Ôn tập phép nhân và phép chia đa thức

Vào trang cá nhân của t mà xem.T vừa làm r.Lười gõ lại lắm T^T

\(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)

\(=\dfrac{x+1}{1998}+\dfrac{x+2}{1997}-\dfrac{x+3}{1996}-\dfrac{x+4}{1995}=0\)

\(=\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1-\dfrac{x+3}{1996}-1-\dfrac{x+4}{1995}-1=0\)

\(=\dfrac{x+1999}{1998}+\dfrac{x+1999}{1998}-\left(\dfrac{x+3}{1996}+1\right)-\left(\dfrac{x+4}{1995}+1\right)=0\)

\(=\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)

\(=\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)

⇔\(x+1999=0\)

Vậy \(x=-1999\)

a) \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)

\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}=\dfrac{x+100}{96}+\dfrac{x+100}{95}\)

\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)

\(\Rightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)

Vì \(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\ne0\) nên \(x+100=0\Leftrightarrow x=-100\)

b) \(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)

\(\Rightarrow\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1=\dfrac{x+3}{1996}+1+\dfrac{x+4}{1995}+1\)

\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}=\dfrac{x+1999}{1996}+\dfrac{x+1999}{1995}\)

\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)

\(\Rightarrow\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)

Vì \(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\ne0\) nên \(x+1999=0\Leftrightarrow x=-1999\)

c) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)

\(\Rightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)

\(\Rightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)

\(\Rightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)

Vì \(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\ne0\) nên \(300-x=0\Leftrightarrow x=300\)

trả lời giúp mình

A=,x²-4xy-4y² +3

= (x-2y)²+3

do ( x-2y)²\(\ge0\forall x;y\)

=> (x-2y)²+3\(\ge3\)

=> A\(\ge3\)

vậy A \(>0\) với mọi số thực x;y

Bổ sung câu trả lời của bạn kuroba kaito

Khi và chỉ khi x - 2y =0

x =2y

Ta có: x² - 4xy + 4y² +3 > 0 với \(\forall\) x,y

\(\Leftrightarrow\) (x² - 4xy + 4y²) + 3 > 0 với \(\forall\) x,y

\(\Leftrightarrow\) (x - y)² +3 > 0 với \(\forall\) x,y

Ta thấy: (x - y)² \(\ge\) 0 với \(\forall\) x,y

3 > 0

\(\Rightarrow\) (x - y)² + 3 > 0 với \(\forall\) x,y

\(\Rightarrow\) x² - 4xy + 4x² + 3 > 0 với \(\forall\) x,y

Đặt \(A=4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(2A=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(2A=\left(3^8-1\right)\left(3^8+1\right)\)

\(2A=3^{16}-1\)

\(A=\dfrac{3^{16}-1}{2}\)

Đề

16x--3=-(5x²-16x+3)=-(5x²-15x-x+3)=-[5x(x-3)-(x-3)]=-[(x-3)(5x-1)]

@Nguyễn Tùng Dương Quang Hào

\(B=\left(\dfrac{1}{x^2-xy}-\dfrac{3y^2}{x^4-xy^3}-\dfrac{y}{x^2+x^2y+xy^2}\right).\left(y+\dfrac{x^2}{x+y}\right)\)

\(B=\left(\dfrac{1}{x\left(x-y\right)}-\dfrac{3y^2}{x\left(x^3-y^3\right)}-\dfrac{y}{x\left(x^2+xy+y\right)}\right).\left(y+\dfrac{x^2}{x+y}\right)\)

\(B=\left(\dfrac{1}{x\left(x-y\right)}-\dfrac{3y^2}{x\left(x-y\right)\left(x^2+xy+y^2\right)}-\dfrac{y}{x\left(x^2+xy+y^2\right)}\right).\left(y+\dfrac{x^2}{x+y}\right)\)

\(B=\left(\dfrac{x^2+xy+y^2}{x\left(x-y\right)\left(x^2+xy+y^2\right)}-\dfrac{3y^2}{x\left(x-y\right)\left(x^2+xy+y^2\right)}-\dfrac{y\left(x-y\right)}{x\left(x^2+xy+y^2\right)}\right).\left(y+\dfrac{x^2}{x+y}\right)\)

\(B=\left(\dfrac{x^2+xy+y^2-3y^2-xy+y^2}{x\left(x-y\right)\left(x^2+xy+y^2\right)}\right).\left(y+\dfrac{x^2}{x+y}\right)\)

\(B=\dfrac{x^2+2y^2-3y^2}{x\left(x-y\right)\left(x^2+xy+y^2\right)}.\left(y+\dfrac{x^2}{x+y}\right)\)

\(B=\dfrac{x^2+2y^2-3y^2}{x\left(x-y\right)\left(x^2+xy+y^2\right)}.\left(\dfrac{y\left(x+y\right)}{x+y}+\dfrac{x^2}{x+y}\right)\)

\(B=\dfrac{x^2+2y^2-3y^2}{x\left(x-y\right)\left(x^2+xy+y^2\right)}.\dfrac{x^2+xy+y^2}{x+y}\)

\(B=\dfrac{x^2+2y^2-3y^2}{x\left(x^2-y^2\right)}\)

P(x)=ax^+bx^2+cx+b

p(x):(x-1) .Q(X)+6

p(x):(x-2).Q(x)+6

p(x):(x-3).Q(x)+6

vì P(X) là bậc 3 và p(X) chia (x-1),(x-2),(x-3) đều dư 6

=>p(x)-6 :(x-1) (p/s: dấu '':'' là chia hết==tớ k viết đc)

p(x)-6: (x-2)

p(x)-6: (x-3)

p(X)-6=m.(x-1)(x-2)(x-3)

mặt khác x=-1 có p(-1)=-18

=>-18-6=m(-1-1)(-1-2)((-1-3)

=>-24=m.(-2)(-3)(-4)

=>-24=m.-24

=>m=1

=>p(x)=x^3-6x^2+11x

bạn có thể nói cho tớ biết bài này bạn lấy ở đâu k ạ??