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a: \(=\dfrac{\left(x^4-y^4\right)^2}{x^2+y^2}=\left(x^2-y^2\right)^2\cdot\left(x^2+y^2\right)\)
b: \(=\dfrac{\left(4x+3\right)\left(16x^2-12x+9\right)}{16x^2-12x+9}=4x+3\)
a: \(=2x^{2n+1-2n}-2\cdot x^{2n}\cdot3\cdot x^{2-2n}+3\cdot x^{2n-1+1-2n}-9\cdot x^{2n-1+2-2n}\)
\(=2x-6x^2+3-9x\)
\(=-6x^2-7x+3\)
b: \(=\left(5x\right)^3-\left(2y\right)^3=125x^3-8y^3\)
a) \(\left( {6{x^3} - 7{x^2} - x + 2} \right):\left( {2x + 1} \right)\)
b) $(x^4-x^3+x^2+3x):(x^2-2x+3)$
c) \(\left( {{x^2} + {y^2} + 6x + 9} \right):\left( {x + y + 3} \right)\)
\(=\left( {{x^2} + 6x + 9 - {y^2}} \right)\left( {x + y + 3} \right)\)
\(=\left[ {\left( {{x^2} + 2x.3 + {3^2}} \right) - {y^2}} \right]:\left( {x + y + 3} \right)\)
\(=\left[ {{{\left( {x + 3} \right)}^2} - {y^2}} \right]:\left( {x + y + 3} \right)\)
\(=\left( {x + 3 - y} \right)\left( {x + 3 + y} \right):\left( {x + y + 3} \right)\)
$= x + 3 - y$
$= x - y + 3$
(6x3 - 7x2 - x + 2) : (2x + 1)
= (6x3 + 3x2 - 10x2 - 5x + 4x + 2) : (2x + 1)
= [(6x3 + 3x2) - (10x2 + 5x) + (4x + 2)] : (2x + 1)
= [3x2(2x + 1) - 5x(2x + 1) + 2(2x + 1)] : (2x + 1)
= (3x2 - 5x + 2)(2x + 1) : (2x + 1)
= 3x2 - 5x + 2
(x4 - x3 + x2 + 3x) : (x2 - 2x + 3)
= (x4 + x3 - 2x3 - 2x2 + 3x2 + 3x) : (x2 - 2x + 3)
= [(x4 + x3) - (2x3 + 2x2) + (3x2 + 3x)] : (x2 - 2x + 3)
= [x3(x + 1) - 2x2(x + 1) + 3x(x + 1)] : (x2 - 2x + 3)
= (x3 - 2x2 + 3x)(x + 1) : (x2 - 2x + 3)
= x(x2 - 2x + 3)(x + 1): (x2 - 2x + 3)
= x(x + 1)
= x2 + x
(x2 - y2 + 6x + 9) : (x + y + 3)
= [(x2 + 6x + 9) - y2] : (x + y + 3)
= [(x + 3)2 - y2] : (x + y + 3)
= (x + 3 + y)(x + 3 - y) : (x + y + 3)
= (x + y + 3)(x - y + 3) : (x + y + 3)
= x - y + 3
CHÚC BN HOK TỐT 
b)\(\frac{9x^4-6x^3+15x^2+2x+1}{3x^2-2x+5}=\frac{3x^2.\left(3x^2-2x+5\right)+2x+1}{3x^2-2x+5}=3x^2+\frac{2x+1}{3x^2-2x+5}\)
=> đa thức dư trong phép chia là 2x+1
\(\frac{x^3+2x^2-3x+9}{x+3}=\frac{x^3+9x^2+27x+27-7x^2-30x-18}{x+3}=\frac{\left(x+3\right)^3-7x^2-30x-18}{x+3}\)
\(\left(x+3\right)^2-\frac{7x^2+21x+9x+18}{x+3}=\left(x+3\right)^2-\frac{7x.\left(x+3\right)+9.\left(x+3\right)-9}{x+3}\)
\(=\left(x+3\right)^2-\frac{\left(7x+9\right).\left(x+3\right)-9}{x+3}=\left(x+3\right)^2-\left(7x+9\right)-\frac{9}{x+3}\)
=> đa thức dư trong phép chia là 9
p/s: t mới lớp 7_sai sót mong bỏ qua :>
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
1) Ta có: \(\left(3-x^2\right)+6-2x=0\)
\(\Leftrightarrow3-x^2+6-2x=0\)
\(\Leftrightarrow-x^2-2x+9=0\)
\(\Leftrightarrow x^2+2x-9=0\)
\(\Leftrightarrow x^2+2x+1=10\)
\(\Leftrightarrow\left(x+1\right)^2=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{10}\\x+1=-\sqrt{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{10}-1\\x=-\sqrt{10}-1\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{10}-1;-\sqrt{10}-1\right\}\)
2) Ta có: \(5\left(2x-1\right)+7=4\left(2-x\right)+2\)
\(\Leftrightarrow10x-5+7=8-4x+2\)
\(\Leftrightarrow10x+4x=8+2+5-7\)
\(\Leftrightarrow14x=8\)
\(\Leftrightarrow x=\dfrac{4}{7}\)
Vậy: \(S=\left\{\dfrac{4}{7}\right\}\)
a)\(x^4-6x^2+2x+28\)
\(=\left(x^4-x^3\right)+\left(x^3-x^2\right)-\left(5x^2-5x\right)-\left(3x-3\right)+25\)
\(=\left(x-1\right)\left(x^3+x^2-5x-3\right)+25\)
=> số dư là 25
b) Cách làm tương tự câu a nhé