Chọn C

Chọn A

Đáp án D

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

Chọn D.

Xét  I = ∫ 0 1 f ' x d x   Đặt  t = x → t 2 = x → 2 t d t = d x

Đổi cận   x = 0 → t = 0 x = 1 → t = 1 . Khi đó  I = 2 ∫ 0 1 t f ' ( t ) d t = 2 A

Tính   A = ∫ 0 1 t f ' ( t ) d t . Đặt  u = t d v = f ' t d t → d u = d t v = f t

Khi đó 

Kẻ IH vuông góc AB

=>H là trung điểm của AB

\(d\left(I;\left(d\right)\right)=IH=\dfrac{\left|1\cdot1+\left(-2\right)\cdot\left(-3\right)-17\right|}{\sqrt{1^2+\left(-3\right)^2}}=\dfrac{10}{\sqrt{10}}=\sqrt{10}\)

\(S_{IAB}=\dfrac{1}{2}\cdot IH\cdot AB=10\)

=>\(\dfrac{1}{2}\cdot\sqrt{10}\cdot2\cdot AI=10\)

=>\(AI=\sqrt{10}\)

\(R=\sqrt{\left(\sqrt{10}\right)^2\cdot2}=10\sqrt{2}\)

=>(C): \(\left(x-1\right)^2+\left(y+2\right)^2=200\)

\(I=\int\limits^0_{\frac{-1}{2}}\frac{dx}{\left(x+1\right)\sqrt{3+2x-x^2}}=\int\limits^0_{\frac{-1}{2}}\frac{dx}{\left(x+1\right)\left(\sqrt{\left(x+1\right)\left(3-x\right)}\right)}\)

                                   \(=\int\limits^0_{\frac{-1}{2}}\frac{dx}{\left(x+1\right)^2\sqrt{\frac{3-x}{x+1}}}\)

Đặt \(t=\sqrt{\frac{3-x}{x+1}}\Rightarrow\frac{dx}{\left(x+1\right)^2}=-\frac{1}{2}\)

Đổi cận : \(x=-\frac{1}{2}\Rightarrow t=\sqrt{7};x=0\Rightarrow t=\sqrt{3}\)

\(I=-\frac{1}{2}\int\limits^{\sqrt{3}}_{\sqrt{7}}dt=\frac{1}{2}\left(\sqrt{7}-\sqrt{3}\right)\)

Bài 3:

a: =>(2x-7)(x-2)=0

=>x=7/2 hoặc x=2

b: =>(x-1)(x+2)=0

=>x=1 hoặc x=-2

d: =>2x+3=0

hay x=-3/2

2/

a/ \(25x^2-1=0\)

<=> \(\left(5x\right)^2-1=0\)

<=> \(\left(5x-1\right)\left(5x+1\right)=0\)

<=> \(\orbr{\begin{cases}5x-1=0\\5x+1=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{1}{5}\\x=-\frac{1}{5}\end{cases}}\)

b/ \(4\left(x-1\right)^2-9=0\)

<=> \(\left[2\left(x-1\right)\right]^2-3^2=0\)

<=> \(\left(2x-2\right)^2-3^2=0\)

<=> \(\left(2x-2-3\right)\left(2x-2+3\right)=0\)

<=> \(\left(2x-5\right)\left(2x+1\right)=0\)

<=> \(\orbr{\begin{cases}2x-5=0\\2x+1=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{1}{2}\end{cases}}\)

c/ \(\frac{1}{4}-9\left(x+1\right)^2=0\)

<=> \(\left(\frac{1}{2}\right)^2-\left[3\left(x-1\right)\right]^2=0\)

<=> \(\left(\frac{1}{2}\right)^2-\left(3x-3\right)^2=0\)

<=> \(\left(\frac{1}{2}-3x+3\right)\left(\frac{1}{2}+3x-3\right)=0\)

<=> \(\left(\frac{7}{2}-3x\right)\left(-\frac{5}{2}+3x\right)=0\)

<=> \(\orbr{\begin{cases}\frac{7}{2}-3x=0\\-\frac{5}{2}+3x=0\end{cases}}\)<=> \(\orbr{\begin{cases}3x=\frac{7}{2}\\3x=\frac{5}{2}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{7}{6}\\x=\frac{5}{6}\end{cases}}\)

d/ \(\frac{1}{16}-\left(2x+\frac{3}{4}\right)^2=0\)

<=> \(\left(\frac{1}{4}\right)^2-\left(2x+\frac{3}{4}\right)^2=0\)

<=> \(\left(\frac{1}{4}-2x-\frac{3}{4}\right)\left(\frac{1}{4}+2x+\frac{3}{4}\right)=0\)

<=> \(\left(-\frac{1}{2}-2x\right)\left(1+2x\right)=0\)

<=> \(2\left(-\frac{1}{4}-x\right)\left(1+2x\right)=0\)

<=> \(\orbr{\begin{cases}-\frac{1}{4}-x=0\\1+2x=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{1}{2}\end{cases}}\)