a) -3x = 15

x = 15 : -3

x = -5

b) 6x + 1 = -23

6x = -23 + 1 = -22

( Đề bài sai hay tôi làm sai nhể :v )

x=15: -3

x= -5

6x= -23-1

6x=-24

x=-24:6

x=-4

a) x= -5
b) x= -4

Lời giải:
a.

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=\frac{x-y}{2-\frac{3}{2}}=\frac{15}{\frac{1}{2}}=30\)

\(\Rightarrow \left\{\begin{matrix} x=60\\ y=45\\ z=40\end{matrix}\right.\)

b)

Từ đkđb suy ra \(\frac{10x}{1}=\frac{5y}{\frac{1}{3}}=\frac{z}{\frac{1}{6}}=\frac{10x-5y+z}{1-\frac{1}{3}+\frac{1}{6}}=\frac{25}{\frac{5}{6}}=30\)

\(\Rightarrow \left\{\begin{matrix} x=3\\ y=2\\ z=5\end{matrix}\right.\)

a. \(3\sqrt{x}=15\)

<=> \(\sqrt{x}=\dfrac{15}{3}\)

<=> \(\sqrt{x}=5\)

<=> x = \(25\)

b. \(-2\sqrt{x}=-10\)

<=> \(\sqrt{x}=\dfrac{-10}{-2}\)

<=> \(\sqrt{x}=5\)

<=> \(x=25\)

c. \(\sqrt{x}>6\)

<=> \(\left(\sqrt{x}\right)^2>6^2\)

<=> x > 36

d. \(\sqrt{x}< 5\)

<=> \(\left(\sqrt{x}\right)^2=5^2\)

<=> x < 25

a)\(3\sqrt{x}=15\)⇒\(\sqrt{x}=5\)⇒x=25

b)\(-2\sqrt{x}=-10\)⇒\(\sqrt{x}=5\)⇒x=25

c)\(\sqrt{x}>6\)⇒x>36

d)\(\sqrt{x}< 5\)⇒x<25

a: =>y/15=-2/3

hay y=-10

b: 2/x=x/18

nên \(x^2=36\)

hay \(x\in\left\{6;-6\right\}\)

c: x/9=16/x

nên \(x^2=144\)

hay \(x\in\left\{12;-12\right\}\)

a)x=-10
b)x=-6

c)x=-12 nhé

`a)4x(x+1)+(3-2x)(3+2x)=15`

`<=>4x^2+4x+9-4x^2=15`

`<=>4x=6`

`<=>x=3/2`

Vậy `S={3/2}`

`b)3x(x-20012)-x+20012=0`

`<=>3x(x-20012)-(x-20012)=0`

`<=>(x-20012)(3x-1)=0`

`<=>` $\left[\begin{matrix} x=20012\\ x=\dfrac{1}{3}\end{matrix}\right.$

Vậy `S={1/3;20012}`

a) 4x(x + 1) + (3 – 2x)(3 + 2x) = 15

⇔4x² + 4x + (9 – 4x²) = 15

⇔ 4x² + 4x + 9 – 4x² = 15

⇔4x = 15 – 9

⇔x=1,5

b)3x(x – 2001²) – x + 2001² = 0

⇔3x(x – 2001²) – (x – 2001²) = 0

⇔(x – 2001²)(3x – 1) = 0

⇔x – 2001² = 0 hay 3x – 1 = 0

⇔x = 2001² hoặc x = \(\dfrac{1}{2}\)

a: x=67

b: x=44

c: x=-2 hoặc x=5

a: x=67

b: x=44

c: x=-2 hoặc x=5

Lời giải:
a. 

$(-2)x-(-21)=15$

$-2x+21=15$

$-2x=15-21=-6$

$x=(-6):(-2)=3$

b.

$(3x-2^2).7^3=7^4$
$3x-2^2=7^4:7^3=7$

$3x-4=7$

$3x=11$

$x=\frac{11}{3}$

\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)

Bài 2:

a) \(=x^2-4-x^2-2x-1=-2x-5\)

b) \(=8x^3-1-8x^3-1=-2\)

Bài 3:

a) \(\Rightarrow x^3+8-x^3+2x=15\)

\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)

b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)

\(\Rightarrow7x=14\Rightarrow x=2\)

a) (x-2)^x-3(x+1)(x-1)+6x^2=5

<=> \(x^2-4x+4-3(x^2-1)+6x^2-5=0\)

<=>\(x^2-4x+4-3x^2+3+6x^2-5=0\)

<=>\(4x^2-4x+2=0\)

<=> \(4x^2-4x+1+1=0\)

<=>\((2x-1)^2+1=0\)

\(ta\) có \((2x-1)^2 > hoặc = 0\)

             1>0

=> \((2x-1)^2+1=0 (vô lí)\)

=> phuơng trình vô nghiêm S={ rỗng }

a: \(3\left(x-3\right)-6x=0\)

=>\(3x-9-6x=0\)

=>-3x-9=0

=>3x+9=0

=>3x=-9

=>\(x=-\dfrac{9}{3}=-3\)

b: Đề thiếu vế phải rồi bạn

c: \(2\left(x-3\right)+3x=9\)

=>2x-6+3x=9

=>5x-6=9

=>5x=6+9=15

=>x=15/5=3

d: \(x\left(x-11\right)+2\left(x-11\right)=0\)

=>\(\left(x-11\right)\left(x+2\right)=0\)

=>\(\left[{}\begin{matrix}x-11=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-2\end{matrix}\right.\)

e: \(x\left(x+2\right)+8=x^2\)

=>\(x^2+2x+8=x^2\)

=>2x+8=0

=>2x=-8

=>x=-8/2=-4

f: \(8\left(x+1\right)+2x=-2\)

=>\(8x+8+2x=-2\)

=>10x=-2-8=-10

=>\(x=-\dfrac{10}{10}=-1\)

g: 12-3(x+2)=0

=>3(x+2)=12

=>x+2=12/3=4

=>x=4-2=2