Thay \(x+y=1\) vào biểu thức \(A=x^2+2xy+y^2+2022\) ta được:

\(A=\left(1\right)^2+2.1.1+\left(1\right)^2+2022\) 

\(A=1+2.1.1+1+2022\)

\(A=1+2+1+2022\) \(=2026\)

Vậy: \(x+y=1\) thì biểu thức \(A=x^2+2xy+y^2+2022\) là \(2026\)

Bài 2

A =  (x + 2)^ 2 - x - 3 x (x + 1) = x² + 4x + 4 - x² + 2x + 3 = 6x + 7

B = x^3 - 2x² + 5x - 10 = x² x (x - 2) + 5 x (x - 2) = (x - 2) x (x² + 5)

Vậy x^3 - 2x² + 5x - 10 : (x - 2) = x² + 5

a) (x-y)(2x+3y)=2x²+3xy-2xy+3y²=2x²+xy+3y²

b) (2x-1)²-(2x-1)=0

<=> 2x-1=0 <=> x=\(\dfrac{1}{2}\)

a) Ta có: (x-y)(2x+3y)

\(=2x^2+3xy-2xy-3y^2\)

\(=2x^2+xy-3y^2\)

b) Ta có: \(4x^2-4x+1=2x-1\)

\(\Leftrightarrow4x^2-4x+1-2x+1=0\)

\(\Leftrightarrow4x^2-4x-2x+2=0\)

\(\Leftrightarrow4x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{1;\dfrac{1}{2}\right\}\)

Bài 1: 

a: 12,3x45=553,5

b: 35,4x88=3115,2

c: 46,5x97=4510,5

d: 57,6x78=4492,8

e: 68,7x49=3366,3

a: \(=\dfrac{x+1+3}{x-3}=\dfrac{x+4}{x-3}\)

c: \(=4x^2-12x+9-4x^2+x=-11x+9\)

\(a,=2x^2-10x+x^2+x-6=3x^2-9x-6\\ b,=x^2+4x+4-x^2+8x-15=12x-11\\ c,=4x^2-12x+9-4x^2+x=-11x+9\)

\(a,=4x^2-4x+1-4x^2+4-x^2-x+6=-x^2-5x+11\\ b,=8x^3+27-8x^3+72x=72x+27\)

a) \(=4x^2-4x+1-4\left(x^2-1\right)-\left(x^2-2x+3x-6\right)=4x^2-4x+1-4x^2+4-x^2-x+6=-x^2-5x+11\)

b) \(=8x^3+27-8x\left(x^2-9\right)=8x^3+27-8x^3+72x=72x+27\)

a.5/14+1/2=6/7

b.5/3-1=2/3

c.4/3-1/3=1

a) 5/14+ 1/2= 6/7

b) 5/3 - 1= 2/3

c) 4/3- 1/3 =1

a.5/14+1/2=6/7

b.5/3-1=2/3

c.4/3-1/3=1

a, \(1+\dfrac{3}{4}=\dfrac{7}{4}\)

b, \(\dfrac{4}{5}-\dfrac{3}{8}=\dfrac{32-15}{40}=\dfrac{17}{40}\)

c, \(1:\dfrac{2}{3}=\dfrac{1.3}{2}=\dfrac{3}{2}\)

d, \(\dfrac{2}{5}.\dfrac{5}{2}=1\)

a)1 + 3/4=7/4

   b)4/5 -3/8=17/40

  c)1 : 2/3=3/2

  d)2/5 x 5/2=1

Đáp án:

a)1 + 3/4=7/4

b)4/5 -3/8=17/40

c)1 : 2/3=3/2

d)2/5 x 5/2=1