Ké ạ

b) Ta có: \(x^3+4x+5=0\)

\(\Leftrightarrow x^3-x+5x+5=0\)

\(\Leftrightarrow x\left(x^2-1\right)+5\left(x+1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)+5\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x+5\right)=0\)

mà \(x^2-x+5>0\forall x\)

nên x+1=0

hay x=-1

Vậy: S={-1}

a)x²-(x+3)(3x+1)=9

⇔(x-3)(x+3)-(x+3)(3x+1)=0

⇔x+3=0 hoặc 3x+1=0 

1.x+3=0 ⇔x=-3

2.3x+1=0⇔x=-1/3

phương trình có 2 nghiệm x=-3 và x=-1/3

Chắc đợi mai lm nha nhiều z tết năm sau mới lm xong

MAI LÀM LÀ XONG LUÔN RỒI!!

Cậu để chữ cứ từ lm đê lm j z

Xem hình sau:

a) 187 + (-54) = 133 b) (-203) + 349 = 146 c) (-175) + (-213) = -388

Hướng dẫn bấm nút trên máy tính:

a) 187 + (-54) = 133

b) (-203) + 349 = 146

c) (-175) + (-213) = -388

a,133

b,-552

c,-388

nhiều thế, đăng ít một thôi bạn

a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)

e) ta dể dàng thấy được : \(a^2+b^2=\left(a+b\right)^2-2ab\)

\(\Rightarrow E=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)

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

\(=4\left(a+b\right)^2-2\left(\left(a+b\right)^2-c^2\right)-2\left(a+b\right)^2\)

\(=4\left(a+b\right)^2-2\left(a+b\right)^2+2c^2-2\left(a+b\right)^2=2c^2\)

g) củng sử dụng cái trên ta có : \(G=\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\)

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

\(=4\left(a+b\right)^2+4\left(a-b\right)^2-2\left(\left(a+b\right)^2-\left(c+d\right)^2\right)-2\left(\left(a-b\right)^2-\left(c-d\right)^2\right)\)

\(=4\left(\left(a+b\right)^2+\left(a-b\right)^2\right)-2\left(\left(a+b\right)^2+\left(a-b\right)^2\right)+2\left(\left(c+d\right)^2+\left(c-d\right)^2\right)\)

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

\(=2\left(4a^2-2\left(a^2-b^2\right)\right)+2\left(4c^2-2\left(c^2-d^2\right)\right)\)

\(=2\left(2a^2+2b^2\right)+2\left(2c^2+2d^2\right)=4\left(a^2+b^2+c^2+d^2\right)\)

bn đăng nhiều quá nên mk làm câu nào hay câu đó nha

mà nè mấy câu a;b;c;d hình như trên mạng có bn lên đó tìm nha .

-1 nha bạn

\(2CH_4^{\left[A\right]}\xrightarrow[1500^{\circ}C]{lln}C_2H_2^{\left[B\right]}+2H_2^{\left[C\right]}\)

\(C_2H_2^{\left[B\right]}+2AgNO_3+2NH_3\xrightarrow[]{}2NH_4NO_3^{\left[E\right]}+C_2Ag_2^{\left[D\right]}\downarrow\)

\(C_2Ag_2^{\left[D\right]}+2HCl^{\left[F\right]}\xrightarrow[]{}C_2H_2^{\left[B\right]}+2AgCl^{\left[G\right]}\downarrow\)

\(2C_2H_2^{\left[B\right]}\xrightarrow[t^\circ]{NH_4Cl,CuCl}CH\equiv C-CH=CH_2^{\left[H\right]}\)

\(CH\equiv C-CH=CH_2^{\left[H\right]}+H_2^{\left[C\right]}\xrightarrow[t^\circ]{Pd\text{ / }PbCO_3}CH_2=CH-CH=CH_2^{\left[I\right]}\)

\(nCH_2=CH-CH=CH_2\xrightarrow[t^{\circ},p]{xt}\left(-CH_2-CH=CH-CH_2\right)_n^{\left[-\left(I\right)_n-\right]}\)

Gọi tên:

(A): Metan

(B): Axetylen

(C): Hiđro

(D): Bạc axetylua

(E): Amoni nitrat

(F): Axit clohiđric

(G): Bạc clorua

(H): Vinylaxetylen

(I): Buta-1,3-đien

-(I)_n-: Cao su buna

\(A:CH_4;B:C_2H_2;C:H_2\\ D:C_2Ag_2;E:NH_4NO_3\\ F:HCl;G:AgCl\\H:C_4H_4;I:C_4H_6\\ 2CH_4-^{1500^0C,làm.lạnh.nhanh}->C_2H_2+3H_2\\ C_2H_2+2AgNO_3+2NH_3->C_2Ag_2+2NH_4NO_3\\ C_2Ag_2+2HCl->2AgCl+C_2H_2\\ 2C_2H_2-^{CuCl_2,NH_4Cl,100^0C}->H_2C=CH-C\equiv CH\\ H_2C=CH-C\equiv CH+H_2-^{Pd.PbCO_3,t^0}->H_2C=CH-CH=CH_2\\ H_2C=CH-CH=CH_2-^{t^0,p,xt}->-\left(H_2C-CH=CH-CH_2\right)-_n\)