mn ơi  giúp em

Bài 3:

\(a,=3x\left(y-4x+6y^2\right)\\ b,=5xy\left(x^2-6x+9\right)=5xy\left(x-3\right)^2\\ d,=\left(x+y\right)\left(x-12\right)\\ f,=2x\left(x-y\right)\left(5x-4y\right)\\ g,=\left(x-2\right)\left(x-2+3x\right)=\left(x-2\right)\left(4x-2\right)=2\left(x-2\right)\left(2x-1\right)\\ h,=x^2\left(1-5x\right)+3xy\left(5x-1\right)=x\left(1-5x\right)\left(x-3y\right)\\ i,=x\left(x-2\right)+4\left(x-2\right)=\left(x+4\right)\left(x-2\right)\\ j,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ k,=4x^2-12x+3x-9=\left(x-3\right)\left(4x+3\right)\\ l,=\left(x+5\right)^2-y^2=\left(x-y+5\right)\left(x+y+5\right)\\ m,=x^2-\left(2y-6\right)^2=\left(x-2y+6\right)\left(x+2y-6\right)\\ n,=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\\ =\left(x^2+5x+5\right)^2-1-24\\ =\left(x^2+5x+5\right)^2-25\\ =\left(x^2+5x\right)\left(x^2+5x+10\right)\\ =x\left(x+5\right)\left(x^2+5x+10\right)\)

Bài 3:

a)3xy-12x²+24xy²=3x(y-4x+8y²)

b)5x³y-30x²y+45xy=5xy(x²-6x+9)=5xy(x-6x+3²)=5xy(x-3)²

d)x(x+y)-12x-12y=x(x+y)-(12x+12y)=x(x+y)-12(x+y)=(x+y)(x-12)

f)10x²(x-y)-8xy(x-y)=(x-y)(10x²-8xy)

g)(x-2)³+3x²-6x=(x-2)³+(3x²-6x)=(x-2)³+3x(x-2)=(x-2)[(x-2)+3x]=(x-2)(x-2+3x)=(x-2)(4x-2)

h)x²+15x²y-3xy-5x³=(x²-3xy)+(15x²y-5x³)=x(x-3y)+5x²(3y-x)=x(x-3y)-5x²(x-3y)=(x-3y)(x-5x²)

i)x²-2x-8+4x=(x²-2x)+(4x-8)=x(x-2)+4(x-2)=(x-2)(x+4)

k)4x²-9x-9=4x²+3x-12x-9=(4x²+3x)-(12x+9)=x(4x+3)-3(4x+3)=(4x+3)(x-3)

l)x²-y²+10x+25=(x²+10x+25)-y²=(x²+10x+5²)-y²=(x+5)²-y²=[(x+5)+y][(x+5)-y]=(x+5+y)(x+5-y)

m)x²-4y²+24y-36=x²-(4y²-24y+36)=x²-[(2y)²-24y+6²]=x²-(2y-6)²=[x+(2y-6)][x-(2y-6)]=(x+2y-6)(x-2y+6)

n)(không biết làm)

b: Thay x=-1 và y=-3 vào (d1), ta được:

-3=-1+2

=>-3=1(loại)

=>A ko thuộc (d1)

Thay x=-1 và y=1 vào (d1), ta đc:

-1+2=1

=>1=1

=>B thuộc (d1)

c: Tọa độ C là:

x+2=-1/2x+2 và y=x+2

=>x=0 và y=2

Bài 6:

Vì \(m^2+1>0\) nên hs nghịch biến trong khoảng \(\left(-\infty;2m\right)\)

Bài 3:

6: \(x< 0\) nên \(y=\sqrt[3]{x}\) nghịch biến

\(x^6+x^4-3x^2-4x+6\)

\(=\left(x^6+2x^5+4x^4+6x^3+5x^2\right)-\left(2x^5+4x^4+8x^3+12x^2+10x\right)+\left(x^4+2x^3+4x^2+6x+5\right)+1\)

\(=x^2\left(x^4+2x^3+4x^2+6x+5\right)-2x\left(x^4+2x^3+4x^2+6x+5\right)+\left(x^4+2x^3+4x^2+6x+5\right)+1\)

\(=\left(x^4+2x^3+4x^2+6x+5\right)\left(x^2-2x+1\right)+1\)

\(=\left[\left(x^4+2x^3+x^2\right)+3\left(x^2+2x+1\right)+2\right]\left(x-1\right)^2+1\)

\(=\left[\left(x^2+x\right)^2+3\left(x+1\right)^2+2\right]\left(x-1\right)^2+1\ge1\)

Dấu "=" xảy ra khi \(x=1\)

làm giúp e bài 4 câu b với ạ

\(b,B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\left(x\ge0;x\ne4;x\ne9\right)\\ B=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)

\(c,B< A\Leftrightarrow\dfrac{\sqrt{x}-4}{\sqrt{x}-2}< \dfrac{\sqrt{x}+1}{\sqrt{x}-2}\Leftrightarrow\dfrac{\sqrt{x}-4}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}< 0\\ \Leftrightarrow\dfrac{-5}{\sqrt{x}-2}< 0\Leftrightarrow\sqrt{x}-2>0\left(-5< 0\right)\\ \Leftrightarrow x>4\\ d,P=\dfrac{B}{A}=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}:\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=\dfrac{\sqrt{x}-4}{\sqrt{x}+1}=1-\dfrac{5}{\sqrt{x}+1}\in Z\\ \Leftrightarrow5⋮\sqrt{x}+1\Leftrightarrow\sqrt{x}+1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-6;-2;0;4\right\}\\ \Leftrightarrow x\in\left\{0;16\right\}\left(\sqrt{x}\ge0\right)\)

\(e,P=1-\dfrac{5}{\sqrt{x}+1}\)

Ta có \(\sqrt{x}+1\ge1,\forall x\Leftrightarrow\dfrac{5}{\sqrt{x}+1}\ge5\Leftrightarrow1-\dfrac{5}{\sqrt{x}+1}\le-4\)

\(P_{max}=-4\Leftrightarrow x=0\)

\(=ab\left(a-b\right)\left(a+b\right)+c^3\left(a-b\right)-c\left(a^3-b^3\right)\)

\(=\left(a-b\right)\left(a^2b+ab^2\right)+c^3\left(a-b\right)-\left(a-b\right)\left(a^2c+abc+b^2c\right)\)

\(=\left(a-b\right)\left(a^2b+ab^2+c^3-a^2c-abc-b^2c\right)\)

\(=\left(a-b\right)\left[ab\left(a-c\right)+b^2\left(a-c\right)-c\left(a^2-c^2\right)\right]\)

\(=\left(a-b\right)\left[ab\left(a-c\right)+b^2\left(a-c\right)-\left(a-c\right)\left(ac+c^2\right)\right]\)

\(=\left(a-b\right)\left(a-c\right)\left(ab+b^2-ac-c^2\right)\)

\(=\left(a-b\right)\left(a-c\right)\left[a\left(b-c\right)+\left(b-c\right)\left(b+c\right)\right]\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)

bài nào zậy bạn

Câu 3 và caau4 bài giải phương trình nhé

Bài 3. Đặt ẩn phụ là 

\(a=2x-\frac{5}{x}\\\)

\(b=x-\frac{1}{x}\)

pt <=> \(b-a=\sqrt{a}-\sqrt{b}\\ \)

\(\left(\sqrt{b}-\sqrt{a}\right)\left(\sqrt{b}+\sqrt{a}\right)=-\left(\sqrt{b}-\sqrt{a}\right)\\ \)

\(\left(\sqrt{b}-\sqrt{a}\right)\left(\sqrt{b}+\sqrt{a}+1\right)=0\)

tới đây xét 2 TH bạn tự giải nhé