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

= x^3 + 3x^2y + 3xy^2 + y^3 - x^3 + 3x^2y - 3xy^2 + y^3

= 6x^2y + 2y^3

= 2y( 3x^2 + y^2)

=> ĐPCM

Đáp án:

$P=\pm 36$

Giải thích các bước giải:

Ta có:

$\begin{array}{c}{x}^2+y^2+z^2=16\\ xy-yz+zx=-10\\ \Rightarrow \left(x^2+y^2+z^2\right)-2.\left(xy-yz+zx\right)=16-2.\left(-10\right)\\ \iff x^2+y^2+z^2-2xy+2yz-2zx=36\\ \iff \left(x^2-2xy+y^2\right)+z^2+2yz-2zx=36\\ \iff {\left(x-y\right)}^2+2z\left(y-x\right)+z^2=36\\ \iff {\left(x-y\right)}^2-2.\left(x-y\right).z+z^2=36\\ \iff {\left(x-y-z\right)}^2=36\\ \iff x-y-z=\pm 6\\ P=x^3-y^3-z^3-3xyz\\ =\left(x^3-3x^{2}y+3x{y}^2-y^3\right)-z^3+3x^{2}y-3x{y}^2-3xyz\\ ={\left(x-y\right)}^3-z^3+3x^{2}y-3x{y}^2-3xyz\\ =\left[\left(x-y\right)-z\right].\left[{\left(x-y\right)}^2+\left(x-y\right).z+z^2\right]+3xy\left(x-y-z\right)\\ =\left(x-y-z\right).\left(x^2-2xy+y^2+xz-yz+z^2+3xy\right)\\ =\left(x-y-z\right).\left(x^2+y^2+z^2+xy-yz+zx\right)\\ \text{Trường hợp 1:}x-y-z=6\\ \Rightarrow P=6.\left(16+\left(-10\right)\right)=36\\ \text{Trường hợp 2:}x-y-z=-6\\ \Rightarrow P=\left(-6\right).\left(16+\left(-10\right)\right)=-36\end{array}$

Vậy $P=\pm 36$.

MÌNH CHỈ BIẾT LÀM B7 THÔI NHA

P= 811^3+ 812^3+815^3+3.811.812.(-815)=  31694

K ĐÚNG HỘ TỚ NHA

( x + 2 )( x² - x + 1 ) - ( x³ + 2x² )

= x( x² - x + 1 ) + 2( x² - x + 1 ) - x³ - 2x²

= x³ - x² + x + 2x² - 2x + 2 - x³ - 2x²

= -x² - x + 2 = ( 1 - x )( x + 2 )

Phương trình hóa học phản ứng 

CaCl₂ + Na₂CO₃ ---> CaCO₃ + 2NaCl

1        :      1           :         1       :      2  (1)

Ta có \(n_{CaCl_2}=\frac{m}{M}=\frac{11,1}{111}=0,1\)(mol) (2) 

Vì phản ứng xảy ra hoàn toàn , kết hơp (1) và (2) ta được

\(n_{Na_2CO_3}=0,1\left(mol\right)\)

=> \(m_{Na_2CO_3}=n.M=0,1.106=10,6\left(g\right)\)

Vậy giá trị của M là 10,6 => Chọn A

A. 10,6 nha

......

.......

Ta có C = x² - 4x + y² - y + 5 

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

= \(\left(x-2\right)^2+\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

=> Min C = 3/4

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2=0\\y-\frac{1}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}\)

Vậy Min C = 3/4 <=> x = 2 ; y = 1/2 

C = ( x² - 4x + 4 ) + ( y² - y + 1/4 ) + 3/4

= ( x - 2 )² + ( y - 1/2 )² + 3/4 ≥ 3/4 ∀ x.y 

Dấu "=" xảy ra <=> x = 2 ; y = 1/2 . Vậy MinC = 3/4

a) A = (2x + 6)(4x² − 12x + 36) − 8x³ + 10.

=8x³+216-8x³+10

=226

b) B = (2x − 1)(4x² + 2x + 1) − 8(x³ + 1). 

=8x³-1-8x³-8

=-9

c) C = (2 + a)(2 − a)(4 + 2a + a² )(a² − 2a + 4). 

=[(2+a)(a² − 2a + 4)] [((2 − a)(4 + 2a + a² )]

=[(a+2)(a² − 2a + 4)] [((2 − a)(4 + 2a + a² )]

=(a³+8)(8-a³)

=8a³-a⁶+64-8a³

=-a⁶+64

=64-a⁶

=(8-a³)(8+a³)

d) D = (a³ b³ − 1)(a³ b³ + 1) − a³ b³ .

=a⁶b⁶-1-a³b³

a, ( 2x - 3 )²- (2x + 1)² = -3

4x²-12x+9-4x²+4x-1=-3

-8x-1=-3

-8x=-2

x=\(\frac{1}{4}\)

b, (5x - 1) ² - (5x + 4)(5x - 4) = 7

25x²-10x+1-25x²+16=7

-10x+17=7

-10x=-10

x=1

c, ( x- 5)² + (x-3)(x+3) - 2(x + 1)²=0

x²-10x+25+x²-9-2x²-4x-2=0

-14x+14=0

-14(x-1)=0

=>x-1=0

x=1

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

\(\Leftrightarrow4x^2-12x+9-4x^2-4x-1=-3\)

\(\Leftrightarrow-16x+8=-3\)

\(\Leftrightarrow-16x=-11\)

\(\Leftrightarrow x=\frac{11}{16}\)

b)\(\left(5x-1\right)^2-\left(5x+4\right)\left(5x-4\right)=7\)

\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)

\(\Leftrightarrow-10x+17=7\)

\(\Leftrightarrow-10x=-10\)

\(\Leftrightarrow x=1\)

c)\(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+1\right)^2=0\)

\(\Leftrightarrow x^2-10x+25+x^2-9-2\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow2x^2-10x-16-2x^2-4x-2=0\)

\(\Leftrightarrow-14x-18=0\)

\(\Leftrightarrow-14x=18\)

\(\Leftrightarrow x=-\frac{9}{7}\)

#H

Ta có (x - y)² - (x + y)² 

= (x - y + x + y)(x - y - x - y) 

= 2x.(-2y)

= -4xy (đpcm) 

VT=(x-y)²-(x+y)²

=x²-2xy+y²-x²-2xy-y²

=-4xy=VP (đccm)

#H

a) B = x - x² + 2

= \(-\left(x^2-x+\frac{1}{4}-\frac{1}{4}-2\right)=-\left(x-\frac{1}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\)

=> Max B = 9/4 

Dấu "=" xảy ra <=> x - 1/2 = 0 <=> x = 1/2

Vậy Max B = 9/4 <=> x = 1/2

d) Ta có P = \(x-x^2-1=-\left(x^2-x+\frac{1}{4}-\frac{1}{4}+1\right)=-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}\le-\frac{3}{4}\)

=> Max P = -3/4 

Dấu "=" xảy ra <=> x -1/2 = 0 <=> x = 1/2

Vậy Max P = -3/4 <=> x = 1/2 

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