a: x^3=7^3

=>x^3=343

=>\(x=\sqrt[3]{343}=7\)

b: x^3=27

=>x^3=3^3

=>x=3

c: x^3=125

=>x^3=5^3

=>x=5

d: (x+1)^3=125

=>x+1=5

=>x=4

e: (x-2)^3=2^3

=>x-2=2

=>x=4

f: (x-2)^3=8

=>x-2=2

=>x=4

h: (x+2)^2=64

=>x+2=8 hoặc x+2=-8

=>x=6 hoặc x=-10

j: =>x-3=2 hoặc x-3=-2

=>x=1 hoặc x=5

k:

9x^2=36

=>x^2=36/9

=>x^2=4

=>x=2 hoặc x=-2

l:

(x-1)^4=16

=>(x-1)^2=4(nhận) hoặc (x-1)^2=-4(loại)

=>x-1=2 hoặc x-1=-2

=>x=3 hoặc x=-1

a) \(x^2-xz-9y^2+3yz\)

\(=\left(x^2-9y^2\right)-\left(xz-3yz\right)\)

\(=\left[x^2-\left(3y\right)^2\right]-z\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x+3y-z\right)\)

b) \(x^3-x^2-5x+125\)

\(=\left(x^3+125\right)-\left(x^2+5x\right)\)

\(=\left(x^3+5^3\right)-x\left(x+5\right)\)

\(=\left(x+5\right)\left(x^2-5x+5^2\right)-x\left(x+5\right)\)

\(=\left(x+5\right)\left(x^2-5x+5^2-x\right)\)

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

c) \(x^3+2x^2-6x-27\)

\(=\left(x^3-27\right)-\left(2x^2-6x\right)\)

\(=\left(x^3-3^3\right)-2x\left(x-3\right)\)

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

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

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

e) \(4x^4+4x^3-x^2-x\)

\(=4x^3\left(x+1\right)-x\left(x+1\right)\)

\(=\left(x+1\right)\left(4x^3-x\right)\)

f) \(x^6-x^4-9x^3+9x^2\)

\(=x^4\left(x^2-1\right)-9x^2\left(x-1\right)\)

\(=x^4\left(x-1\right)\left(x+1\right)-9x^2\left(x-1\right)\)

\(=\left(x-1\right)\left[x^4\left(x+1\right)-9x^2\right]\)

\(=\left(x-1\right)\left(x^5+x^4-9x^2\right)\)

a: \(27-27x+9x^2-x^3=\left(3-x\right)^3\)

b: \(125-x^3=\left(5-x\right)\left(25+5x+x^2\right)\)

1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)

2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)

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

5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)

7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)

8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

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

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)

b) \(x^3-15x^2+75x-125\)

= \(\left(x-5\right)^3\)

Thay x = 35 vào ta đc:

\(\left(35-5\right)^3\) = 27000

c) \(x^3+12x^2+48x+65\)

= \(x^3+5x^2+7x^2+13x+65\)

= \(x^2\left(x+5\right)+7x\left(x+5\right)+13\left(x+5\right)\)

= \(\left(x+5\right)\left(x^2+7x+13\right)\)

Thay x = 6 vào ta đc:

\(\left(6+5\right)\left(6^2+7.6+13\right)\)=1001

2)

Ta có: a+b+c=0

=> \(\left(a+b+c\right)^3=0\)

=> \(a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3a^2c+3ac^2+6abc=0\)=> \(a^3+b^3+c^3+\left(3a^2b+3ab^2+3abc\right)+\left(3b^2c+3bc^2+3abc\right)+\left(3a^2c+3ac^2+3abc\right)-3abc=0\)

=> \(a^3+b^3+c^3+3b\left(a+b+c\right)+3bc\left(a+b+c\right)+3ac\left(a+b+c\right)=3abc\)

Do a+b+C = 0

=> \(a^3+b^3+c^3=3abc\) (đpcm)

\(a,A=\left(x+5\right)^3\)

\(b,B=\left(x-3\right)^3\)

\(c,C=\frac{x^3}{8}+\frac{x^2y}{4}+\frac{xy^2}{6}+\frac{y^3}{27}=\left(\frac{x}{2}+\frac{y}{3}\right)^3\)

Mk nghĩ đề bài phần c fải như trên ,cn đâu bn tự thay số vào nha.

Bài 1: 

a: \(x^3-6x^2+11x-6\)

\(=x^3-x^2-5x^2+5x+6x-6\)

\(=\left(x-1\right)\left(x^2-5x+6\right)\)

\(=\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

b: \(x^3-6x^2-9x+14\)

\(=x^3-7x^2+x^2-7x-2x+14\)

\(=\left(x-7\right)\left(x^2+x-2\right)\)

\(=\left(x-7\right)\left(x+2\right)\left(x-1\right)\)

c: \(x^3+6x^2+11x+6\)

\(=x^3+3x^2+3x^2+9x+2x+6\)

\(=\left(x+3\right)\left(x^2+3x+2\right)\)

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

a) Đk \(x\ge\pm5\)\(\sqrt{x^2-25}=2.\sqrt{x-5}\Leftrightarrow x^2-25=4\left(x-5\right)\Leftrightarrow x+5=4\Leftrightarrow x=-1\)

b) \(\sqrt{x}+\frac{1}{\sqrt{x}}=2\Leftrightarrow x+1-2\sqrt{x}=0\)\(\Leftrightarrow\left(\sqrt{x}-1\right)^2=0\Leftrightarrow x=1\)

c) \(\sqrt{x-5}+\frac{1}{3}\sqrt{9x-45}=\frac{1}{5}\sqrt{25x-125}+6\)

\(\Leftrightarrow\sqrt{x-5}+\frac{1}{3}\sqrt{9\left(x-5\right)}=\frac{1}{5}\sqrt{25\left(x-5\right)}+6\)

\(\Leftrightarrow\left(\sqrt{x}-5\right)\left(1+1-1-6\right)=0\Leftrightarrow-5\left(\sqrt{x}-5\right)=0\Rightarrow\sqrt{x}-5=0\Leftrightarrow x=25\)

a) A=(x+5)³ Thay x= -10 vào ta được A=(-10+5)³= -125

b) B=(x-3)³ Thay x=13 vào ta được B=(13-3)³=1000

c) C=(x/2 - y/3)³ Thay x=-8 và y=6 ta được C=(-8/2 - 6/3)³= -216