a: \(\left(x^2-2x+1\right)\left(x^2+bx+c\right)\)

\(=x^4+bx^3+cx^2-2x^3-2b\cdot x^2-2x\cdot c+x^2+bx+c\)

\(=x^4+x^3\left(b-2\right)+x^2\left(c-2b+1\right)+x\left(-2+b\right)+c\)

Theo đề, ta có: b-2=-2; c-2b+1=2; b-2=-2; a=c

=>b=0; c=1; a=c=1

b: (x-2)(x^2+bx+c)+a

\(=x^3+bx^2+cx-2x^2-2bx-2c+a\)

\(=x^3+x^2\left(b-2\right)+x\left(c-2b\right)-2c+a\)

Theo đề ta có: b-2=3; c-2b=-1; -2c+a=-3

=>b=5; c=-1+2b=-1+10=9; a=-3+2c=-3+2*9=15

a: ta có: \(\dfrac{\left(x+2\right)^2}{2}+\dfrac{\left(2x+1\right)^2}{4}+\dfrac{\left(2x-1\right)^2}{8}-\left(x+1\right)^2=0\)

\(\Leftrightarrow4\left(x^2+4x+4\right)+2\left(4x^2+4x+1\right)+4x^2-4x+1-8\left(x+1\right)^2=0\)

\(\Leftrightarrow4x^2+16x+16+8x^2+8x+2+4x^2-4x+1-8\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow16x^2+20x+19-8x^2-16x-8=0\)

\(\Leftrightarrow8x^2+4x+11=0\)

\(\text{Δ}=4^2-4\cdot8\cdot11=-336< 0\)

Vì Δ<0 nên phương trình vô nghiệm

b.

PT \(\Leftrightarrow \frac{x^2+2x+1}{2}-\frac{4x^2-4x+1}{3}+\frac{4x^2+4x+1}{4}-\frac{x^2-10x+25}{6}=0\)

\(\Leftrightarrow \left(\frac{x^2+2x+1}{2}+\frac{4x^2+4x+1}{4}\right)-\left(\frac{4x^2-4x+1}{3}+\frac{x^2-10x+25}{6}\right)=0\)

\(\Leftrightarrow \frac{6x^2+8x+3}{4}-\frac{9x^2-18x+27}{6}=0\)

\(\Leftrightarrow \frac{3(6x^2+8x+3)-2(9x^2-18x+27)}{12}=0\)

$\Leftrightarrow 5x-\frac{15}{4}=0$

$\Leftrightarrow x=\frac{3}{4}$

c.

PT $\Leftrightarrow (x^3+9x^2+27x+27)-(3x^3+12x^2)+(x^3+6x^2+12x+8)=(-x^3+3x^2-3x+1)-8$

$\Leftrightarrow 42x+42=0$

$\Leftrightarrow x=-1$

\(a.\Leftrightarrow x^2+x-6+2x^2+4x+2=x^2-6x+9-2x^2+4x\)

\(\Leftrightarrow4x^2+7x-13=0\)(pt vô nghiệm)

\(b.\Leftrightarrow x^3+3x^2+3x+1-x^2+2x+8=x^3-8+2x^2\)

\(\Leftrightarrow5x=-17\Rightarrow x=\frac{-17}{5}\)

Đặt \(t=x^2+2x+2\left(t\ge1\right)\)

\(c.\Leftrightarrow\frac{t-1}{t}+\frac{t}{t+1}=\frac{7}{6}\)\(\Leftrightarrow\frac{t^2-1+t^2}{t^2+t}=\frac{7}{6}\)\(\Leftrightarrow12t^2-6=7t^2+7t\)

\(\Leftrightarrow5t^2-7t-6=0\Rightarrow\orbr{\begin{cases}t=2\left(tm\right)\\t=\frac{-3}{5}\left(l\right)\end{cases}}\)

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

1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

=>-8x=6-14=-8

=>x=1

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

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`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

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`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

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`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

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`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

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`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

Bạn xem lại đề nhé.

a) \(A=x^2+5y^2+2xy-4x-8y+2015\)

\(A=x^2-4x+4-2y\left(x-2\right)+y^2+2011+4y^2\)

\(A=\left(x-2\right)^2-2y\left(x-2\right)+y^2+2011+4y^2\)

\(A=\left(x-2-y\right)^2+4y^2+2011\)

Vì \(\left(x-y-2\right)^2\ge0;4y^2\ge0\)

\(\Rightarrow A_{min}=2011\)

Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}x-y-2=0\\4y^2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)

b) \(B=\left(x-2012\right)^2+\left(x+2013\right)^2\)

\(B=x^2-4024x+2012^2+x^2+4026x+2013^2\)

\(B=2x^2+2x+2012^2+2013^2\)

\(B=2\left(x^2+x+\dfrac{1}{4}\right)+2012^2+2013^2-\dfrac{1}{2}\)

\(B=2\left(x+\dfrac{1}{2}\right)^2+2012^2+2013^2-\dfrac{1}{2}\)

\(\Rightarrow B_{min}=2012^2+2013^2-\dfrac{1}{2}\)

Dấu bằng xảy ra : \(\Leftrightarrow x=-\dfrac{1}{2}\)

`a)3x(2x^2-3x+4)`

`=6x^3-9x^2+12x`

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`b)(x+3)^2+(3x-2)(x+4)`

`=x^2+6x+9+3x^2+12x-2x-8`

`=4x^2+16x+1`

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`c)[2x-4]/[x-1]+[2x+2]/[x^2-1]`       `ĐK: x \ne +-1`

`=[(2x-4)(x+1)+2x+2]/[(x-1)(x+1)]`

`=[2x^2+2x-4x-4+2x+2]/[(x-1)(x+1)]`

`=[2x^2-2]/[x^2-1]`

`=2`

hếp

a) \(x-2=\left(x-2\right)^2\)

\(\left(x-2\right)^2-\left(x-2\right)=0\)

\(\left(x-2\right)\left(x-2-1\right)=0\)

\(\left(x-2\right)\left(x-3\right)=0\)

\(\Rightarrow x-2=0\) hoặc \(x-3=0\)

*) \(x-2=0\)

\(x=2\)

*) \(x-3=0\)

\(x=3\)

Vậy \(x=2;x=3\)

b) \(x+5=2\left(x+5\right)^2\)

\(2\left(x+5\right)^2-\left(x+5\right)=0\)

\(\left(x+5\right)\left[2\left(x+5\right)-1\right]=0\)

\(\left(x+5\right)\left(2x+10-1\right)=0\)

\(\left(x+5\right)\left(2x+9\right)=0\)

\(\Rightarrow x+5=0\) hoặc \(2x+9=0\)

*) \(x+5=0\)

\(x=-5\)

*) \(2x+9=0\)

\(2x=-9\)

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

Vậy \(x=-5;x=-\dfrac{9}{2}\)

c) \(\left(x^2+1\right)\left(2x-1\right)+2x=1\)

\(\left(x^2+1\right)\left(2x-1\right)+2x-1=0\)

\(\left(x^2+1\right)\left(2x-1\right)+\left(2x-1\right)=0\)

\(\left(2x-1\right)\left(x^2+1+1\right)=0\)

\(\left(2x-1\right)\left(x^2+2\right)=0\)

\(\Rightarrow2x-1=0\) hoặc \(x^2+2=0\)

*) \(2x-1=0\)

\(2x=1\)

\(x=\dfrac{1}{2}\)

*) \(x^2+2=0\) 

\(x^2=-2\) (vô lí)

Vậy \(x=\dfrac{1}{2}\)

d) Sửa đề:

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

\(\left(x^2+3\right)\left(x+1\right)+\left(x+1\right)=0\)

\(\left(x+1\right)\left(x^2+3+1\right)=0\)

\(\left(x+1\right)\left(x^2+4\right)=0\)

\(\Rightarrow x+1=0\) hoặc \(x^2+4=0\)

*) \(x+1=0\)

\(x=-1\)

*) \(x^2+4=0\)

\(x^2=-4\) (vô lí)

Vậy \(x=-1\)