#h24cfs_20022023

b này y hệt mình=)) chỉ khác là b vẫn muốn kết bạn với mn còn mình thì mặc kệ luôn

ĐK: \(x\ge0,x\ne4\)

\(\dfrac{1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-4}{2\sqrt{x}-x}=\dfrac{-1}{2-\sqrt{x}}+\dfrac{5\sqrt{x}-4}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{-\sqrt{x}+5\sqrt{x}-4}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(2-\sqrt{x}\right)}\)

\(\dfrac{2+\sqrt{x}}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{2+\sqrt{x}}{\sqrt{x}}+\dfrac{\sqrt{x}}{2-\sqrt{x}}=\dfrac{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{4-x+x}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{4}{\sqrt{x}\left(2-\sqrt{x}\right)}\)

\(\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-4}{2\sqrt{x}-x}\right):\left(\dfrac{2+\sqrt{x}}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(2-\sqrt{x}\right)}:\dfrac{4}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(2-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(2-\sqrt[]{x}\right)}{4}=\sqrt{x}-1\)

ĐK: \(x\ge0,x\ne4\)

\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-2\sqrt{x}\left(\sqrt{x}-2\right)-\left(5\sqrt{x}+2\right)}{x-4}=\dfrac{x+3\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2}{x-4}=\dfrac{-x+2\sqrt{x}}{x-4}=\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{-\sqrt{x}}{\sqrt{x}+2}\)\(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{3\sqrt{x}-x}{x+4\sqrt{x}+4}=\dfrac{-\sqrt{x}}{\sqrt{x}+2}:\dfrac{-\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)^2}=\dfrac{-\sqrt{x}}{\sqrt{x}+2}.\dfrac{\left(\sqrt{x}+2\right)^2}{-\sqrt{x}\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

Vậy...

a) \(\dfrac{39}{41}=\dfrac{39.71}{41.71}=\dfrac{2769}{2911},\dfrac{69}{71}=\dfrac{69.41}{71.41}=\dfrac{2829}{2911}\)

\(2769< 2829\Rightarrow\dfrac{2769}{2911}< \dfrac{2829}{2911}\Rightarrow\dfrac{39}{41}< \dfrac{69}{71}\)

b) \(\dfrac{59}{-58}=\dfrac{59.\left(-68\right)}{-58.\left(-68\right)}=\dfrac{-4012}{3944},\dfrac{-69}{68}=\dfrac{-69.58}{68.58}=\dfrac{-4002}{3944}\)

\(-4012< -4002\Rightarrow\dfrac{-4012}{3944}< \dfrac{-4002}{3944}\Rightarrow\dfrac{59}{-58}< \dfrac{-69}{68}\)

c) \(\dfrac{-49}{50}=\dfrac{-49.9}{50.9}=\dfrac{-441}{450},\dfrac{91}{-90}=\dfrac{91.\left(-5\right)}{-90.\left(-5\right)}=\dfrac{-455}{450}\)

\(-441>-455\Rightarrow\dfrac{-441}{450}>\dfrac{-455}{450}\Rightarrow\dfrac{-49}{50}>\dfrac{91}{-90}\)

d) \(\dfrac{1399}{-1411}=\dfrac{1399.\left(-1151\right)}{-1411.\left(-1151\right)}=\dfrac{-1610249}{1624061},\dfrac{1169}{-1151}=\dfrac{1169.\left(-1411\right)}{-1151\left(-1411\right)}=\dfrac{-1649459}{1624061}\)

\(-1610249>-1649459\Rightarrow\dfrac{-1610249}{1624061}>\dfrac{-1649459}{1624061}\Rightarrow\dfrac{1399}{-1411}>\dfrac{1169}{-1151}\)

a) \(M+\left(12x^2-6xy\right)=3x^2+4xy-2y^2\Rightarrow M=3x^2+4xy-2y^2-12x^2+6xy=-9x^2-2y^2+10xy\)

b) \(M-\left(5xy-6y^2\right)=x^2-87xy+6y^2\Rightarrow M=x^2-87xy+6y^2+5xy-6y^2=x^2-82xy\)

c) \(\left(-20x^2y+15xy-y^3\right)-M=21x^2y-2y^3\Rightarrow M=-20x^2y+15xy-y^3-21x^2y+2y^3=-41x^2y+15xy+y^3\)

d) \(M+\left(12x^4-15x^2y+2xy^2+7\right)=0\Rightarrow M=-12x^4+15x^2y-2xy^2-7\)

a) Các hạng tử của A: 5xy², -2x³y², -6xy, -10

Hạng tử có bậc cao nhất là -2x³y² (bậc 5)

b) \(A+B=5xy^2-2x^3y^2-6xy-10+2x^3y^2-3xy^2-10xy+2=(5xy^2-3xy^2)+(-2x^3y^2+2x^3y^2)+(-6xy-10xy)-10+2=2xy^2+0-16xy-8=2xy^2-16xy-8\)

Đa thức A+B có bậc là 3

c) \(A-B=5xy^2-2x^3y^2-6xy-10-\left(2x^3y^2-3xy^2-10xy+2\right)=5xy^2-2x^3y^2-6xy-10-2x^3y^2+3xy^2+10xy-2=\left(-2x^3y^2-2x^3y^2\right)+\left(5xy^2+3xy^2\right)+\left(-6xy+10xy\right)-10-2=-4x^3y^2+8xy^2+4xy-12\)

Thay x=1, y=2 vào biểu thức ta được: \(-4.1^32^2+8.1.2^2+4.1.2-12=-16+32+8-12=12\)

d) Để A+M=\(-3x^3y^2+2xy^2+8xy-1\) thì \(M=-3x^3y^2+2xy^2+8xy-1-A=-3x^3y^2+2xy^2+8xy-1-\left(5xy^2-2x^3y^2-6xy-10\right)=-3x^3y^2+2xy^2+8xy-1-5xy^2+2x^3y^2+6xy+10=-x^3y^2-3xy^2+14xy+9\)

Vậy...

6A

\(\left(-3x^2y^3z+9x^3y^2z^2-12x^4yz^2\right):\left(-3xyz\right)=\dfrac{-3x^2y^3z}{-3xyz}+\dfrac{9x^3y^2z^2}{-3xyz}+\dfrac{-12x^4yz^2}{-3xyz}=xy^2-3x^2yz+4x^3z\)

7A\(f\left(x\right)-g\left(x\right)=x^4-3x^3+x^2-x+10-\left(2x^4-2x^2+x-5\right)=x^4-3x^3+x^2-x+10-2x^4+2x^2-x+5=-x^4-3x^3+3x^2-2x+15\)

8B

\(A=\left(4x-6\right)\left(x+7\right)-4x\left(x+5\right)-2x=4x^2+22x-42-4x^2-20x-2x=-42\)

\(-42< -1\Rightarrow A< -1\)

9C

Thay x=1,y=1 ta được: \(1^{n-1}\left(1+1\right)-1\left(1^{n-1}+1^{n-1}\right)=1.2-\left(1+1\right)=0\)

Phần mấu chốt là \(1^{n-1}=1\) với mọi n

10D

\(15x^3yz^5.\left(-3y^3z^2\right)=-45x^3y^4z^7\)