a) \(6xy+4x-9y-7=0\)

  \(\Leftrightarrow2x.\left(3y+2\right)-9y-6-1=0\)

\(\Leftrightarrow2x.\left(3y+x\right)-3.\left(3y+2\right)=1\)

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

Mà \(x,y\in Z\Rightarrow2x-3;3y+2\in Z\)

Tự làm típ

\(A=x^3+y^3+xy\)

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

\(A=x^2-xy+y^2+xy\)( vì \(x+y=1\))

\(A=x^2+y^2\)

Áp dụng bất đẳng thức Bunhiakovxky ta có :

\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x\cdot1+y\cdot1\right)^2=\left(x+y\right)^2=1\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge1\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)

Hay \(x^3+y^3+xy\ge\frac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)

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Ta có 64=4^3

=> x+1=4

=>x=3

(x+1)³=4³

x+1=4

x=3

\(\left(x+1\right)^3=64\)

\(\left(x+1\right)^3=4^3\)

\(\Rightarrow x+1=4\)

\(x=4-1\)

\(x=3\)

xy + 3y - 5x = 9 nhé...mình viết nhầm ạ

11=1x11=11x1=-1x-11=-11x-1

TH1:

2x-1=1                            y+4=11

2x=2                                y=7

x=1

TH2:

2x-1=11                            y+4=1

2x=12                                y=-5

x=6

TH3:

2x-1=-1                            y+4=-11

2x=-2                                y=-15

x=-1

TH4:

2x-1=-11                            y+4=-1

2x=-10                                y=-5

x=-5

a)(2x-1)(y+4)=11

          Ta có:11=1.11=11.1=(-1).(-11)=(-11).(-1)

                 Do đó ta có bảng sau:

          Vậy các cặp (x;y) TM là:(0;-15)(-5;-5)(6;-3)(1;7)

TK: Tìm Min (x^4 + 1) (y^4 + 1) với x + y = căn10 ; x , y > 0 - Thanh Truc

4:

\(P=\left(x+4\right)\left(x^2-4x+16\right)-\left(64-x^3\right)\)

\(=x^3+64-64+x^3=2x^3\)

Khi x=-20 thì \(P=2\cdot\left(-20\right)^3=-16000\)

=>Chọn C

2: Đề khó hiểu quá bạn ơi

\("="\Leftrightarrow x=1;y=3\)

Bạn thêm vào dòng cuối nhé :v Mình quên ghi :v

\(\frac{-7x^2+42x-64}{x^2-6x+10}\)

\(\Rightarrow7+\frac{6}{\left(x-3\right)^2+1}=y^2+2y\)

\(\Rightarrow\frac{6}{\left(x-3\right)^2+1}=\left(y-1\right)^2+6\)

\(\Rightarrow6=\left[\left(y-1\right)^2+6\right]\left[\left(x-3\right)^2+1\right]\)

\(\Rightarrow0=\left(y-1\right)^2\left(x-3\right)^2+6\left(x-3\right)^2+\left(y-1\right)^2\)

ta có x+y=\(\sqrt{10}\)=>(x+y)^2=10

=x^4.y^4+1+x^4+y^4+2x^2.y^2-2x^2.y^2

=x^4.y^4+1+(x^2+y^2)^2-2x^y^2=x^4.y^4+1+[(x+y)^2-2xy]

=x^4.y^4+1+(10-2xy)-2x^2.y^2

=x^4.y^4+1+100-40xy+4.x^2.y^2-2x^2.y^2

=x^4.y^4+101-40xy+2.x^2.y^2

=(x^4.y^4-8.x^2.y^2+16)+(10.x^2.y^2-40xy+40)+45

=(x^2.y^2-4)^2+10.(xy-2)^2+45\(\ge\)0

dấu = xảy ra \(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y=\sqrt{10}\\x.y=2\end{matrix}\right.\)

vậy Min A=45

\(\left\{{}\begin{matrix}x+y=\sqrt{10}\\x.y=2\end{matrix}\right.\)là nghiệm pt x^2-\(\sqrt{10}\)x+2

=>\(\Delta\)=(-\(\sqrt{10}\))^2-4.2=2>0

=>\(\left\{{}\begin{matrix}x=\dfrac{\sqrt{10}-\sqrt{2}}{2}\\y=\dfrac{\sqrt{10}+\sqrt{2}}{2}\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=\dfrac{\sqrt{10}-\sqrt{2}}{2}\\y=\dfrac{\sqrt{10}+\sqrt{2}}{2}\end{matrix}\right.\)

ta có x+y==>(x+y)^2=10

=x^4.y^4+1+x^4+y^4+2x^2.y^2-2x^2.y^2

=x^4.y^4+1+(x^2+y^2)^2-2x^y^2=x^4.y^4+1+[(x+y)^2-2xy]

=x^4.y^4+1+(10-2xy)-2x^2.y^2

=x^4.y^4+1+100-40xy+4.x^2.y^2-2x^2.y^2

=x^4.y^4+101-40xy+2.x^2.y^2

=(x^4.y^4-8.x^2.y^2+16)+(10.x^2.y^2-40xy+40)+45

=(x^2.y^2-4)^2+10.(xy-2)^2+45

dấu = xảy ra {x+y=√10x.y=2{x+y=10x.y=2

vậy Min A=45