Ai lm đc câu nào thì giúp mk với , cảm ơn !!

\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)

a: ĐKXĐ: x>=0; x<>9

\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{-3\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\frac{-3}{\sqrt{x}+3}\)

b: \(P<\frac12\)

=>\(-\frac{3}{\sqrt{x}+3}-\frac12<0\)

=>\(\frac{-6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}<0\)

=>\(\frac{-\sqrt{x}-9}{2\left(\sqrt{x}+3\right)}<0\)

=>\(\frac{\sqrt{x}+9}{2\left(\sqrt{x}+3\right)}>0\) (luôn đúng)

=>x>=0 và x<>9

1: \(y=\frac{x+m}{x-1}\)

=>y'=\(\frac{\left(x+m\right)^{\prime}\left(x-1\right)-\left(x+m\right)\left(x-1\right)^{\prime}}{\left(x-1\right)^2}=\frac{\left(x-1\right)-\left(x+m\right)}{\left(x-1\right)^2}=\frac{-m-1}{\left(x-1\right)^2}\)

Khi x=2 thì \(y=\frac{2+m}{2-1}=m+2\)

Khi x=4 thì \(y=\frac{4+m}{4-1}=\frac{m+4}{3}\)

TH1: Hàm số đồng biến trên khoảng [2;4]

=>y'>0

=>-m-1>0

=>-m>1

=>m<-1

Khi Hàm số đồng biến trên khoảng [2;4] thì \(y_{\min}=y\left(2\right)=m+2\)

=>m+2=3

=>m=1(loại)

TH2: Hàm số nghịch biến trên khoảng [2;4]

=>y'<0

=>-m-1<0

=>-m<1

=>m>-1

Khi Hàm số nghịch biến trên khoảng [2;4] thì \(y_{\min}=y\left(4\right)=\frac{m+4}{3}\)

Theo đề, ta có: \(\frac{m+4}{3}=3\)

=>m+4=9

=>m=5(nhận)

TH3: Hàm số là hàm hằng

=>-m-1=0

=>m+1=0

=>m=-1

=>\(y=\frac{x-1}{x-1}=1<>3\)

=>Loại

2: \(y=2x^3-3x^2-m\)

=>y'=\(2\cdot3x^2-3\cdot2x=6x^2-6x=6x\left(x-1\right)\)

Đặt y'=0

=>6x(x-1)=0

=>x=0(nhận) hoặc x=1(nhận)

Khi x=-1 thì \(y=2\cdot\left(-1\right)^3-3\cdot\left(-1\right)^2-m=-2-3-m=-m-5\)

Khi x=0 thì \(y=2\cdot0^3-3\cdot0^2-m=-m\)

Khi x=1 thì \(y=2\cdot1^3-3\cdot1^2-m=2-3-m=-1-m\)

Vì - m-5<-m-1<-m

nên y min=-m-5

=>-5-m=1

=>m=-5-1=-6

a) ĐK:\(x\ge0;x\ne9\)

\(P=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)

b)\(P=-\dfrac{3}{\sqrt{x}+3}\) 

Có \(\sqrt{x}+3\ge3;\forall x\ge0\)

\(\Leftrightarrow-\dfrac{3}{\sqrt{x}+3}\ge-\dfrac{1}{3}\)

\(P_{min}=-\dfrac{1}{3}\Leftrightarrow x=0\)

a) Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{-3}{\sqrt{x}+3}\)

\(a,P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ P=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ P=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\dfrac{-3}{\sqrt{x}+3}\\ b,P=\dfrac{-3}{\sqrt{x}+3}\ge\dfrac{-3}{0+3}=-1\\ P_{min}=-1\Leftrightarrow x=0\)

Bạn kiểm tra lại đề, \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\) hay \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\)

1: \(y=\frac{2x+m}{x+1}\)

=>y'=\(\frac{\left(2x+m\right)^{\prime}\left(x+1\right)-\left(2x+m\right)\left(x+1\right)^{\prime}}{\left(x+1\right)^2}=\frac{2\left(x+1\right)-\left(2x+m\right)}{\left(x+1\right)^2}=\frac{2-m}{\left(x+1\right)^2}\)
Khi x=0 thì y=\(\frac{2\cdot0+m}{0+1}=\frac{m}{1}=m\)

Khi x=1 thì \(y=\frac{2\cdot1+m}{1+1}=\frac{m+2}{2}\)

TH1: Hàm số đồng biến trên khoảng [0;1]

=>y'>0

=>2-m>0

=>m<2

Khi hàm số đồng biến trên khoảng [0;1] thì y max=y(1)=(m+2)/2

y max=2

=>m+2=4

=>m=2(loại)

TH2: Hàm số đồng biến trên khoảng [0;1]

=>y'<0

=>2-m<0

=>m>2

Khi hàm số nghịch biến trên khoảng [0;1] thì y max=y(0)=m

y max=2

=>m=2(loại)

TH3: Hàm số là hàm hằng

=>2-m=0

=>m=2

Khi m=2 thì \(y=\frac{2\cdot x+2}{x+1}=2\)

=>\(y_{\max}=2\) khi m=2