a: =>7x=63

hay x=9

b: =>3x=-15

hay x=-5

d: =>-6x=-16

hay x=8/3

a) \(7x=63\Leftrightarrow x=9\)

b) \(3x=-15\Leftrightarrow x=-5\)

c) \(2x-5=0\Leftrightarrow2x=5\Leftrightarrow x=\dfrac{5}{2}\)

d) \(-6x=-16\Leftrightarrow x=\dfrac{8}{3}\)

Ta có: -2 – 7x > (3 + 2x) – (5 – 6x) ⇔ -2 – 7x > 3 + 2x – 5 + 6x

⇔ -7x – 2x – 6x > 3 – 5 + 2

⇔ -15x > 0 ⇔ x < 0

Vậy tập nghiệm của bất phương trình là: {x|x < 0}

a: =>(x-1)(x+1)(x-2)(x+2)=0

hay \(x\in\left\{1;-1;2;-2\right\}\)

b: \(\Leftrightarrow\sqrt{x}-6=0\)

hay x=36

c: =>(2x+1)(2x-1)=0

hay \(x\in\left\{-\dfrac{1}{2};\dfrac{1}{2}\right\}\)

a: =>10x-14=15-9x

=>19x=29

hay x=29/19

b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)

=>30x+9=36+32x+24

=>30x+9=32x+60

=>-2x=51

hay x=-51/2

c: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

=>35x-5+60x=96-6x

=>101x=101

hay x=1

d: \(\Leftrightarrow12\left(\dfrac{1}{2}-\dfrac{3}{2}x\right)=-5x+6\)

\(\Leftrightarrow6-18x+5x-6=0\)

=>-13x=0

hay x=0

\(a,\dfrac{5x-7}{3}=\dfrac{5-3x}{2}\\ \Leftrightarrow2\left(5x-7\right)=3\left(5-3x\right)\\ \Leftrightarrow10x-14=15-9x\\ \Leftrightarrow10x-14-15+9x=0\\ \Leftrightarrow19x-19=0\\ \Leftrightarrow x=1\)

\(b,\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\\ \Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\\ \Leftrightarrow30x+9=36+24+32x\\ \Leftrightarrow36+24+32x-30x-9=0\\ \Leftrightarrow2x+51=0\\ \Leftrightarrow x=-\dfrac{51}{2}\)

\(c,\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\\ \Leftrightarrow\dfrac{7x-1+12x}{6}=\dfrac{16-x}{5}\\ \Leftrightarrow5\left(19x-1\right)=6\left(16-x\right)\\ \Leftrightarrow95x-5=96-6x\\ \Leftrightarrow95x-5-96+6x=0\\ \Leftrightarrow101x-101=0\\ \Leftrightarrow x=1\)

\(d,4\left(0,5-1,5x\right)=-\dfrac{5x-6}{3}\\ \Leftrightarrow12\left(0,5-1,5x\right)=6-5x\\ \Leftrightarrow6-18x=6-5x\\ \Leftrightarrow6-5x-6+18x=0\\ \Leftrightarrow13x=0\\ \Leftrightarrow x=0\)

giup mình voi huhu

a) Đặt \(x^2=a\left(a\ge0\right)\)

Ta có: \(2x^4-7x^2+4=0\)

Suy ra: \(2a^2-7a+4=0\)

\(\Delta=49-4\cdot2\cdot4=49-32=17\)

Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}a_1=\dfrac{7-\sqrt{17}}{4}\left(nhận\right)\\a_2=\dfrac{-7+\sqrt{17}}{4}\left(loại\right)\end{matrix}\right.\)

Suy ra: \(x^2=\dfrac{7-\sqrt{17}}{4}\)

\(\Leftrightarrow x=\pm\dfrac{\sqrt{7-\sqrt{17}}}{2}\)

Vậy: \(S=\left\{\dfrac{\sqrt{7-\sqrt{17}}}{2};-\dfrac{\sqrt{7-\sqrt{17}}}{2}\right\}\) 

câu f là 9+3x hay 9-3x vậy???

a) 

$2x+6=0$

$2x=-6$

$x=-3$

b) $4x+20=0$

$4x=-20$

$x=-5$

c) 

$2(x-1)=5x-7$

$2x-2=5x-7$

$3x=5$

$x=\frac{5}{3}$

d) $2x-3=0$

$2x=3$

$x=\frac{3}{2}$

e) 

$3x-1=x+3$

$2x=4$

$x=2$

f) 

$15-7x=9-3x$

$6=4x$

$x=\frac{3}{2}$

g) $x-3=18$

$x=18+3=21$

h) 

$2x+1=15-5x$

$7x=14$

$x=2$