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 : \(\Leftrightarrow x=-\dfrac{1}{2}\)

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`

\(A=\left(x-1\right)\left(x-8\right)\left(x-4\right)\left(x-5\right)+2002\)

\(\Leftrightarrow A=\left(x^2-9x+8\right)\left(x^2-9x+20\right)+2002\)

Đặt \(x^2-9x+14=y\)

\(\Rightarrow A=\left(y-6\right)\left(y+6\right)+2002\)

\(\Leftrightarrow A=y^2-36+2002\)

\(\Leftrightarrow A=y^2+1966\ge1966\)

Dấu "=" xảy ra khi

 \(x^2-9x+14=0\)

\(\Leftrightarrow x=2,7\)

bó tay

-100 taij x=0

a) \(3x\left(2x+1\right)=5\left(2x+1\right)\)

\(3x=5\)

\(x=\frac{5}{3}\)

b) \(\left(3x-8\right)^2=\left(2x-7\right)^2\)

\(3x-8=2x-7\)

\(x=1\)

c) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)

\(\left(4x^2-3x-18\right)^2=\left(4x^2+3x\right)^2\)

\(4x^2-3x-18=4x^2+3x\)

\(6x=-18\)

\(x=-3\)

d) Sai đề

e) ko bt

a) \(\left|x+2\right|+\left|x-3\right|=7\)

Lập bảng xét dấu:

* Nếu \(x< -2\) thì pttt:

\(-x-2-x+3=7\)

\(\Leftrightarrow-2x+1=7\)

\(\Leftrightarrow-2x=6\)

\(\Leftrightarrow x=-3\left(tm\right)\)

* Nếu \(-2\le x\le3\) thì pttt:

\(x+2-x+3=7\)

\(\Leftrightarrow5=7\) ( vô lí )

* Nếu \(x>3\) thì pttt:

\(x+2+x-3=7\)

\(\Leftrightarrow2x-1=7\)

\(\Leftrightarrow2x=8\)

\(\Leftrightarrow x=4\left(tm\right)\)

Vậy phương trình có tập nghiệm \(S=\left\{-3;4\right\}\)

b) \(\left|x+2\right|-6x=1\)

* Nếu \(x+2>0\Leftrightarrow x>2\) thì pttt:

\(x+2-6x=1\)

\(\Leftrightarrow-6x=-1\)

\(\Leftrightarrow x=1\left(ktm\right)\)

* Nếu \(x+2< 0\Leftrightarrow x< 2\) thì pttt:

\(-x-2-6x=1\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=-\dfrac{3}{7}\left(tm\right)\)

Vậy pt có tập nghiệm \(S=\left\{\dfrac{-3}{7}\right\}\)

c) \(\left|2x+1\right|+\left|x+8\right|=4x\)

Lập bảng xét dấu: (Dấu .. là khoảng cách, không cần phải ghi)

* Nếu \(x< -8\) thì pttt:

\(-2x-1-x-8=4x\)

\(\Leftrightarrow-2x-x-4x=8+1\)

\(\Leftrightarrow-7x=9\)

\(\Leftrightarrow x=\dfrac{-9}{7}\left(ktm\right)\)

* Nếu \(-8< x< \dfrac{-1}{2}\) thì pttt:

\(-2x-1+x+8=4x\)

\(\Leftrightarrow-5x=-7\)

\(\Leftrightarrow x=\dfrac{7}{5}\left(ktm\right)\)

* Nếu \(x>\dfrac{-1}{2}\) thì pttt:

\(2x+1+x+8=4x\)

\(\Leftrightarrow2x+x-4x=-8-1\)

\(\Leftrightarrow-x=-9\)

\(\Leftrightarrow x=9\left(tm\right)\)

Vậy phương trình có tập nghiệm \(S=\left\{9\right\}\)