Bài 9:

a) Ta có: \(A=\left(2x+y\right)^2-\left(2x+y\right)\left(2x-y\right)+y\left(x-y\right)\)

\(=4x^2+4xy+y^2-4x^2+y^2-xy-y^2\)

\(=3xy-y^2\)

\(=3\cdot\left(-2\right)\cdot3-3^2=-18-9=-27\)

b) Ta có: \(B=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)

\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)

\(=-13ab+2a+b-2\)

\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)+2\cdot\dfrac{1}{2}+\left(-3\right)-2\)

\(=\dfrac{31}{2}\)

Bài 7: 

a) \(498^2=\left(500-2\right)^2=250000-2000+4=248004\)

b) \(93\cdot107=100^2-7^2=10000-49=9951\)

c) \(163^2+74\cdot163+37^2=\left(163+37\right)^2=200^2=40000\)

d) \(1995^2-1994\cdot1996=1995^2-1995^2+1=1\)

e) \(9^8\cdot2^8-\left(18^4-1\right)\left(18^4+1\right)\)

\(=18^8-18^8+1=1\)

f) \(125^2-2\cdot125\cdot25+25^2=\left(125-25\right)^2=100^2=10000\)

Bài 8:

a) Ta có: \(\left(x^2+3x+1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)

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

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

\(=x^4+4x^2+4\)

b) Ta có: \(\left(3x^3+3x+1\right)\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)

\(=\left(3x^3+1\right)^2-9x^2-\left(3x^3+1\right)^2\)

\(=-9x^2\)

c) Ta có: \(\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)-\left(2x^2+1\right)^2\)

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

\(=-4x^2\)

Lời giải :

1. \(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)

\(=\frac{a^3}{8}+\frac{3a^2b}{4}+\frac{3ab^2}{2}+b^3+\frac{a^3}{8}-\frac{3a^2b}{4}+\frac{3ab^2}{2}-b^3\)

\(=\frac{a^3}{4}+3ab^2\)

Lời giải :

2. \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy...

3. \(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(=64x^3-48x^2+12x-1-64x^3+48x^2-12x+9\)

\(=8\)

1) \(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)

\(=\left(\frac{a}{2}+b\right)^2+\left(\frac{a}{2}-b\right)^2\)

\(=\left(\frac{a}{2}+b\right)\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{b}b+b^2\right]+\left(\frac{a}{2}-b\right)\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]\)

\(=\frac{a}{2}\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{2}b+b^2\right]+b\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{2}b+b^2\right]+\frac{a}{2}\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]\)\(-b\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]\)

\(=\frac{a^3}{8}+\frac{a^2b}{2}+\frac{ab^2}{2}+\frac{ba^2}{4}+b^2a+b^3+\frac{a^3}{8}-\frac{a^2b}{2}+\frac{ab^2}{2}-\frac{ba^2}{4}+b^2a-b^3\)

\(=\frac{a^3}{4}+3ab^2\)

2) \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow x^3-3x^2.1+3.x.1^2-1^3=0\)

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

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=0-1\)

\(\Rightarrow x=-1\)

3) \(A=\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(A=64x^3-32x^2+4x-16x^2+8x-1-64x^3-12x+48x^2+9\)

\(A=8\)

Vậy: biểu thức không phụ thuộc vào biến

1) \(\left(x+5\right)^3-x^3-125\)

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

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

\(=x^3+10x^2+25x+5x^2+50x+125-x^3-125\)

\(=15x^2+75x\)

2) \(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)

\(\Leftrightarrow x^3-4x^2+4x-2x^2+8x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow24x+10=0\)

\(\Leftrightarrow24x=0-10\)

\(\Leftrightarrow24x=-10\)

\(\Leftrightarrow x=-\frac{10}{24}=-\frac{5}{12}\)

\(\Rightarrow x=-\frac{5}{12}\)

3) \(\left(x-1\right)^3-x^3+3x^2-3x+1\)

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

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

\(=x^3-2x^2+x-x^2+2x-1-x^3-3x^2-3x+1\)

\(=0\)

Vậy: biểu thức không phụ thuộc vào biến

125 : 5^2 . (x-3) = 3

125 : 25 .(x-3) = 3

5 (x-3) =3

x-3 = 3 : 5

Đề bài sai hay sao ý?

2x -5 = 3

2x = 3+5 =8

x =4

 (19 .x+2.5^2 ): 14 = (13-8)^2 - 4^2

19x + 50) : 14 = 25 - 16

(19x+50) : 14 = 9

19x+50 = 3*14 = 126

 19x = 126-50 = 76

x = 76 / 19 = 4

\(a\)) \(125:5^2.\left(x-3\right)=3\)

\(5^3:5^2.\left(x-3\right)=3\)

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

\(5x-15=3\)

\(5x=3+15\)

\(5x=18\)

\(x=\frac{18}{5}\)\(.\)Vậy \(x=\frac{18}{5}\)

\(b\)) \(\left(2x-5\right)^3=27\)

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

\(\Rightarrow2x-5=3\)

\(\Rightarrow2x=8\)

\(\Rightarrow x=4\)\(.\)Vậy \(x=4\)

\(c\))\(\left(19.x+2.5^2\right):14=\left(13-8\right)^2-4^2\)

\(\left(19.x+2.25\right):14=5^2-16\)

\(\left(19.x+2.25\right):14=25-16\)

\(\left(19.x+50\right):14=9\)

\(19.x+50=126\)

\(19.x=126-50\)

\(19.x=76\)

\(x=76:19\)

\(x=4\)\(.\)Vậy \(x=4\)

1/ a) \(2.3.12.12.3=2.3.2^2.3.2^2.3.3=2^5.3^4\)
    b) \(3.5.27.125=3.5.3^3.5^3=3^4.5^4=\left(3.5\right)^4\)
2/ a) \(\left(27^3\right)^4=27^{3.4}=27^{12}\)
Vậy \(\left(27^3\right)^4=27^{12}\)
b) \(5^{36}=\left(5^6\right)^6\)         và          \(11^{24}=\left(11^4\right)^6\)
Do đó \(5^6=15625\)      và        \(11^4=14641\)
Vì 15625>14641 nên\(\left(5^6\right)^6>\left(11^4\right)^6hay5^{36}>11^{24}.\)
3/ a) \(x^3=125=>x=5\)
b) \(\left(3x-14\right)^3=2^5.5^2+200\)
    \(\left(3x-14\right)^3=1000\)
    \(3x-14=10^3\)
    \(3x=10^3+14\)
    \(3x=1014\)
       \(x=\frac{1014}{3}=338\)
c) \(\left(2x-1\right)^4=81\)
    \(\left(2x-1\right)^4=3^4\)
    \(2x-1=3\)
     \(2x=3+1\)
        \(x=\frac{4}{2}=2\)
d) \(5x+3^4=2^2.7^2\)
    \(5x+3^4=\left(2.7\right)^2=14^2\)
    \(5x+81=196\)
    \(5x=196-81\)
    \(5x=115\)
       \(x=\frac{115}{5}=23\)
e) \(4^x=1024=>x=5\).

So sánh

A 5 mũ 36 và 11 mũ 24

B 7.2 mũ 13 và 2 mũ 16

tìm x

x mũ 2=121, 2 mũ x+3=1024, 5 mũ x+1=625

  Bài 1:

2^\(x\) = 4

2\(^x\) = 2²

 \(x=2\)

Vậy \(x=2\)

Bài 2:

2\(^x\) = 8

2\(^x\) = 2³

\(x=3\)

Vậy \(x=3\)

Bài 3

2\(^x\) = 16

2\(^x\) = 2⁴

  \(x=4\)

Vậy \(x=4\)

a) \(\left(12x-4^3\right).8^3=4.8^4\)

\(12x-4^3=32\)

12x  = 96

x = 8

b) \(\left(3x-2^4\right).7^3=2.7^4\)

3x - 2⁴ = 14

3x = 30

x = 10

Thanks you ! Mình biết làm rồi !