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

\(=9x^2-18x+4-2\left(x^2-2x+1\right)\)

\(=9x^2-18x+4-2x^2+4x-2\)

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

\(\left(3x-2\right)^2-2\left(x-1\right)^2=9x^2-12x+4-2x^2+4x-2=7x^2-8x+2\)

\(=9x^2-12x+4-2x^2+4x-2\\ =7x^2-8x+2\)

Ta có: \(\left(3x-4\right)^2-9\left(x-1\right)\left(x-3\right)=13\)

\(\Leftrightarrow\left(3x-4\right)^2-9\left(x^2-4x+3\right)=13\)

\(\Leftrightarrow9x^2-24x+16-9x^2+36x-27=13\)

\(\Leftrightarrow12x=24\)

hay x=2

P(x) = 9x⁴ + 5x³ + 8x² - 15x³ - 4x² - x⁴ + 15 - 7x²

        = (9x⁴ - x⁴) + (5x³ - 15x³) + (8x² - 4x² - 7x²) + 15

        = 8x⁴ - 10x³ - 3x² + 15

Ta có: P(1) = 8. 1⁴ - 10. 1³ - 3. 1² + 15 = 8 - 10 - 3 + 15 = 10

          P(0) = 8. 0⁴ - 10. 0³ - 3. 0² + 15 = 0 - 0 - 0 + 15 = 15

          P(-1) = 8.(-1)⁴ - 10(-1)³ - 3(-1)² + 15 = -8 - (-10) - (-3) + 15 = 20

a,

(x²-x+1)(x+1)-x³+3x=15

x³-x²+x+x²-x+1-x³+3x=15

x³-x³-x²+x²+x-x+3x+1=15

3x+1=15

3x=15-1

3x=14

x=14/3

b,

(x+3)(x-2)+3x=\(\frac{4}{x+\frac{3}{4}}\)

x²-2x+3x-6+3x=\(\frac{4}{x+\frac{3}{4}}\)

x²-2x+3x+3x-6=\(\frac{4}{x+\frac{3}{4}}\)

Tới đây hết biết , đề có gì sai sai sao ý !

c,

(x²-5)(x+2)+5x=2x²+17

x³+2x²-5x-10+5x=2x²+17

x³+2x²-5x+5x-10=2x²+17

x³+2x²-10=2x²+17

x³-10=17

x³=17+10

x³=27

\(\Rightarrow x=3\)(Vì : 3³=27)

_k_ nhé bn

Nhân ra thôi bạn, có hằng đẳng thức gì đâu !

a) \(\left(x^2-x+1\right)\left(x+1\right)-x^3+3x=15\)

\(\Leftrightarrow\left(x^2-x+1\right)\cdot x+x^2-x+1-x^3+3x=15\)

\(\Leftrightarrow x^3-x^2+x+x^2-x+1-x^3+3x=15\)

\(\Leftrightarrow1+3x=15\Leftrightarrow3x=14\Leftrightarrow x=\frac{14}{3}\)

b) \(\left(x+3\right)\left(x-2\right)+3x=4\cdot\left(x+\frac{3}{4}\right)\)

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

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

\(\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)

c) \(\left(x^2-5\right)\left(x+2\right)+5x=2x^2+17\)

\(\Leftrightarrow x^3-5x+2x^2-10+5x=2x^2+17\)

\(\Leftrightarrow x^3=27\Leftrightarrow x=3\)

a: (x+1)^3-x(x-2)^2+x-1=0

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

=>x^3+3x^2+4x-x^3+4x^2-4x=0

=>7x^2=0

=>x=0

b: =>x^3-3x^2+3x-1-x^3-27+3x^2-12=2

=>3x=2+1+27+12=39+3=42

=>x=14

a)\(\left(x^2-x+1\right).\left(x+1\right)-x^3+3x=15\)

\(x^3+x^2-x^2-x+x+1-x^3+3x=15\)

\(1+3x=15\)

\(3x=15-1\)

\(3x=14\)

\(x=\frac{14}{3}\)
 

b) \(\left(x+3\right).\left(x-2\right)+3x=4\left(x+\frac{3}{4}\right)\)

\(x^2-2x+3x-6+3x=4x+3\)

\(x^2-2x+3x+3x-4x=6+3\)

\(x^2=9\)

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

 vậy x=3 hoặc x=-3

a)  \(\left(x^2-x+1\right)\cdot\left(x+1\right)-x^3+3x=15\)

\(x^3-x^2+x+x^2-x+1-x^3+3x=15\)

\(3x+1=15\)

\(x=\frac{14}{3}\)

b) \(\left(x+3\right)\left(x-2\right)+3x=4\left(x+\frac{3}{4}\right)\)

\(x^2+3x-2x-6+3x=4x+3\)

\(x^2=9\)

\(x=+-3\)

c) \(\left(x^2-5\right)\left(x+2\right)+5x=2x^2+17\)

\(x^3-5x+2x^2-10+5x=2x^2+17\)

\(x^3=27\)

\(x=3\)

Chúng ta sẽ sử dụng hằng đẳng thức em nhé :)

a. \(x^3+1-x^3+3x=15\Leftrightarrow3x=14\Leftrightarrow x=\frac{14}{3}\)

b. \(x^2+x-6+3x=4x+3\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

c. \(x^3+2x^2-5x-10+5x=2x^2+17\Leftrightarrow x^3=27\Leftrightarrow x=3\)