bó tay dù sao mk cũng muốn bạn tick cho mk nha

đa thức chia có bậc 2 nên đa thức dư có bậc không quá 1. vậy đa thức dư có bậc nhất dạng ax+b

Ta có: \(x^{67}+x^{47}+x^{27}+x^7+x+1=\left(x^2-1\right).Q\left(x\right)+ax+b\)

Cho x=1 rồi x=-1 ta được: \(\hept{\begin{cases}1+1+1+1+1+1=a+b\\-1-1-1-1-1+1=-a+b\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=6\\-a+b=-4\end{cases}\Leftrightarrow\hept{\begin{cases}a=5\\b=1\end{cases}}}\)

Vậy dư trong phép chia trên là 5x+1

a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)

\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{-2}{3}\)

b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)

\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{-11}{20}\)

c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)

\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)

\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)

\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{-4}{5}\)

d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)

\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{2}{9}\)

e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)

\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)

\(\Rightarrow x=\dfrac{-7}{2}\)

Ta có :

\(\frac{x-3}{97}+\frac{x-27}{73}+\frac{x-67}{33}+\frac{x-73}{27}=4\)

\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)

\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)

Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}>0\) Nên \(x-100=0\)

\(\Leftrightarrow x=100\)

Vậy \(x=100\)

\(\Leftrightarrow\frac{x-3}{87}+\frac{x-27}{79}+\frac{x-67}{33}+\frac{x-73}{27}-4=0\)

\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)

\(\Leftrightarrow\left(\frac{x-3-97}{97}\right)+\left(\frac{x-27-73}{73}\right)+\left(\frac{x-67-33}{33}\right)+\left(\frac{x-73-27}{27}\right)=0\)

\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)

Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\ne0\)

\(\Rightarrow x-100=0\Leftrightarrow x=100\)

\(\frac{x-3}{97}+\frac{x-27}{73}+\frac{x-67}{33}+\frac{x-73}{27}=4\)    

\(\frac{x-3}{97}-1+\frac{x-27}{73}-1+\frac{x-67}{33}-1+\frac{x-73}{27}-1=4-1-1-1-1\)     

\(\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)    

\(\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)   

\(x-100=0\) 

\(x=100\)

`a)2x^2+3(x-1)(x+1)=5x(x+1)`

`<=>2x^2+3x^2-3=5x^2+5x`

`<=>5x=-3`

`<=>x=-3/5`

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`b)(x-3)^3+3-x=0` nhỉ?

`<=>(x-3)^3-(x-3)=0`

`<=>(x-3)(x^2-1)=0`

`<=>[(x=3),(x^2=1<=>x=+-1):}`

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`c)5x(x-2000)-x+2000=0`

`<=>5x(x-2000)-(x-2000)=0`

`<=>(x-2000)(5x-1)=0`

`<=>[(x=2000),(x=1/5):}`

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`d)3(2x-3)+2(2-x)=-3`

`<=>6x-9+4-2x=-3`

`<=>4x=2`

`<=>x=1/2`

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`e)x+6x^2=0`

`<=>x(1+6x)=0`

`<=>[(x=0),(x=-1/6):}`

(x+1)(x+3)(x+5)(x+7)+2002

=(x+1)(x+7)(x+3)(x+5)+2004

=(x^2+8x+7)(x^2+8x+15)+2004

đặt x^2+8x+11=t

=> (t-4)(t+4)+2004

=t^2-16+2004

=t^2+1988

=x^2+8x+11+1988

=x^2+8x+1999

(x^2+8x+1999 ):(x^2+8x+1)=1 dư 1998 (chia đa thức )

vậy số dư là 1998

có j ko hiểu thì cứ hỏi nha ^^

Bạn ơi bạn đặt t = x² + 8x + 11

chứ có phải t² = x² + 8x + 11 

đâu bạn 

1) \(A=36x^2+12x+1=\left(6x+1\right)^2\ge0\)

\(minA=0\Leftrightarrow x=-\dfrac{1}{6}\)

2) \(B=9x^2+6x+1=\left(3x+1\right)^2\ge0\)

\(minB=0\Leftrightarrow x=-\dfrac{1}{3}\)

4) \(D=x^2-4x+y^2-8y+6=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\) 

\(minD=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

3) \(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)

\(minC\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

5) \(E=\left(x-8\right)^2+\left(x+7\right)^2=2x^2-2x+113=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\)

\(minE=\dfrac{225}{2}\Leftrightarrow x=\dfrac{1}{2}\)

e: Ta có: \(E=\left(x-8\right)^2+\left(x+7\right)^2\)

\(=x^2-16x+64+x^2+14x+49\)

\(=2x^2-2x+113\)

\(=2\left(x^2-x+\dfrac{113}{2}\right)\)

\(=2\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{225}{4}\right)\)

\(=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)