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1) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)\)
\(=x^3-16x-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x^3-16x-x^4+1\)
b) \(7x\left(4y-x\right)+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)
\(=28xy-7x^2+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)
\(=28xy-7x^2+4y^2-28xy-4y^2+7x\)
\(=-7x^2+7x\)
c) \(\left(3x-1\right)\left(2x-5\right)-4\left(2x^2-5x+2\right)\)
\(=6x^2-17x+5-4\left(2x^2-5x+2\right)\)
\(=6x^2-17x+5-8x^2+20x-8\)
\(=-2x^2+3x-3\)
a) x(x+4)(x-4)-(x2+1)(x2-1)
=>x(x2-42)-(x4-12)
=>x3-16x-x4+1
=>-x4-x3-15x
b) 7x(4y-x)+4y(y-7x)-2(2y2-3.5x)
=>28xy-7x2+4y2-28xy-4y2+30x
=>-7x2+30x
c) (3x+1)(2x-5)-4(2x2-5x+2)
=>6x2-15x+2x-5-8x2+20x-8
=>-2x2+7x-13
Giải phương trình:
a) (x+2)3 - (x-2)3 = 12x(x-1) - 8
<=> (x2 + 3.x2.2 + 3.x.22 + 23) - (x2 - 3.x2.2 + 3.x.22 - 23) - [12x(x-1) - 8] = 0
<=> (x3 + 6x2 + 12x + 8) - (x3 - 6x2 + 12x - 8) - (12x2 - 12x - 8) = 0
<=> x3 + 6x2 + 12x + 8 - x3 + 6x2 - 12x + 8 - 12x2 + 12x + 8 = 0
<=> 12x +32 = 0
<=> x = \(\frac{-32}{12}\) = \(-2\frac{2}{3}\)
Vậy phương trình có nghiệm duy nhất là \(-2\frac{2}{3}\)
b) (3x-1)2 - 5(2x+1)2 + (6x-3)(2x+1) = (x-1)2
<=> (9x2 - 6x + 1) - 5(4x2 + 4x + 1) + 3(2x - 1)(2x + 1) - (x2 - 2x +1) = 0
<=> 9x2 - 6x + 1 - 20x2 - 20x - 5 + 3(4x2 - 1) - x2 + 2x -1 = 0
<=> 9x2 - 6x + 1 - 20x2 - 20x - 5 + 12x2 - 3 - x2 + 2x -1 = 0
<=> -24x - 8 = 0
<=> x = \(\frac{-8}{24}\) = \(\frac{-1}{3}\)
Vậy phương trình có nghiệm duy nhất là \(\frac{-1}{3}\)
a) (x - 1)3 + (2 - x)(4 + 2x + x2) + 3x(x + 2) = 12
<=> x3 - 2x2 + x - x2 + 2x - 1 + 8 + 4x + 2x2 - 4x - 2x2 + 3x2 + 6x = 17
<=> 9x + 7 = 17
<=> 9x = 17 - 7
<=> 9x = 10
<=> x = \(\frac{10}{9}\)
b) (x + 2)(x2 - 2x + 4) - x(x2 - 2) = 15
<=> x3 - 2x2 + 4x + 2x2 - 4x + 8 - x3 + 2x = 15
<=> 2x + 8 = 15
<=> 2x = 15 - 8
<=> 2x = 7
<=> x = \(\frac{7}{2}\)
c) (x - 3)3 - (x - 3)(x2 + 3x + 9) + 9(x2 + 1)2 = 15
<=> x3 + 45x - 18 - x3 - 3x2 - 9x + 3x2 + 9x + 27 = 15
<=> 45x + 9 = 15
<=> 45x = 15 - 9
<=> 45x = 6
<=> x = \(\frac{6}{45}\)
d) x(x - 5)(x + 5) - (x + 2)(x2 - 2x + 4) = 3
<=> x3 - 25x - x3 + 2x2 - 4x - 8 = 3
<=> -25x - 8 = 3
<=> -25x = 3 + 8
<=> -25x = 11
<=> x = \(-\frac{11}{25}\)
a) (x - 2)(x + 3) = 6
=> x2 + 3x - 2x - 6 = 6
=> x2 + x - 6 - 6 = 0
=> x2 + x - 12 = 0
=> x2 + 4x - 3x - 12 = 0
=> x(x + 4) - 3(x + 4) = 0
=> (x - 3)(x + 4) = 0
=> \(\orbr{\begin{cases}x-3=0\\x+4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
b) (2x - 3)(x + 2) = 4
=> 2x2 + 4x - 3x - 6 = 4
=> 2x2 + x - 6 - 4 = 0
=> 2x2 + x - 10 = 0
=> 2x2 + 5x - 4x - 10 = 0
=> x(2x + 5) - 2(2x + 5) = 0
=> (x - 2)(2x + 5) = 0
=> \(\orbr{\begin{cases}x-2=0\\2x+5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)
a, \(12-2\left(1-x\right)^2=\left(3x-2\right)\left(2x-3\right)\)
\(< =>12-2\left(1-2x+x^2\right)=6x^2-9x-4x+6\)
\(< =>12-2+4x-2x^2=6x^2-13x+6\)
\(< =>10+4x-2x^2-6x^2+13x-6=0\)
\(< =>-8x^2+17x+4=0< =>\orbr{\begin{cases}x=\frac{17-\sqrt{417}}{16}\\x=\frac{17+\sqrt{417}}{16}\end{cases}}\)
b, \(10x+3-5x=4x+12< =>5x+3-4x-12=0\)
\(< =>x-9=0< =>x=9\)
c, \(11x+42-2x=100-9x-22< =>9x+42-100+9x+22=0\)
\(< =>18x+64-100=0< =>18x-36=0< =>x=\frac{36}{18}=2\)
d, \(2x-\left(3-5x\right)=4\left(x+3\right)< =>2x-3+5x=4x+12\)
\(< =>7x-3-4x-12=0< =>3x-15=0< =>x=\frac{15}{3}=5\)
e, \(2\left(x-3\right)+5x\left(x-1\right)=5x^2< =>2x-6+5x^2-5=5x^2\)
\(< =>2x-11+5x^2-5x^2=0< =>2x-11=0< =>x=\frac{11}{2}\)
f, \(-6\left(1,5-2x\right)=3\left(-15+2x\right)< =>-6\left(\frac{3}{2}-2x\right)=3\left(2x-15\right)\)
\(< =>-9+12x-6x+45=0< =>6x+36=0< =>x=-6\)
g, \(14x-\left(2x+7\right)=3x+12x-13< =>14x-2x-7=15x-13\)
\(< =>12x-7-15x+13=0< =>-3x+6=0< =>x=-2\)
h, \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)
\(< =>x^2-16-6x+4=x^2-8x+16\)
\(< =>x^2-6x-12-x^2+8x-16=0\)
\(< =>2x-28=0< =>x=\frac{28}{2}=14\)
q, \(4\left(x-2\right)-\left(x-3\right)\left(2x-5\right)=?\)thiếu đề
a, \(\frac{1+2x-5}{6}=\frac{3-x}{4}\)
\(\frac{4+8x-20}{24}=\frac{18-6x}{24}\)
\(-16-8x=18-6x\)
\(-16-8x-18+6x=0\)
\(-34-2x=0\)
\(2x=-34\Leftrightarrow x=-17\)
b, \(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)ĐKXĐ : x \(\ne\)-1 ; 0
\(\frac{x^2+3x}{x^2+x}+\frac{x^2-x-2}{x^2+x}=\frac{2x^2+2x}{x^2+x}\)
\(x^2+3x+x^2-x-2=2x^2+2x\)
\(2x^2+2x-2=2x^2+2x\)
\(2x^2+2x-2x^2-2x-2=0\)
\(-2\ne0\) Nên phuwong trình vô nghiệm. (xem lại hộ)
a ) \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x^2-3x^2\right)+\left(6x+3x\right)+\left(8-1\right)=17\)
\(\Leftrightarrow9x+7=17\)
\(\Leftrightarrow9x=10\)
\(\Leftrightarrow x=\dfrac{10}{9}\)
Vậy nghiệm của p/t là : \(\dfrac{10}{9}\)
b ) \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=3\)
\(\Leftrightarrow x^3-25x-x^3-8=3\)
\(\Leftrightarrow-25x-8=3\)
\(\Leftrightarrow-25x=11\)
\(\Leftrightarrow x=-\dfrac{11}{25}\)
Vậy nghiệm của p/t là : \(-\dfrac{11}{25}\)
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