(y=1-\left(sin2x+cos2x\right)^3\)

\(=1-\left[\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\right]^3\)\(=1-2\sqrt{2}.sin^3\left(2x+\dfrac{\pi}{4}\right)\)

Có \(-1\le sin\left(2x+\dfrac{\pi}{4}\right)\le1\)

\(\Leftrightarrow-1\le sin^3\left(2x+\dfrac{\pi}{4}\right)\le1\) \(\Leftrightarrow1+2\sqrt{2}\ge y\ge1-2\sqrt{2}\)

\(\Rightarrow y_{min}=1-2\sqrt{2}\Leftrightarrow sin\left(2x+\dfrac{\pi}{4}\right)=1\)\(\Leftrightarrow x=\dfrac{\pi}{8}+k\pi\left(k\in Z\right)\)

\(\Rightarrow y_{max}=1+2\sqrt{2}\Leftrightarrow sin\left(2x+\dfrac{\pi}{4}\right)=-1\)\(\Leftrightarrow x=\dfrac{-3\pi}{8}+k\pi\left(k\in Z\right)\)

Ý B

Câu 6: Hàm số có TXĐ: D=R

\(y=\sqrt{4-2sin^52x}-8\) 

Có \(-1\le sin2x\le1\)

\(\Leftrightarrow-1\le sin^52x\le1\)

\(\Leftrightarrow2\ge-2sin^52x\ge-2\)

\(\Leftrightarrow\)\(\sqrt{6}-8\ge y\ge\sqrt{2}-8\) 

Ý A

Câu 7: TXĐ: D=R

\(y=\dfrac{3}{3-\sqrt{1-cosx}}\)

Có \(-1\le cosx\le1\) \(\Leftrightarrow2\ge1-cosx\ge0\) \(\Leftrightarrow3-\sqrt{2}\ge3-\sqrt{1-cosx}\ge3\)

\(\Leftrightarrow\dfrac{3}{3-\sqrt{2}}\le y\le1\)

Vậy \(y_{min}=\dfrac{3}{3-\sqrt{2}}\Leftrightarrow x=\pi+k2\pi\) (k nguyên)

\(y_{max}=1\Leftrightarrow x=k2\pi\)  (k nguyên)

nhìu dzậy ai mà lèm nổi ;-;

Đường thẳng d có 1 vtpt là \(\left(1;-2\right)\)

Đường thẳng \(d'\) vuông góc d nên có 1 vtpt là (2;1) (đảo thứ tự tọa độ vtpt của d và đảo dấu 1 trong 2 vị trí tùy thích)

Phương trình d':

\(2\left(x+1\right)+1\left(y-1\right)=0\Leftrightarrow2x+y+1=0\)

Ta có

 \(a^2+1=a^2+ab+bc+ca=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right).\left(a+c\right)\\ Cmtt:b^2+1=\left(b+a\right).\left(b+c\right)\\ c^2+1=\left(c+a\right).\left(c+b\right)\)

Nên

 \(\dfrac{b-c}{a^2+1}+\dfrac{c-a}{b^2+1}+\dfrac{a-b}{c^2+1}\\ =\dfrac{\left(b-c\right)}{\left(a+b\right)\left(a+c\right)}+\dfrac{\left(c-a\right)}{\left(b+c\right)\left(b+a\right)}+\dfrac{\left(a-b\right)}{\left(c+a\right)\left(c+b\right)}\\ =\dfrac{\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)+\left(a-b\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\\ =\dfrac{b^2-c^2+c^2-a^2+a^2-b^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\\ =0\)

\(\dfrac{b-c}{a^2+1}+\dfrac{c-a}{b^2+1}+\dfrac{a-b}{c^2+1}\)

\(=\dfrac{b-c}{a^2+ab+bc+ac}+\dfrac{c-a}{b^2+ab+bc+ca}+\dfrac{a-b}{c^2+ab+bc+ca}\)

\(=\dfrac{b-c}{a\left(a+b\right)+c\left(a+b\right)}+\dfrac{c-a}{b\left(a+b\right)+c\left(a+b\right)}+\dfrac{a-b}{c\left(c+a\right)+b\left(a+c\right)}\)

\(=\dfrac{b-c}{\left(a+c\right)\left(a+b\right)}+\dfrac{c-a}{\left(b+c\right)\left(a+b\right)}+\dfrac{a-b}{\left(b+c\right)\left(a+c\right)}\)

\(=\dfrac{\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(a+c\right)+\left(a-b\right)\left(a+b\right)}{\left(a+c\right)\left(a+b\right)\left(b+c\right)}\)

\(=\dfrac{b^2-c^2+c^2-a^2+a^2-b^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\) 

Bài 5:

a: 2x-(3-5x)=4(x+3)

=>2x-3+5x=4x+12

=>7x-3=4x+12

=>3x=15

=>x=5

b: =>5/3x-2/3+x=1+5/2-3/2x

=>25/6x=25/6

=>x=1

c: 3x-2=2x-3

=>3x-2x=-3+2

=>x=-1

d: =>2u+27=4u+27

=>u=0

e: =>5-x+6=12-8x

=>-x+11=12-8x

=>7x=1

=>x=1/7

f: =>-90+12x=-45+6x

=>12x-90=6x-45

=>6x-45=0

=>x=9/2