ĐKXĐ: $x\notin\{-5;-4;-3\}$

$\frac{1}{x+5}+\frac{2}{x+4}+\frac{3}{x+3}+3=0$

$\Leftrightarrow \left(\frac{1}{x+3}+1\right)+\left(\frac{2}{x+4}+1\right)+\left(\frac{3}{x+3}+1\right)=0$

$\Leftrightarrow \frac{x+6}{x+5}+\frac{x+6}{x+4}+\frac{x+6}{x+3}=0$

$\Leftrightarrow (x+6)\left(\frac{1}{x+5}+\frac{1}{x+4}+\frac{1}{x+3}\right)=0$

+, TH1: $x+6=0\Leftrightarrow x=-6$ (tm ĐKXĐ)

+, TH2: $\frac{1}{x+5}+\frac{1}{x+4}+\frac{1}{x+3}=0$

$\Leftrightarrow \frac{(x+4)(x+3)+(x+3)(x+5)+(x+4)(x+5)}{(x+3)(x+4)(x+5)}=0$

$\Rightarrow 3x^2+24x+47=0$

$\Leftrightarrow \left[\begin{array}{}x =\frac{-12+\sqrt3}{3}(TM) \\x=\frac{-12-\sqrt3}{3}(TM) \end{array}\right. $

$\text{#}Toru$

ĐKXĐ: \(x\notin\left\{-5;-4;-3\right\}\)

\(\dfrac{1}{x+5}+\dfrac{2}{x+4}+\dfrac{3}{x+3}+3=0\)

=>\(\left(\dfrac{1}{x+5}+1\right)+\left(\dfrac{2}{x+4}+1\right)+\left(\dfrac{3}{x+3}+1\right)=0\)

=>\(\dfrac{x+6}{x+5}+\dfrac{x+6}{x+4}+\dfrac{x+6}{x+3}=0\)

=>\(\left(x+6\right)\left(\dfrac{1}{x+5}+\dfrac{1}{x+4}+\dfrac{1}{x+3}\right)=0\)

=>x+6=0

=>x=-6(nhận)

12 - 4 = 8

12 - (-4) = 12 + 4 = 16

-12 - (-4) = -12 + 4 = -8

12 + (-4) = 12 - 4 = 8

15 - (-5) = 15 + 5 = 20

12 - 4 = 8

12 - ( -4) = 12 + 4 = 16

-12 - ( -4) = -12 + 4 = - ( 12 - 4 ) = -8

12 + ( -4) = 12 - 4 =8

15 - ( -5) = 15 + 5 =20

90TV

3².5 - 2².7 + 83

= 9.5 - 4.7 + 83

= 45 - 28 + 83

= 17 + 83

= 100

3².5 - 2².7 + 83

= 9.5 - 4.7 + 83

= 45 - 28 + 83

= 17 + 83

= 100

9+2 = 11

12 + 3 = 15

10 - 1 = 9

21 - 11 = 10

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

9+2=11

12+3=15

10-1=9

21-11=10

(-2)-(-3)=(0-2)+3=-2+3=1

a: \(-\dfrac{9}{4}=-2,25\)

mà -2,25<-2,12

nên \(-\dfrac{9}{4}< -2,12\)

b: \(-1\dfrac{1}{5}=-1,2\)

mà -1,2>-1,75

nên \(-1\dfrac{1}{5}>-1,75\)

c:

DẠ TOÁN LỚP 7 Ạ, EM GHI NHẦM LỚP 6

Bài 1

a) 3,6 : 1,2 . 3 + 5 . 10 : 2 - 4

= 3 . 3 + 50 : 2 - 4

= 9 + 25 - 4

= 34 - 4

= 30

b)

\(\left(\dfrac{1}{2}x+2\right)\left(x^2+\dfrac{1}{4}\right)=0\)

mà \(x^2+\dfrac{1}{4}>=\dfrac{1}{4}>0\forall x\)

nên \(\dfrac{1}{2}x+2=0\)

=>\(\dfrac{x}{2}=-2\)

=>x=-4

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