1: x^2+y^2+6x-2y=0

=>x^2+6x+9+y^2-2y+1=10

=>(x+3)^2+(y-1)^2=10

=>R=căn 10; I(-3;1)

Vì (d1)//(d) nên (d1): x-3y+c=0

Theo đề, ta có: d(I;(d1))=căn 10

=>\(\dfrac{\left|-3\cdot1+1\cdot\left(-3\right)+c\right|}{\sqrt{1^2+\left(-3\right)^2}}=\sqrt{10}\)

=>|c-6|=10

=>c=16 hoặc c=-4

a:Sửa đề: y=x^3-3x^2+2

y'=3x^2-3*2x=3x^2-6x

y=2

=>x^3-3x^2=0

=>x=0 hoặc x=3

=>y'=0 hoặc y'=3*3^2-6*3=27-18=9

A(0;2); y'=0; y=2

Phương trình tiếp tuyến có dạng là;

y-2=0(x-0)

=>y=2

A(3;2); y'=9; y=2

Phương trình tiếp tuyến có dạng là:

y-2=9(x-3)

=>y=9x-27+2=9x-25

b: Tiếp tuyến tại M song song với y=6x+1

=>y'=6

=>3x^2-6x=6

=>x^2-2x=2

=>x=1+căn 3 hoặc x=1-căn 3

=>y=0 hoặc y=0

M(1+căn 3;0); y=0; y'=6

Phương trình tiếp tuyến là:

y-0=6(x-1-căn 3)=6x-6-6căn3

M(1-căn 3;0); y=0; y'=6

Phương trình tiếp tuyến là:

y-0=6(x-1+căn 3)

=>y=6x-6+6căn 3

1.

Pt hoành độ giao điểm: \(\frac{2x-3}{x+3}=x-1\)

\(\Leftrightarrow2x-3=x^2+2x-3\)

\(\Leftrightarrow x=0\Rightarrow y=-1\)

Vậy tung độ giao điểm là \(-1\)

2.

\(y'=4x^3+4x\Rightarrow\left\{{}\begin{matrix}y'\left(1\right)=8\\y\left(1\right)=3\end{matrix}\right.\)

Pttt: \(y=8\left(x-1\right)+3=8x-5\)

3.

\(y'=3x^2-6x\)

Lấy y chia y' và lấy phần dư ta được pt đường thẳng là: \(y=-2x+1\)

a,

\(\left|x+\dfrac{9}{2}\right|\ge0\forall x\\ \left|y+\dfrac{4}{3}\right|\ge0\forall y\\ \left|z+\dfrac{7}{2}\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x,y,z\)

Mà

\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-9}{2}\\y=\dfrac{-4}{3}\\z=\dfrac{-7}{2}\end{matrix}\right.\)

Vậy \(x=\dfrac{-9}{2};y=\dfrac{-4}{3};z=\dfrac{-7}{2}\)

d,

\(\left|x+\dfrac{3}{4}\right|\ge0\forall x\\ \left|y-\dfrac{1}{5}\right|\ge0\forall y\\ \left|x+y+z\right|\ge0\forall x,y,z\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x,y,z\)

Mà

\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-3}{4}+\dfrac{1}{5}+z=0\end{matrix}\right.\\\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-11}{20}+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\z=\dfrac{11}{20}\end{matrix}\right.\)

Bạn mới hỏi ở dưới rồi :v

b,

\(\left|x+\dfrac{3}{4}\right|\ge0\forall x\\ \left|y-\dfrac{2}{5}\right|\ge0\forall y\\ \left|z+\dfrac{1}{2}\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\ge0\forall x,y,z\\ \)

Mà \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{2}{5}\right|=0\\\left|z+\dfrac{1}{2}\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{2}{5}=0\\z+\dfrac{1}{2}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{2}{5}\\z=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy ...

c,

\(\left|x+\dfrac{19}{5}\right|\ge0\forall x\\ \left|y+\dfrac{1890}{1975}\right|\ge0\forall y\\ \left|z-2004\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x,y,z\)

Mà

\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{1890}{1975}\right|=0\\\left|z-2004\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-1890}{1975}=\dfrac{-378}{395}\\z=2004\end{matrix}\right. \)

Vậy ...

Đáp án A

Ta có 3 x + y − 4 = 0 ⇔ y = 4 − 3 x

y 1 = − 2 y ' 1 = − 3 ⇔ 1 + b a − 2 = − 2 − 2 − a b a − 2 2 = − 3

⇔ b = 3 − 2 a − 2 − a 3 − 2 a = − 3 a 2 − 4 a + 4

⇔ b = 3 − 2 a a = 1 a = 2 ⇔ a = 1 b = 1 a = 2 b = − 1 L

Vậy  a = 1 ; b = 1 ⇒ a + b = 2

\(3.\)

\(-2x^2+3x+2\ge0\)

\(\Leftrightarrow-\left(x-2\right)\left(2x+1\right)\ge0\)

\(\Leftrightarrow\left(2-x\right)\left(2x+1\right)\ge0\)

Giải bất phương trình ra được: \(\frac{-1}{2}\le x\le2\)

Vậy \(x\in\left\{\frac{-1}{2};2\right\}\)

\(5.\)

Đường thẳng đã cho song song với đường thẳng \(2x+y+2020=0\)

<=> Đường thẳng đã cho có véc tơ pháp tuyến là \(n\left(2;1\right)\)

Mà đường thẳng đã cho đi qua \(M\left(3;0\right)\)nên ta có phương trình:

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

\(2x+y-6=0\)