e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

f: Ta có: \(2x^2+7x+3=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

1) ĐKXĐ: \(16x^2-25\ge0\)

\(\Leftrightarrow x^2\ge\dfrac{25}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{4}\\x\le-\dfrac{5}{4}\end{matrix}\right.\)

2) ĐKXĐ: \(4x^2-49\ge0\Leftrightarrow x^2\ge\dfrac{49}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{7}{2}\\x\le-\dfrac{7}{2}\end{matrix}\right.\)

3) ĐKXĐ: \(8-x^2\ge0\Leftrightarrow x^2\le8\)

\(\Leftrightarrow-2\sqrt{2}\le x\le2\sqrt{2}\)

4) ĐKXĐ: \(x^2-12\ge0\Leftrightarrow x^2\ge12\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge2\sqrt{3}\\x\le-2\sqrt{3}\end{matrix}\right.\)

5) ĐKXĐ: \(x^2+4\ge0\left(đúng\forall x\right)\)

a) \(3^x-27.3^5.3^2=0\)

\(3^x-3^3.3^5.3^3=0\)

\(3^x=3^{13}\)

\(x=13\)

b) \(\left(-5+3x\right):4=16\)

\(-5+3x=64\)

\(3x=69\)

\(x=23\)

a) -5 . (2 - x) + 4(x - 3) = 10x - 15

-10 + 5x + 4x -12 = 10x - 15

5x + 4x - 10x = -15 + 10 + 12

-x = 7

x = -7

b) 5 . (3 - 2x) + 5 . (x - 4) = 6 - 4x

15 - 10x + 5x - 20 = 6 - 4x

-10x + 5x + 4x = 6 - 15 + 20

-x = 11

x = -11

c) - 7 . (3x - 5) + 2 . (7x - 14) = 28

-21x + 35 + 14x - 28 = 28

-21x + 14x = 28 - 35 + 28

-7x = 21

x = 21 : (-7)

x = -3

d) 4 . (x - 5) - 3 . (x + 7) = 5 . (-4)

4x - 20 - 3x - 21 = -20

4x - 3x = -20 + 20 + 21

x = 21

e) 5 . (4 - x) - 7. (-x + 2) = 4 - 9 + 3

20 - 5x + 7x - 14 = -2

-5x + 7x = -2 - 20 + 14

2x = -8

x = -8 : 2

x = -4

Đúng 100%

câu c

- 7 ( 3x - 5 ) + 2 ( 7x - 14 ) = 28

- 21x + 35 + 14x - 28 = 28

 21x - 14x = 35 - 28 - 28

7x = - 21

x = ( - 21) : 7

x = - 3 

a, -5 .(2- x) +4 (x-3) =10x -15
<=> -19 +5x +4x-12 -10x +15=0
<=> -x=16
<=>x=16
b, 5(3-2x) +5.(x-4)=6-4x
<=> 15-10x +5x-20 -6+4x=0
<=> -x=-11
<=> x=11
c, -7.(3x-5) +7.(7x -14)=28
<=> -21x +35+49x -98-28=0
<=>28x=-91
<=>x= -\(\frac{13}{4}\)
d, 4.(x-5) -3.(x+7)=5.(-4)
<=>4x-20-3x-21=-20
<=>x=21
e,5.(4-x)-7.(-x+2)=4-9+3
<=>20x -5x +7x -14= -2
<=>22x=12
<=>x=\(\frac{6}{11}\)
P tự kết luận nhé k hiểu hay mk sai chỗ nào p pảo mk nhé

a,(-12).x= 60-12

(-12).x=48

x=48:(-12)

x = -4

b,(-5).x + 5 = -24+6

(-5).x + 5 = 30

(-5).x      = 30-5

(-5).x      = 25

x = 25 : (-5)

x = -5

c,

$3x+36=-7x-64$

$10x=-100$

$x=-10$

$b,-5x-178=14x+145$

$-19x=323$

$x=-17$

Ta có: \(H=\left(\sqrt{4x^2-12x+9}+\sqrt{4x^2+4x+1}\right)\)

     \(\Leftrightarrow H=\left(\sqrt{\left(2x-3\right)^2}+\sqrt{\left(2x+1\right)^2}\right)\)

    \(\Leftrightarrow H=\left|2x-3\right|+\left|2x+1\right|\)

Xét tính chất về trị tuyệt đối sau: \(\left|a\right|+\left|b\right|\ge ab\) với \(ab\ge0\)

Ta viết lại \(H=\left|3-2x\right|+\left|2x+1\right|\ge\left|\left(3-2x\right)+\left(2x+1\right)\right|=4\) khi \(\left(3-2x\right)\left(2x+1\right)\ge0\)

\(\Rightarrow H\ge4\)khi \(3-2x\ge0\)và \(2x+1\ge0\) hoặc \(3-2x\le0\) và \(2x+1\le0\)

\(\Leftrightarrow x\le\frac{3}{2}\) và \(x\ge\frac{-1}{2}\)hoặc \(x\ge\frac{3}{2}\)và \(x\le\frac{-1}{2}\)(vô lý)

Vậy \(GTNN\left(H\right)=4\) khi \(\frac{-1}{2}\le x\le\frac{3}{2}\)

Mình có giải thích hơi dài nha cậu tick mình nha

a: Ta có: \(7x+25=144\)

\(\Leftrightarrow7x=119\)

hay x=17

b: Ta có: \(33-12x=9\)

\(\Leftrightarrow12x=24\)

hay x=2

c: Ta có: \(128-3\left(x+4\right)=23\)

\(\Leftrightarrow3\left(x+4\right)=105\)

\(\Leftrightarrow x+4=35\)

hay x=31

d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)

\(\Leftrightarrow5\left(726-3x\right)=2175\)

\(\Leftrightarrow726-3x=435\)

\(\Leftrightarrow3x=291\)

hay x=97

e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)

\(\Leftrightarrow41-\left(2x+5\right)=18\)

\(\Leftrightarrow2x+5=23\)

\(\Leftrightarrow2x=18\)

hay x=9

Bn cần bài nào vậy

f: Ta có: \(10-4x+120:8=16+1\)

\(\Leftrightarrow4x=17-25=-8\)

hay x=-2

g: Ta có: \(x+9x+7x+5x=2244\)

\(\Leftrightarrow22x=2244\)

hay x=102

h: Ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)

\(\Leftrightarrow100x+5050=5750\)

\(\Leftrightarrow100x=700\)

hay x=7

A.

^{_{\(\Leftrightarrow\)}} 9x - 2x - 6 = 3x + 1

\(\Leftrightarrow\) 4x = 7

\(\Leftrightarrow\) x = \(\dfrac{7}{4}\)

B.

\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-13}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\) 5x + 15 - 4x +12 = x - 13

\(\Leftrightarrow\) 0x = -40 ( phương trình vô nghiệm)

C.

\(\Leftrightarrow\) 7x + 8 \(\ge\) 3x -3

\(\Leftrightarrow\) 4x \(\ge\) - 11

\(\Leftrightarrow\)\(x\ge\dfrac{-11}{4}\)