a) Để phương trình có nghiệm thì: m²≠0 =>m≠0

b) Vì phương trình có nghiệm bằng -2m

=>-(m+2)2m-m²=0 ⇔-2m²-4m-m²=0 ⇔-3m²-4m=0 ⇔-m(3m+4)=0 

⇔m=0 hay m=\(\dfrac{-4}{3}\)mà m phải khác 0 nên m=\(\dfrac{-4}{3}\).

c) -(m+2)x-m²=0 ⇔x=\(\dfrac{m^2}{m+2}\)>0 ⇔m+2>0 ⇔m>-2.

d)  -(m+2)x-m²=0 ⇔x=\(\dfrac{m^2}{m+2}\).

Để x nguyên thì m² ⋮ m+2.

⇔ m²-4+4 ⋮ m+2

⇔ 4 ⋮ m+2

⇔ m∈{-1;-3;0;-4;2;-6} mà m khác 0 nên m∈{-1;-3;-4;2;-6}

ko bt

a: \(Q=\dfrac{2x^2-4x+x-3-6}{\left(x-3\right)\left(x-2\right)}\cdot\dfrac{x-2}{x^2+1}=\dfrac{2x^2-3x-9}{x-3}\cdot\dfrac{1}{x^2+1}\)

\(=\dfrac{2x^2-6x+3x-9}{x-3}\cdot\dfrac{1}{x^2+1}=\dfrac{2x+3}{x^2+1}\)

b: Để Q>0 thì 2x+3>0

hay x>-3/2

Bài 2:

a) Q=(\(\dfrac{2x}{x-3}-\dfrac{1}{2-x}-\dfrac{6}{x^2-5x+6}\)):\(\dfrac{x^2+1}{x-2}\)

       =(\(\dfrac{2x}{x-3}+\dfrac{1}{x-2}-\dfrac{6}{x^2-2x-3x+6}\)):\(\dfrac{x^2+1}{x-2}\)

        =[\(\dfrac{2x}{x-3}+\dfrac{1}{x-2}-\dfrac{6}{x\left(x-2\right)-3\left(x-2\right)}\)]:\(\dfrac{x^2+1}{x-2}\)

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

        =\(\dfrac{2x^2-4x+x-9}{\left(x-2\right)\left(x-3\right)}\):\(\dfrac{x^2+1}{x-2}\)

        =\(\dfrac{2x^2-3x-9}{\left(x-2\right)\left(x-3\right)}\):\(\dfrac{x^2+1}{x-2}\)

        =\(\dfrac{2x^2-6x+3x-9}{\left(x-2\right)\left(x-3\right)}\):\(\dfrac{x^2+1}{x-2}\)

        =\(\dfrac{2x\left(x-3\right)+3\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}\):\(\dfrac{x^2+1}{x-2}\)

        =\(\dfrac{\left(x-3\right)\left(2x+3\right)}{\left(x-2\right)\left(x-3\right)}\):\(\dfrac{x^2+1}{x-2}\)

        =\(\dfrac{\left(2x+3\right)}{\left(x-2\right)}\):\(\dfrac{x^2+1}{x-2}\)

         =\(\dfrac{2x+3}{x^2+1}\)

em ăn nhiều bánh nhất

em ăn \(\dfrac{1}{2}=\dfrac{3}{6}\) cái bánh

chị ăn \(\dfrac{2}{3}=\dfrac{4}{6}\) cái bánh

⇒chị ăn nhiều bánh hơn

(cách mình trình bày chắc chắn là không đúng nên bạn trình bày lại theo cách cô chỉ nha!)

Ta có:

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

\(4-4x=-4\left(x-1\right)\)

\(x^2-1=\left(x-1\right)\left(x+1\right)\)

\(\Rightarrow\) MTC là \(3.\left(-4\right).\left(x-1\right)\left(x+1\right)=-12\left(x-1\right)\left(x+1\right)\)

Do đó:

\(\dfrac{11x}{3x-3}=\dfrac{11x}{3\left(x-1\right)}=\dfrac{11x.\left(-4\right).\left(x+1\right)}{3\left(x-1\right).\left(-4\right)\left(x+1\right)}=\dfrac{-44x\left(x+1\right)}{-12\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{5}{4-4x}=\dfrac{5}{-4\left(x-1\right)}=\dfrac{5.3\left(x+1\right)}{-4\left(x-1\right).3\left(x+1\right)}=\dfrac{15\left(x+1\right)}{-12\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{2x}{x^2-1}=\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=\dfrac{2x.\left(-12\right)}{-12\left(x-1\right)\left(x+1\right)}=\dfrac{-24x}{-12\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{2}{5}-9-\dfrac{7}{2}\)

\(=\dfrac{2}{5}-\dfrac{9}{1}-\dfrac{7}{2}\)

\(=\left(\dfrac{2}{5}-\dfrac{9}{1}\right)-\dfrac{7}{2}\)

\(=\left(\dfrac{2}{5}-\dfrac{45}{5}\right)-\dfrac{7}{2}\)

\(=-\dfrac{43}{5}-\dfrac{7}{2}\)

\(=-\dfrac{86}{10}-\dfrac{35}{10}\)

\(=-\dfrac{121}{10}\)

\(\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\\ =\dfrac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\dfrac{2+\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\\ =\dfrac{2-\sqrt{3}+2+\sqrt{3}}{2^2-\left(\sqrt{3}\right)^2}\\ =\dfrac{2+2}{4-3}\\ =4\)

Ta có: \(\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)

\(=2-\sqrt{3}+2+\sqrt{3}\)

=4

b) Ta có: \(\sqrt{150}-\sqrt{1.6}\cdot\sqrt{60}+4.5\cdot\sqrt{2\dfrac{2}{3}}-\sqrt{6}\)

\(=5\sqrt{6}-4\sqrt{6}-\sqrt{6}+\dfrac{9}{2}\cdot\sqrt{\dfrac{8}{3}}\)

\(=\dfrac{9}{2}\cdot\dfrac{2\sqrt{2}}{\sqrt{3}}\)

\(=3\sqrt{6}\)

\(\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4.5\sqrt{2\dfrac{2}{3}}-\sqrt{6}\\ =5\sqrt{6}+4\sqrt{6}+3\sqrt{6}-\sqrt{6}\\ =11\sqrt{6}\)

2) nH2=0,1(mol)

a) PTHH: Fe +2  HCl -> FeCl2 + H2

0,1______0,2______0,1____0,1(mol)

nFe=nH2=0,1(mol)

=>mFe=nFe.M(Fe)=0,1.56=5,6(g)

=> mFeO=mX-mFe= 9,2-5,6=3,6(g)

=> nFeO=mFeO/M(FeO)=3,6/72=0,05(mol)

PTHH: FeO +2 HCl -> FeCl2 + H2

0,05_________0,1___0,05__0,05(mol)

b) Sao lại mỗi oxit a, có một oxit thôi mà :(  Chắc % KL mỗi chất.

%mFeO=(mFeO/mhh).100%=(3,6/9,2).100=39,13%

=>%mFe=100%-%mFeO=100%-39,13%=60,87%

c) nHCl(tổng)= 2.nFe +2.nFeO=2.0,1+2.0,05=0,3(mol)

=>mHCl=nHCl.M(HCl)=0,3.36,5=10,95(g)

=>mddHCl=(mHCl.100%/C%ddHCl=(10,95.100)/7,3=150(g)

d) - Dung dich thu được chứa FeCl2.

mFeCl2=nFeCl2(tổng) . M(FeCl2)= (0,1+0,05).127=19,05(g)

mddFeCl2=mddHCl+mhh-mH2=150+9,2-0,1.2=159(g)

=> C%ddFeCl2=(mFeCl2/mddFeCl2).100%=(19,05/159).100=11,981%

Câu 10:

12/24=50%

Câu 11:

6/12=50%

Câu 12: 

1/4=25%

Câu 13: 

=320x15:100=48(kg)

Câu 10:

12/24=50%

Câu 11:

6/12=50%

Câu 12: 

1/4=25%

Câu 13: 

=320x15:100=48(kg)

10: 12:24=0,5=50%

11. 6:12=0,5=50%

12. 1:4=0,25=25%

13. 320x15:100=48kg