3x+x3=121 =>x{3+3}=121 =>x9=121 =>sai 

đề sai

\(3x+3x=11\cdot11\)

\(6x=121\)

\(x=\frac{121}{6}\)

X = 8  nhe ban !

trả lời hộ mk nha nhớ ghi cách lam vào nhé 

X=8,kb và kick mình nha bạn

\(A=6x^2+23x+21-\left(6x^2+23x-55\right)=76\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ C=x^4+x^3-3x^2-2x-\left(x^4+x^3-x^2-2x^2-2x+2\right)\\ =-2\)

3x + x3 = 11 x 11

3x + x3 = 121

Mà 121 = 38 + 83

Vậy x = 8

x = 8

vì 11 x 11 = 121

mà 1 = 8 + 3 nên x = 8

quý vị khán giả thấy đúng thì k nhá

3x + x3 = 11x11

3x + x3 = 121

Ta đặt phép tính như sau:

3x + x3 = 121 (cái này bạn đặt tính dọc nhé!)

- NX : Hàng đơn vị : x + 3 = 1 => x+3 có nhớ => x = 8

Vậy x =8

Thế thôi, k nhé?

cu the thay vao la tinh duoc

chk hoc tot

Không tên

3x + x3 = 11 x 11 

=> 30 + x + x.10 + 3 = 121

=> 33 + x.11 = 121

=> x.11 = 121 - 33 = 88

=> x = 88 : 11 = 8 

a) \(\dfrac{1}{x+2}+\dfrac{5}{2x^2+3x-2}\)

\(=\dfrac{1}{x+2}+\dfrac{5}{2x^2+4x-x-2}\)

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

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

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

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

\(=\dfrac{2}{2x-1}\)

\(---\)

b) \(\dfrac{-3x^2}{x^3+1}+\dfrac{1}{x^2-x+1}+\dfrac{1}{x+1}\) (sửa đề)

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

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

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

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

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

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

\(---\)

c) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}\)

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

\(=\dfrac{1+x+1-x}{1^2-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}\)

\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}\)

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

\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}\)

\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}\)

\(=\dfrac{4\left(1+x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\dfrac{4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}\)

\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}\)

\(=\dfrac{8}{1-x^8}\)

#\(Toru\)

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

__

`x^3+1` chứ cậu nhỉ?

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

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

__

A = (3x + 7)(2x + 3) – (3x – 5)(2x + 11)

= 3x.2x + 3x.3 + 7.2x + 7.3 – (3x.2x + 3x.11 – 5.2x – 5.11)

=   6 x 2   +   9 x   +   14 x   +   21   –   ( 6 x 2   +   33 x   –   10 x   –   55 )     =   6 x 2   +   23 x   +   21   –   6 x 2   –   33 x   +   10 x   +   55   =   76

B   =   x ( 2 x   +   1 )   –   x 2 ( x   +   2 )   +   x 3   –   x   +   3     =   x . 2 x   +   x   –   ( x 2 . x   +   2 x 2 )   +   x 3   –   x   +   3     =   2 x 2   +   x   –   x 3   –   2 x 2   +   x 3   –   x   +   3   =   3

Từ đó ta có A = 76; B = 3 mà 76 = 25.3 + 1 nên A = 25B + 1

Đáp án cần chọn là: C

1) Ta có: \(\dfrac{3x+7}{11}=\dfrac{5x-7}{9}\)

\(\Leftrightarrow9\left(3x+7\right)=11\left(5x-7\right)\)

\(\Leftrightarrow27x+63=55x-77\)

\(\Leftrightarrow27x-55x=-77-63\)

\(\Leftrightarrow-28x=-140\)

hay x=5

Vậy: S={5}

1, 3x+7/11= 5x-7/9

3x-5x=-7/11-7/9

-2x=-140/99

x= 70/99

vậy ...

2, x-7/4=9/-7+x

x-x=7/4-9/7

x không có gt thỏa mãn 

3, x-1/-2=8/1-x 

x+x =-1/2+8/1 

2x =15/2 

x =15/4 

vậy ...