a) (-2)+ (-5) = -7

Vì: -7< -5

=> (-2)+ (-5) < -7

b) (-3)+ (-8)= -11

Vì: (-10) > (-11)

=> -10> (-3)+ (-8)

a)(-2)+(-5)<(-5)

b)(-10)>(-3)+(-8)

a) \(\left(-2\right)+\left(-5\right)..........\left(-5\right)\)

\(\left(-7\right)< \left(-5\right)\)

Vậy \(\left(-2\right)+\left(-5\right)< \left(-5\right)\)

b) \(\left(-10\right)...........\left(-3\right)+\left(-8\right)\)

\(\left(-10\right)>\left(-11\right)\)

Vậy \(\left(-10\right)>\left(-3\right)+\left(-8\right)\)

a) (-5) . (-4) + (-5) . 14 = (-5 ) . [ (-4) + 14 ] = -50

b ) 3 . ( 5 + 8 ) = 13 . ( -3 ) + 13 . 8 = 65

a, \(\left(-6\right)+\left(-3\right).....\left(-6\right)\)

\(\left(-9\right)< \left(-6\right)\)

Vậy \(\left(-6\right)+\left(-3\right)< \left(-6\right)\)

b,\(\left(-9\right)+\left(-12\right)....\left(-20\right)\)

\(\left(-21\right)< \left(-20\right)\)

\(\Rightarrow\left(-9\right)+\left(-12\right)< \left(-20\right)\)

a)

b)

a)

a, Ta thấy : \(\left\{{}\begin{matrix}\left(2a+1\right)^2\ge0\\\left(b+3\right)^2\ge0\\\left(5c-6\right)^2\ge0\end{matrix}\right.\)\(\forall a,b,c\in R\)

\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\ge0\forall a,b,c\in R\)

Mà \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\le0\)

Nên trường hợp chỉ xảy ra là : \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2=0\)

- Dấu " = " xảy ra \(\left\{{}\begin{matrix}2a+1=0\\b+3=0\\5c-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=-3\\c=\dfrac{6}{5}\end{matrix}\right.\)

Vậy ...

b,c,d tương tự câu a nha chỉ cần thay số vào là ra ;-;

\(\left|4\right|< \left|7\right|\) \(\left|-2\right|>\left|-5\right|\) \(\left|-3\right|>\left|0\right|\)

\(\left|6\right|=\left|-6\right|\)

Nhớ ủng hộ 1 Đúng !

1, <

2, =

3, <

4, >

a) \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(5x+5\right)^2\)

\(A=\left[\left(3x+1\right)-\left(5x+5\right)\right]^2\)

\(A=\left(-2x-4\right)^2\)

A = (3x + 1)² - 2(3x + 1)(5x + 5) + (5x + 5)²

= [(3x + 1)-(5x + 5)]²

= (3x + 1 - 5x - 5)²

= [(-2x) - 4]²

B = (3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ +1)(3³² + 1)

=> (3 - 1)B = (3 - 1)(3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ +1)(3³² + 1)

=>2B = (3² - 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ +1)(3³² + 1)

= (3⁴ - 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ +1)(3³² + 1)

= (3⁸ - 1)(3⁸ + 1)(3¹⁶ +1)(3³² + 1)

= (3¹⁶ - 1)3¹⁶ +1)(3³² + 1)

= (3³² - 1)(3³² + 1)

= 3⁶⁴ - 1

vì 2B = 3⁶⁴ - 1

=> B = \(\dfrac{3^{64}-1}{2}\)

C = a² + b² + c² + 2ab - 2ac - 2bc + a² + b² + c² - 2ab + 2ac - 2bc - 2( b² - 2bc + c²⁾

= 2a² + 2b² + 2c² - 4bc - 2b² + 4bc - 2c²

= 2a²

a/ \(x=\dfrac{-5}{12}\)

b/ \(x\approx-1,9526\)

c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)

d/ \(x=\dfrac{-20}{13}\)

a) (x-2)³+6(x+1)²-x³+12=0

⇒ x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

⇒ x³-6x²+12x-8+6x²+12x+6-x³+12=0

⇒ 24x+10=0

⇒ 24x=-10

⇒ x=-5/12

a.

PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)

b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.

c.

PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$

$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý) 

Do đó pt vô nghiệm.

d.

PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$

$\Leftrightarrow 11x+3=-54x-97$

$\Leftrightarrow 65x=-100$

$\Leftrightarrow x=\frac{-20}{13}$

Điền đúng sai vào chỗ trống (....)

a) (−36):2=−18 Đ

b) 600:(−15)=−4 S

c) 27:(−1)=27 S

d) (−65):(−5)=13 Đ