giải

5x-(4-2x+x^2)(x+2)+x(x-1)(x+1)=0

5x-(4x+8-2x^2-4x+x^3+2x^2)+x(x^2-1)=0

5x-4x-8+2x^2+4x-x^3-2x^2+x^3-1x=0

(5x-4x+4x-1x)+(-8)+(2x^2-2x^2)+(-x^3+x^3)=0

4x+(-8)=0

4x=0+8

4x=8

x=8:4

x=2

D)(4x+1)(16x^2-4x+1)-16x(4x^2-5)=17

64x^3-16x^2+4x+16x^2-4x+1-64x^3+80x=17

80x+1=17

80x=17-1

80x=16

x=1/5

\(5x-\left(4-2x+x^2\right)\left(x+2\right)+x\left(x-1\right)\left(x+1\right)=0\)

\(\Rightarrow5x-\left(4x-2x^2+x^3+8-4x+2x^2\right)+\left(x^2-x\right)\left(x+1\right)=0\)

\(\Rightarrow5x-\left(4x-2x^2+x^3+8-4x+2x^2\right)+\left(x^3+x-x^2-x\right)=0\)

\(\Rightarrow5x-4x+2x^2-x^3-8+4x-2x^2+x^3+x-x^2-x=0\)

\(\Rightarrow4x-8=0\Rightarrow4x=8\Rightarrow x=2\)

a/ \(=x^4+x^2+1+2x^3+2x+2x^2=\left(x^2+x+1\right)^2\)

b/ \(=y^4+\left(-2x^2-34\right)y^2+32xy+x^4-34x^2+225\)

câu này bn coi lại đc k , mk k lm ra 

làm khuyến mại 1 câu;

a) = 12x² -12x² +20x -10x +17 =0

10x = -17

x = -17/10

x/2 - ( 3x/5 - 13/5 ) = -( 7/5 + 7/10x )

a) = 12x2 -12x2 +20x -10x +17 =0

10x = -17

x = -17/10

a) \(8x+56:14=60\)

\(\Rightarrow8x+4=60\)

\(\Rightarrow8x=56\)

\(\Rightarrow x=\dfrac{56}{8}\)

\(\Rightarrow x=7\)

b) Mình làm rồi nhé !

c) \(41-2^{x+1}=9\)

\(\Rightarrow2^{x+1}=41-9\)

\(\Rightarrow2^{x+1}=32\)

\(\Rightarrow2^{x+1}=2^5\)

\(\Rightarrow x+1=5\)

\(\Rightarrow x=4\)

d) \(3^{2x-4}-x^0=8\)

\(\Rightarrow3^{2x-4}-1=8\)

\(\Rightarrow3^{2x-4}=9\)

\(\Rightarrow3^{2x-4}=3^2\)

\(\Rightarrow2x-4=2\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=3\)

g) \(65-4^{x+2}=2014^0\)

\(\Rightarrow65-4^{x+2}=1\)

\(\Rightarrow4^{x+2}=64\)

\(\Rightarrow4^{x+2}=4^3\)

\(\Rightarrow x+2=3\)

\(\Rightarrow x=1\)

i) \(120+2\left(4x-17\right)=214\)

\(\Rightarrow2\left(4x-17\right)=214-120\)

\(\Rightarrow2\left(4x-17\right)=94\)

\(\Rightarrow4x-17=47\)

\(\Rightarrow4x=47+17\)

\(\Rightarrow4x=64\)

\(\Rightarrow x=16\)

a: \(8x+56:14=60\)

=>8x+4=60

=>8x=60-4=56

=>x=56/8=7

b: \(5^{2x-3}-2\cdot5^2=5^2\cdot3\)

=>\(5^{2x-3}=5^2\cdot3+2\cdot5^2=5^3\)

=>2x-3=3

=>2x=6

=>x=3

c: \(41-2^{x+1}=9\)

=>\(2^{x+1}=41-9=32\)

=>x+1=5

=>x=4

d: \(3^{2x-4}-x^0=8\)

=>\(3^{2x-4}-1=8\)

=>\(3^{2x-4}=8+1=9\)

=>2x-4=2

=>2x=6

=>x=3

g: \(65-4^{x+2}=2014^0\)

=>\(65-4^{x+2}=1\)

=>\(4^{x+2}=65-1=64\)

=>x+2=3

=>x=1

i: 120+2(4x-17)=214

=>2(4x-17)=214-120=94

=>4x-17=94/2=47

=>4x=64

=>\(x=\dfrac{64}{4}=16\)

[(10x+4).(2x+1)].[(4x-2).(5x+7)] + 17 = (20x + 18x + 4)(20x + 18x - 14) + 17.

Đến đây ta đặt 20x + 18x - 5 = t, ta được: (t - 9)(t + 9) + 17 = 0 \(\Leftrightarrow\) t² - 81 + 17 = 0

Đến đây bạn tự làm tiếp nhé!

\(x+\frac{11}{17}+x+\frac{2}{17}+x+\frac{4}{17}=4x\)

\(\Rightarrow3x+1=4x\)

\(\Rightarrow x=1\)

Vì vế trái >=0 => vế phải >=0 => x>=0

Khi đó ta có: \(x+\frac{11}{17}+x+\frac{2}{17}+x+\frac{4}{17}=4x\)

             =>  \(3x+1=4x\)

             => \(x=1\)