ĐKXĐ : \(\hept{\begin{cases}2x+1\ne0\\3+x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-\frac{1}{2}\\x\ne-3\end{cases}}\)

Đáp án:

\(x\ne-\frac{1}{2}\)\(;x\ne3\)

HT

a) Ta có (n - 1)(n + 1) - (n - 7)(n - 5) 

= n² - 1 - (n² - 12n + 35)

= n² - 1 - n² + 12n - 35

= 12n - 36 = 12(n - 3) \(⋮12\forall n\inℤ\)

b) Ta có n(2n - 3) - 2n(n + 2) 

= 2n² - 3n - 2n² - 2n 

= - 5n \(⋮5\forall n\inℤ\)

Ta có: 3(2² + 1)(2⁴ + 1).....(2⁶⁴ + 1) + 1

= (2² - 1)(2² + 1)(2⁴ + 1)....(2⁶⁴ + 1) + 1

= (2⁴ - 1)(2⁴ + 1).....(2⁶⁴ + 1) + 1

= (2⁸ - 1)(2⁸ + 1).....(2⁶⁴ + 1) + 1

.....

= (2⁶⁴ - 1)(2⁶⁴ + 1) + 1

= 2¹²⁸ - 1 + 1 = 2¹²⁸

Trả lời:

a. A = 263² + 74 x 263 + 37²

= 263² + 2.263.37 + 37²

= ( 263 + 37 )² 

= 300²

= 90000

b. B = 136² - 92 x 136 + 46²

= 136² - 2.136.46 + 46²

= ( 136 - 46 )²

= 90²

= 8100

1. \(1001^2+1006^2+1011^2+1012^2-1003^2-1004^2-1009^2-1014^2=0\)

\(\Leftrightarrow\left(1001^2-1003^2\right)+\left(1006^2-1004^2\right)+\left(1011^2-1009^2\right)+\left(1012^2-1014^2\right)=0\)

Sử dụng HĐT số 3 a^2-b^2=(a+b)(a-b). Ta được:

\(\Leftrightarrow-2.2004+2.2010+2.2020-2.2026=0\)

\(\Leftrightarrow-2004+2010+2020-2026=0\)

\(\Leftrightarrow0=0\left(lđ\right)\left(đpcm\right)\)

2. \(-9x^2+12x-15< 0\)

\(\Leftrightarrow9x^2-12x+15>0\)

\(\Leftrightarrow\left(3x-2\right)^2+11>0\left(lđ\forall x\in R\right)\)

Trả lời:

g) G = ( 3x + 5 ).( 2x - 1 ) + ( 4x - 1 ).( 3x + 2 ) 

= 6x² - 3x + 10x - 5 + 12x² + 8x - 3x - 2 

= 18x² + 12x - 7

Ta có: | x | = 2 => x = 2 hoặc x = - 2

Thay x = 2 vào G, ta có:

G = 18.2² + 12.2. - 7 = 89

Thay x = - 2 vào G, ta có:

G = 18.(- 2 )² + 12.( - 2 ) - 7 = 41

h) H = ( 2x + y ).( 2z + y ) + ( x - y ).( y - z ) 

= 4xz + 2xy + 2yz + y² + xy - xz - y² + yz

= 3xz + 3xy + 3yz 

Ta có: z = | 1 | = 1

Thay x = 1; y = 1; z = 1 vào H, ta có:

H = 3.1.1 + 3.1.1 + 3.1.1 = 9

Trả lời:

A = ( - 5x + 4 ) ( 3x - 2 ) + ( - 2x + 3 ) ( x - 2 ) 

= - 15x² + 10x + 12x - 8 - 2x² + 4x + 3x - 6

= - 17x² + 29x - 14

Thay x = - 2 vào A, ta có:

A = - 17.( - 2 )² + 29. ( - 2 ) - 14 = - 140

A=(-5x+4)(3x-2)+(-2x+3)(x-2)

A=-15x²+10x+12x-8-2x²+4x+3x-6

A=-17x²+29x-14

Tại x=-2

=> A=-17.(-2)²+29.(-2)-14=-140

#H

(a²+3)(3-a²)

=9-a⁴

#H

a)\(\left(5x-1\right)^2-196=0\)

\(\Leftrightarrow\left(5x-1\right)^2=196\)

\(\Leftrightarrow5x-1=14\)

\(\Leftrightarrow x=3\)

b)\(4x^2+\frac{1}{4}=2x\)

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

\(\Leftrightarrow\left(2x-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow2x+\frac{1}{2}=0\)

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

c)\(x^2-12x=-36\)

\(\Leftrightarrow x^2-12x+36=0\)

\(\Leftrightarrow\left(x-6\right)^2=0\)

\(\Leftrightarrow x-6=0\)

\(\Leftrightarrow x=6\)

#H

a) (5x - 1)² - 196 = 0

<=> (5x - 1 - 14)(5x - 1 + 14) = 0

<=> (5x - 15)(5x + 13) = 0

<=> \(\orbr{\begin{cases}5x-15=0\\5x+13=0\end{cases}}\) <=> \(\orbr{\begin{cases}x=3\\x=-\frac{13}{5}\end{cases}}\)

Vậy S = {3; -13/5}

b) Ta có: 4x² + 1/4 = 2x

<=> 16x² - 8x + 1  = 0

<=> (4x - 1)² = 0

<=> 4x-  1 = 0

<=> x = 1/4

Vậy S = {1/4}

c) x² - 12x = -36

<=> x² - 12x + 36 = 0 

<=> (x - 6)² 0 

<=> x - 6 = 0

<=> x = 6

Vậy S = {6}

\(x^3+3x^2+3x-3=0\)

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

\(\Leftrightarrow x+1=\sqrt[3]{4}\)

\(\Leftrightarrow x=\sqrt[3]{4}-1\)

Vậy ...