Câu 5:

a: Khi m=3 thì \(f\left(x\right)=\left(2\cdot3+1\right)x-3=7x-3\)

\(f\left(-3\right)=7\cdot\left(-3\right)-3=-21-3=-24\)

\(f\left(0\right)=7\cdot0-3=-3\)

b: Thay x=2 và y=3 vào f(x)=(2m+1)x-3, ta được:

\(2\left(2m+1\right)-3=3\)

=>2(2m+1)=6

=>2m+1=3

=>2m=2

=>m=1

c: Thay m=1 vào hàm số, ta được:

\(y=\left(2\cdot1+1\right)x-3=3x-3\)

*Vẽ đồ thị

d: Để hàm số y=(2m+1)x-3 là hàm số bậc nhất thì \(2m+1\ne0\)

=>\(2m\ne-1\)

=>\(m\ne-\dfrac{1}{2}\)

e: Để đồ thị hàm số y=(2m+1)x-3 song song với đường thẳng y=5x+1 thì \(\left\{{}\begin{matrix}2m+1=5\\-3\ne1\end{matrix}\right.\)

=>2m+1=5

=>2m=4

=>m=2

Phương trình nhận \(x=2\) là nghiệm nên:

\(\left(4.2-3y+3\right)\left(2+4y-2\right)=0\)

\(\Leftrightarrow\left(11-3y\right).4y=0\)

\(\Leftrightarrow\left[{}\begin{matrix}11-3y=0\\4y=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{11}{3}\\y=0\end{matrix}\right.\)

\(=\left(\dfrac{1+2\left(x-y\right)\left(2x-2y+1\right)-2x+2y-1}{2x-2y+1}\right):\dfrac{\left(2x-2y\right)\left(2x-2y+1\right)-4x^2+8xy-4y^2}{2x-2y+1}\)

\(=\dfrac{1+\left(2x-2y\right)^2+2x-2y-2x+2y-1}{2x-2y+1}\cdot\dfrac{2x-2y+1}{\left(2x-2y\right)^2+2x-2y-4x^2+8xy-4y^2}\)

\(=\dfrac{\left(2x-2y\right)^2}{4x^2-8xy+4y^2+2x-2y-4x^2+8xy-4y^2}=2x-2y\)

=2(x-y) luôn là số chẵn

a: \(F\left(3\right)=3\left(3-2\right)=3\cdot1=3\)

\(\left[F\left(\dfrac{2}{3}\right)\right]^2=\left[\dfrac{2}{3}\cdot\left(\dfrac{2}{3}-2\right)\right]^2\)

\(=\left[\dfrac{2}{3}\cdot\dfrac{-4}{3}\right]^2=\left(-\dfrac{8}{9}\right)^2=\dfrac{64}{81}\)

\(G\left(-\dfrac{1}{2}\right)=-\left(-\dfrac{1}{2}\right)+6=6+\dfrac{1}{2}=\dfrac{13}{2}\)

b: F(x)=0

=>x(x-2)=0

=>\(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

c: F(a)=G(a)

=>\(a\left(a-2\right)=-a+6\)

=>\(a^2-2a+a-6=0\)

=>\(a^2-a-6=0\)

=>(a-3)(a+2)=0

=>\(\left[{}\begin{matrix}a-3=0\\a+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=3\\a=-2\end{matrix}\right.\)

Bài 2:

a: Sửa đề: \(A=-4x^2-5y^2+8xy+10y+12\)

\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)

\(=-\left(2x-2y\right)^2-\left(y-5\right)^2+37\le37\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}2x-2y=0\\ y-5=0\end{cases}\Rightarrow\begin{cases}y=5\\ x=y=5\end{cases}\)

b: \(B=-x^2-y^2+xy+2x+2y\)

\(=-\frac14\left(4x^2+4y^2-4xy-8x-8y\right)\)

\(=-\frac14\left(4x^2-4xy+y^2-8x+4y+3y^2-12y\right)\)

\(=-\frac14\left\lbrack\left(2x-y\right)^2-4\left(2x-y\right)+4+3y^2-12y+12-16\right\rbrack\)

\(=-\frac14\left\lbrack\left(2x-y-2\right)^2+3\left(y-2\right)^2-16\right\rbrack=-\frac14\left(2x-y-2\right)^2-\frac34\left(y-2\right)^2+4\le4\forall x,y\)

Dấu '=' xảy ra khi y-2=0 và 2x-y-2=0

=>y=2 và 2x=y+2=2+2=4

=>x=2 và y=2

Bài 1:

d: \(D=2x^2+3y^2+4xy-8x-2y\)

\(=2x^2+4xy+2y^2-8x-8y+y^2+6y\)

\(=2\left(x+y\right)^2-8\left(x+y\right)+8+y^2+6y+9-17\)

\(=2\left(x+y-2\right)^2+\left(y+3\right)^2-17\ge-17\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y+3=0\\ x+y-2=0\end{cases}\Rightarrow\begin{cases}y=-3\\ x=-y+2=-\left(-3\right)+2=3+2=5\end{cases}\)

f: \(F=2x^2+8xy+11y^2-4x-2y+6\)

\(=2x^2+8xy+8y^2-4x-8y+3y^2+6y+6\)

\(=2\left(x+2y\right)^2-4\left(x+2y\right)+2+3y^2+6y+3+1\)

\(=2\left(x+2y-1\right)^2+3\left(y+1\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi y+1=0 và x+2y-1=0

=>y=-1 và x=-2y+1=-2*(-1)+1=2+1=3

h: \(H=x^2+y^2-xy-x+y+1\)

\(=\frac14\left(4x^2+4y^2-4xy-4x+4y+4\right)\)

\(=\frac14\left(4x^2-4xy+y^2-4x+2y+3y^2+2y+4\right)\)

\(=\frac14\left\lbrack\left(2x-y\right)^2-2\left(2x-y\right)+1+3y^2+2y+\frac13+\frac83\right\rbrack\)

\(=\frac14\cdot\left\lbrack\left(2x-y-1\right)^2+3\left(y+\frac13\right)^2+\frac83\right\rbrack\ge\frac23\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y+\frac13=0\\ 2x-y-1=0\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ 2x=y+1=-\frac13+1=\frac23\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ x=\frac13\end{cases}\)

Bài 2:

a: Sửa đề: \(A=-4x^2-5y^2+8xy+10y+12\)

\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)

\(=-\left(2x-2y\right)^2-\left(y-5\right)^2+37\le37\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}2x-2y=0\\ y-5=0\end{cases}\Rightarrow\begin{cases}y=5\\ x=y=5\end{cases}\)

b: \(B=-x^2-y^2+xy+2x+2y\)

\(=-\frac14\left(4x^2+4y^2-4xy-8x-8y\right)\)

\(=-\frac14\left(4x^2-4xy+y^2-8x+4y+3y^2-12y\right)\)

\(=-\frac14\left\lbrack\left(2x-y\right)^2-4\left(2x-y\right)+4+3y^2-12y+12-16\right\rbrack\)

\(=-\frac14\left\lbrack\left(2x-y-2\right)^2+3\left(y-2\right)^2-16\right\rbrack=-\frac14\left(2x-y-2\right)^2-\frac34\left(y-2\right)^2+4\le4\forall x,y\)

Dấu '=' xảy ra khi y-2=0 và 2x-y-2=0

=>y=2 và 2x=y+2=2+2=4

=>x=2 và y=2

Bài 1:

d: \(D=2x^2+3y^2+4xy-8x-2y\)

\(=2x^2+4xy+2y^2-8x-8y+y^2+6y\)

\(=2\left(x+y\right)^2-8\left(x+y\right)+8+y^2+6y+9-17\)

\(=2\left(x+y-2\right)^2+\left(y+3\right)^2-17\ge-17\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y+3=0\\ x+y-2=0\end{cases}\Rightarrow\begin{cases}y=-3\\ x=-y+2=-\left(-3\right)+2=3+2=5\end{cases}\)

f: \(F=2x^2+8xy+11y^2-4x-2y+6\)

\(=2x^2+8xy+8y^2-4x-8y+3y^2+6y+6\)

\(=2\left(x+2y\right)^2-4\left(x+2y\right)+2+3y^2+6y+3+1\)

\(=2\left(x+2y-1\right)^2+3\left(y+1\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi y+1=0 và x+2y-1=0

=>y=-1 và x=-2y+1=-2*(-1)+1=2+1=3

h: \(H=x^2+y^2-xy-x+y+1\)

\(=\frac14\left(4x^2+4y^2-4xy-4x+4y+4\right)\)

\(=\frac14\left(4x^2-4xy+y^2-4x+2y+3y^2+2y+4\right)\)

\(=\frac14\left\lbrack\left(2x-y\right)^2-2\left(2x-y\right)+1+3y^2+2y+\frac13+\frac83\right\rbrack\)

\(=\frac14\cdot\left\lbrack\left(2x-y-1\right)^2+3\left(y+\frac13\right)^2+\frac83\right\rbrack\ge\frac23\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y+\frac13=0\\ 2x-y-1=0\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ 2x=y+1=-\frac13+1=\frac23\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ x=\frac13\end{cases}\)

Ta có:

D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18

D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18

D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1

D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1

Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3

Hay x = 5 , y = -3

Đc chx bạn

a) Ta có: 

\(f\left(-2\right)=\left|3\cdot-2-1\right|=\left|-6-1\right|=\left|-7\right|=7\) 

\(f\left(2\right)=\left|3\cdot2-1\right|=\left|6-1\right|=5\)

\(f\left(-\dfrac{1}{4}\right)=\left|3\cdot-\dfrac{1}{4}-1\right|=\left|-\dfrac{3}{4}-1\right|=\left|-\dfrac{7}{4}\right|=\dfrac{7}{4}\) 

b) Ta có: 

\(f\left(x\right)=10\)

\(\Rightarrow\left|3x-1\right|=10\)

Với \(x\ge\dfrac{1}{3}\Rightarrow3x-1=10\)

\(\Rightarrow3x=11\Rightarrow x=\dfrac{11}{3}\left(tm\right)\)

Với \(x< \dfrac{1}{3}\Rightarrow3x-1=-10\)

\(\Rightarrow3x=-9\Rightarrow x=-3\left(tm\right)\) 

_______

\(f\left(x\right)=-3\)

\(\Rightarrow\left|3x-1\right|=-3\)

Mà: \(\left|3x-1\right|\ge0\forall x\) và \(-3< 0\)

\(\Rightarrow\left|3x-1\right|=-3\) (vô lý)

\(\Rightarrow\) không có x thỏa mãn