Câu 1: (2x+y)(y-2x)+4x²=y²-4x²+4x²=y²

Với y=10 giá trị biểu thức trên là 10²=100

Câu 2: 

a. xy-11x=x.(y-11)

b. x²+4y²+4xy-16=(x²+4xy+4y²)-16

=(x+2y)²-16=(x+2y+4)(x+2y-4)

I don know

1. \(B=\left(x-2\right)\left(x+2\right)\left(x+3\right)-\left(x+1\right)^3\)

\(=\left(x^2-4\right)\left(x+3\right)-\left(x^3+3x^2+3x+1\right)\)

\(=x^3+3x^2-4x-12-x^3-3x^2-3x-1\)

\(=-7x-13\)

2. \(64-x^2-y^2+2xy=64-\left(x^2+y^2-2xy\right)\)

\(=64-\left(x-y\right)^2=\left(8+x-y\right)\left(8-x+y\right)\)

3. \(2x^3-x^2+2x-1=0\)

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

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

Vì \(x^2\ge0\)\(\Rightarrow x^2+1>0\)

\(\Rightarrow2x-1=0\)\(\Rightarrow2x=1\)\(\Rightarrow x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\)

Bài 1.

B = ( x - 2 )( x + 2 )( x + 3 ) - ( x + 1 )³

= ( x² - 4 )( x + 3 ) - ( x³ + 3x² + 3x + 1 )

= x³ + 3x² - 4x - 12 - x³ - 3x² - 3x - 1

= -7x - 13

Bài 2.

64 - x² - y² + 2xy

= 64 - ( x² - 2xy + y² )

= 8² - ( x - y )²

= ( 8 -  x + y )( 8 + x - y )

Bài 3.

2x³ - x² + 2x - 1 = 0

<=> ( 2x³ - x² ) + ( 2x - 1 ) = 0

<=> x²( 2x - 1 ) + 1( 2x - 1 ) = 0

<=> ( 2x - 1 )( x² + 1 ) = 0

<=> \(\orbr{\begin{cases}2x-1=0\\x^2+1=0\end{cases}}\Leftrightarrow x=\frac{1}{2}\)( vì x² + 1 ≥ 1 > 0  ∀ x )

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\(\text{Câu 1 thì bạn tự làm chịu khó tí:}y^2-4x^2=\left(y+2x\right)\left(y-2x\right)\)

\(\text{Bài 2:}ab\left(b-a\right)+bc\left(c-b\right)+ac\left(a-c\right)\)

\(=ab\left(b-a\right)-bc\left(b-c\right)+ac\left(a-c\right)\)

\(=ab\left(b-a\right)-bc\left[\left(b-a\right)+\left(a-c\right)\right]+ac\left(a-c\right)\)

\(=\left(ab-bc\right)\left(b-a\right)+\left(ac-bc\right)\left(a-c\right)\)

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

a, \(=\left(x+y\right)^2-2^2=\left(x+y-2\right)\left(x+y+2\right)\)

b, =  \(y^2-4x^2+4x^2=y^2\)

Thay y = 10 vào BT trên, ta có:

\(y^2=10^2=100\)

Vậy giá trị của BT là 100

\(2x\left(x-3y\right)-4y\left(x+2\right)-2\left(x^2-3y-4xy\right)\)

\(=2x^2-6xy-4xy+8y-2x^2-6y-8xy\)

\(=2x^2-10xy+8y-2x^2-14xy\)

\(=10xy+8y-14xy\)

\(=-4xy+8y\)

\(=-4.\left(\frac{-2}{3}.\frac{3}{4}\right)+8.\frac{3}{4}\)

\(=-4.\frac{-1}{2}+6\)

\(=2+6=8\)

\(2x^2-6xy-4xy-8y-2x^2+6y+8xy\)

\(=-2y-2xy\)

thay \(x=\frac{-2}{3};y=\frac{3}{4}\) vào biểu thức ta có

\(-2.\frac{3}{4}-2.\frac{-2}{3}\frac{3}{4}=\frac{-3}{2}+1=\frac{-3+2}{2}=\frac{-1}{2}\)

nếu có sai bn thông cảm

\(2x\left(x-3y\right)-4y\left(x+2\right)-2\left(x^2-3y-4xy\right)\)

\(=2x^2-3y-4xy+8y-2x^2+3y+4xy\)

\(=-2y-2xy\)

Thay x,y ta có:

\(-2y-2xy=-2\left(\frac{3}{4}\right)-2\left(\frac{-2}{3}.\frac{3}{4}\right)\)

\(-2y-2xy=\frac{-3}{2}-2.\frac{-1}{2}\)

\(-2y-2xy=\frac{-3}{2}-\left(-1\right)\)

\(-2y-2xy=\frac{-3}{2}+1=\frac{-3}{2}+\frac{2}{2}=\frac{-1}{2}\)

Vậy biểu thức trên có giá trị bằng \(\frac{-1}{2}\)

khó quá bạn ơi !

2. Ta có: P = 2x² + y² - 4x - 4y + 10

P = 2(x² - 2x + 1) + (y² - 4y + 4) + 4

P = 2(x - 1)² + (y - 2)² + 4 \(\ge\)4 \(\forall\)x;y

=> P luôn dương với mọi biến x;y

3 Ta có:

(2n + 1)(n² - 3n - 1) - 2n³ + 1

= 2n³ - 6n² - 2n + n² - 3n - 1 - 2n³ + 1

= -5n² - 5n = -5n(n + 1) \(⋮\)5 \(\forall\)n \(\in\)Z

1×2=2

\(A=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\\ A=\dfrac{2\left(\dfrac{1}{2}-2\right)}{\dfrac{1}{2}+2}=\dfrac{2\left(-\dfrac{3}{2}\right)}{\dfrac{5}{2}}=\left(-3\right)\cdot\dfrac{2}{5}=-\dfrac{6}{5}\)

\(B=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}=\dfrac{-5}{-5+10}=\dfrac{-5}{5}=-1\)