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a) (x2 – 2x+ 1)(x – 1)
= x2 . x + x2.(-1) + (-2x). x + (-2x). (-1) + 1 . x + 1 . (-1)
= x3 - x2 - 2x2 + 2x + x – 1
= x3 - 3x2 + 3x – 1
b) (x3 – 2x2 + x -1)(5 – x)
= x3 . 5 + x3 . (-x) + (-2 x2) . 5 + (-2x2)(-x) + x . 5 + x(-x) + (-1) . 5 + (-1) . (-x)
= 5 x3 – x4 – 10x2 + 2x3 +5x – x2 – 5 + x
= - x4 + 7x3 – 11x2+ 6x - 5.
Suy ra kết quả của phép nhan:
(x3 – 2x2 + x -1)(x - 5) = (x3 – 2x2 + x -1)(-(5 - x))
= - (x3 – 2x2 + x -1)(5 – x)
= - (- x4 + 7x3 – 11x2+ 6x -5)
= x4 - 7x3 + 11x2- 6x + 5
a) (x2 – 2x+ 1)(x – 1)
= x2 . x + x2.(-1) + (-2x). x + (-2x). (-1) + 1 . x + 1 . (-1)
= x3 - x2 - 2x2 + 2x + x – 1
= x3 - 3x2 + 3x – 1
b) (x3 – 2x2 + x -1)(5 – x)
= x3 . 5 + x3 . (-x) + (-2 x2) . 5 + (-2x2)(-x) + x . 5 + x(-x) + (-1) . 5 + (-1) . (-x)
= 5 x3 – x4 – 10x2 + 2x3 +5x – x2 – 5 + x
= - x4 + 7x3 – 11x2+ 6x - 5.
Suy ra kết quả của phép nhan:
(x3 – 2x2 + x -1)(x - 5) = (x3 – 2x2 + x -1)(-(5 - x))
= - (x3 – 2x2 + x -1)(5 – x)
= - (- x4 + 7x3 – 11x2+ 6x -5)
= x4 - 7x3 + 11x2- 6x + 5
Lời giải:
a)
\((x-2)(x-3)+2x=(x-2)^2-2\)
\(\Leftrightarrow (x-2)(x-2-1)+2x=(x-2)^2-2\)
\(\Leftrightarrow (x-2)^2-(x-2)+2x=(x-2)^2-2\)
\(\Leftrightarrow x+4=0\Rightarrow x=-4\)
b)
\((x-1)^2+3x(x-1)+7=(2x-1)^2+5(x-3)\)
\(\Leftrightarrow (x-1)^2+3x(x-1)+7=x^2+(x-1)^2+2x(x-1)+5(x-3)\)
\(\Leftrightarrow x(x-1)+7=x^2+5(x-3)\)
\(\Leftrightarrow 6x=22\Rightarrow x=\frac{11}{3}\)
c)
\(5(x^2-2x-1)+2(3x-2)=5(x+1)^2=5(x^2-2x+1)\)
\(\Leftrightarrow -5+2(3x-2)=5\)
\(\Leftrightarrow 3x-2=5\Rightarrow x=\frac{7}{3}\)
d)
\((x-1)(x^2+x+1)-2x=x(x-1)(x+1)=x(x^2-1)\)
\(\Leftrightarrow x^3-1-2x=x^3-x\Leftrightarrow -1-x=0\Rightarrow x=-1\)
1,
a,\(2x\left(3x^2-5x+3\right)\)
\(=6x^3-10x^2+6x\)
b,\(-2x\left(x^2+5x-3\right)\)
\(=-2x^3-10x^2+6x\)
c,\(-\dfrac{1}{2}x\left(2x^3-4x+3\right)\)
\(=-x^4+2x^2-\dfrac{3}{2}x\)
Bài 2:
a) \(\left(2x-1\right)\left(x^2-5-4\right)\)
\(=\left(2x-1\right)\left(x^2-9\right)\)
\(=2x^3-18x-x^2+9\)
b) \(-\left(5x-4\right)\left(2x+3\right)\)
\(=-\left(10x^2+15x-8x-12\right)\)
\(=-10x^2-7x+12\)
c) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-y^3\)
\(a,x^2\left(x-2x^3\right)\)
\(=x^3-2x^5\)
\(b,\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
\(c,\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
\(d,\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=\left(6x^2+x-2\right)\left(3-x\right)\)
\(=18x^2+3x-6-6x^3-x^2+2x\)
\(=17x^2+5x-6-6x^3-x^2\)
\(e,\left(x+3\right)\left(x^2+3x-5\right)\)
\(=x^3+3x^2-5x+3x^2+9x-15\)
\(=x^3+6x^2+4x-15\)
\(f,\left(xy-2\right)\left(x^3-2x-6\right)\)
\(=x^4y-2x^2y-6xy-2x^3+4x-12\)
\(g,\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)\)
\(=20x^5-4x^4+8x^3-12x^2-5x^4+x^3-2x^2+3x+10x^3-2x^2+4x-6\)
\(=20x^5-9x^4+19x^3-16x^2+7x-6\)
a. x2(x−2x3)= x3-2x5
b. (x−2)(x−x2+4)= x2-x3+4x-2x+2x2-8= -x3+3x2+2x-8
c. (x2−1)(x2+2x)= x4+2x3-x2-2x
d. (2x−1)(3x+2)(3−x) = (6x2+x-2)(3-x)=18x2-6x3+3x-x2-6+2x =-6x3+17x2+5x-6
e. (x+3)(x2+3x−5)= x3+3x2-5x+3x2+9x-15= x3+6x2+4x-15
f. (xy−2)(x3−2x−6)= x4y-2x2y-6xy-2x3+4x+12
g. (5x3−x2+2x−3)(4x2−x+2)= 20x5-9x4+19x3-12x2+7x-6
Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
a: \(=6x^2-9x+14x-21-4x^2+20x-25-2x\left(x+6\right)+5-31x\)
\(=2x^2-6x-41-2x^2-12x\)
=-18x-41
b: \(=2x^2-6x-2x^2+6x+14=14\)
c: \(=x^3+1-x^3+1=2\)
a); b) Do tích = 0
=> Từng thừa số = 0 và ta nhận xét: \(x^2+2;x^2+3>0\)
=> a) \(\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
và câu b) \(\orbr{\begin{cases}x=\frac{1}{2}\\x=5\end{cases}}\)
a; *x-1=0 <=>x=1
*2x+5=0 <=>x=-2,5
*x2+2=0 <=> ko có x
b; tương tự a
Bài 1:
a)(4x-3)(3x+2)-(6x+1)(2x-5)+1
=12x2-x-6-12x2+28x+5+1
=27x
b)(3x+4)2+(4x-1)2+(2+5x)(2-5x)
=9x2+24x+16+16x2-8x+1+4-25x2
=16x+21
c)(2x+1)(4x2-2x+1)+(2-3x)(4+6x+9x2)-9
=8x3+1+8-27x3-9
=-19x3
Bài 2:
a)3x(x-4)-x(5+3x)=-34
=>3x2-12x-3x2-5x=-34
=>-17x=-34
=>x=2
Vậy x=2
b)(3x+1)2+(5x-2)2=34(x+2)(x-2)
=>9x2+6x+1+25x2-20x+4=34(x2-4)
=>34x2-14x+5-34x2+136=0
=>-14x+141=0
=>-14x=-141
=>x=\(\frac{141}{14}\)
Vậy x=\(\frac{141}{14}\)
c)x3+3x2+3x+28=0
=>x3-x2+7x+4x2-4x+28=0
=>x(x2-x+7)+4(x2-x+7)=0
=>(x+4)(x2-x+7)=0
\(\Rightarrow\left[\begin{array}{nghiempt}x+4=0\\x^2-x+7=0\left(2\right)\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-4\\\left(2\right)\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{27}{4}>0\end{array}\right.\)
=>(2) vô nghiệm
Vậy x=-4