a. x × 3 = 27

x = 27 : 3

x = 9

b. 4 × x = 20

x = 20 : 4

x = 5

c. 10 + x : 2 = 20

x : 2 = 20 – 10

x : 2 = 10

x = 10 × 2

x = 20

d. x × 3 = 27 + 3

x × 3 = 30

x = 30 : 3

x = 10

e. 27 : x = 789 – 780

27 : x = 9

x = 27 : 9

x = 3

Phương pháp giải:

- Muốn tìm một số hạng ta lấy tổng trừ đi số hạng kia.

- Muốn tìm một thừa số ta lấy tích chia cho thừa số kia.

Lời giải chi tiết:

a)

● x + 2 = 8

    x = 8 − 2

    x = 6

● x × 2 = 8

    x = 8 : 2

    x = 4

b)

● x + 3 = 12

    x = 12 − 3

    x = 9

● x × 3 = 12

    x = 12 : 3

    x = 4

c)

● 3 + x = 27

    x = 27 − 3

    x = 24

● 3 × x = 27

    x = 27 : 3

    x = 9

a) x + 2 = 8

x = 8 -2 

x = 6

 x × 2 = 8

x = 8 :2

x = 4

b) x + 3 = 12

x = 12 - 3

x = 9

x × 3 = 12

x = 12 : 3

x = 4

c) 3 + x = 27

x = 27 - 3

x = 24

3 × x = 27

x = 27 :3

x = 9

2(x-5)+3=-7

=>2x-10=-7+3=-4

=>2x=-4+10=6

=>x=6:2=3

4x+(27+2^3)=1

=>4x+(27+8)=1

=>4x+35=1

=>4x=1-35=-34

=>x=-34:4=8.5

lx-3l+5=-27

=>lx-3l=-27-5=-32

mà lx-3l>0 với mọi x

=>k co gt x t/m

-3lx-1l=-27

=>lx-1l=-27:(-3)=9

=>x-1=+9

+)x-1=9=>x=10

+)x-1=-9=>x=-8

a) 3^x=27

3^x=3³

x=3

b.2^x:2^3=1

2^x:8=1

2^x=1x8

2^x=8

2^x=2³

x=3

c.27<3^x<243

3³=27(loại)

3⁴=81 (lấy)

3⁵=243(loại)

vậy x=4

d.4^x-^3=36

   (x + 3)(x² - 3x + 9) - x(x - 2)² = 27

\(\Leftrightarrow\) x³ + 27 - x( x² - 4x + 4) = 27

\(\Leftrightarrow\) x³ + 27 - x³ + 4x² - 4x - 27 = 0

\(\Leftrightarrow\) 4x² - 4x = 0

\(\Leftrightarrow\) 4x ( x - 1) = 0

khi 4x = 0 hoặc x - 1 = 0

\(\Leftrightarrow\)  x = 0        \(\Leftrightarrow\) x = 1

 Chúc bạn học tốt

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)^2=27\\ x.x^2-x.3x+x.9+3.x^2-3.3x+3.9-x.x^2+x.2^2=27\\ x^3-3x^2+9x+3x^2-9x+27-x^3+4x=27\\ 4x+27=27\\ 4x=0\\ x=0\)

x³+27+(x+3).(x-9)=0 nha mk ghi nhầm 

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)

a: Ta có: \(4x^2\left(x-2\right)-x+2=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x^2-1\right)=0\)

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

b: Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=2\end{matrix}\right.\)

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-9\right)=27\\ x.x^2-x.3x+x.9-x.x^2+x.9=27\\ x^3-3x^2+9x-x^3+9x=27\\ 3x^2+18x=27\\ 21x^2=27\\ x^2=\dfrac{9}{7}\\ \Rightarrow x=\sqrt{\dfrac{9}{7}}\)

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-9\right)=27\\ x.x^2-x.3x+x.9+3.x^2-3.3x+3.9-x.x^2+x.9=27\\ x^3-3x^2+9x+3x^2-9x+27-x^3+9x=27\\ 9x+27=27\\ 9x=0\\ x=0\)

\(a,\Leftrightarrow\left(x-2\right)^3-3x\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-2-3x\right)=0\\ \Leftrightarrow\left(x-2\right)\left(-2x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\\left(x-1\right)^2+5=0\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=-3\)

\(x^3+27=-x^2+9\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-2x+6\right)=0\)

hay x=-3