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-3x^3+5x^2-9x+15 -3x-5 x^2 -3x^3-5x^2 - 10x^2-9x+15 -(10/3)x 10x^2+(50/3)x - -(23/3)x+15 +23/9 -(23/3)x-115/9 - 250/9
Chả biết có sai ko @@
x^4-2x^3 +2x-1 x^2-1 x^2-2x x^4 -x^2 - -2x^3+x^2+2x-1 -2x^3 +2x - x^2-1 +1 x^2-1 - 0
Thực hiện phép tính:
a. ( 2x - 6 )( 12x2 + 9x + 36 )
b. ( 2x4 + x3 - 3x2 + 5x - 2 ) : ( 5x2 - 5x + 5 )
Bài 1:
a: \(=\dfrac{4x^3-6x^2+6x^2-9x-10x+15}{2x-3}\)
\(=2x^2+3x-5\)
b: \(=\dfrac{5x^4+5x^3+4x^3+4x^2-6x^2-6x+2x+2-10}{x+1}\)
\(=5x^3+4x^2-6x+2-\dfrac{10}{x+1}\)
c: \(=\dfrac{5x^3+10x^2+4x^2+8x-5x-10+11}{x+2}\)
\(=5x^2+4x-5+\dfrac{11}{x+2}\)
d: \(=\dfrac{\left(x+1\right)^3}{x+1}=\left(x+1\right)^2\)
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
c) Ta có: \(\dfrac{5x^4+9x^3-2x^2-4x-8}{x-1}\)
\(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
\(=\dfrac{5x^3\left(x-1\right)+14x^2\left(x-1\right)+12x\left(x-1\right)+8\left(x-1\right)}{x-1}\)
\(=5x^3+14x^2+12x+8\)
d) Ta có: \(\dfrac{5x^3+14x^2+12x+8}{x+2}\)
\(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}\)
\(=\dfrac{5x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)}{x+2}\)
\(=5x^2+4x+4\)