a: =(x-3)(2x+5)

b: \(\Leftrightarrow\left(x-2\right)\left(x+2+3-2x\right)=0\)

=>(x-2)(5-x)=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

TK

c)=\(\left(x-1\right)^3=0\)=>x=1

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

ĐKXĐ: ...

\(\left(\dfrac{x-1}{x+2}\right)^2-4\left(\dfrac{x+2}{x-3}\right)^2+3\left(\dfrac{x-1}{x-3}\right)=0\)

Đặt \(\left\{{}\begin{matrix}\dfrac{x-1}{x+2}=a\\\dfrac{x+2}{x-3}=b\end{matrix}\right.\)

\(\Rightarrow a^2-4b^2+3ab=0\Leftrightarrow\left(a-b\right)\left(a+4b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\a+4b=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}-\dfrac{x+2}{x-3}=0\\\dfrac{x-1}{x+2}+\dfrac{4x+8}{x-3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x-3\right)-\left(x+2\right)^2=0\\\left(x-\right)\left(x-3\right)+4\left(x+2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(\frac{x}{2016}+\frac{x-1}{2015}+\frac{x-2}{2014}+\frac{x-3}{2013}=4\)

\(\Leftrightarrow\left(\frac{x}{2016}-1\right)+\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)+\left(\frac{x-3}{2013}-1\right)=0\)

\(\Leftrightarrow\frac{x-2016}{2016}+\frac{x-2016}{2015}+\frac{x-2016}{2014}+\frac{x-2016}{2013}=0\)

\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}\right)=0\)

Dễ thấy cái vế sau > 0 nên x=2016

Câu b có cách nào hay hơn bằng cách phá ko ta,hóng quá:)

\(125x^3=\left(2x+1\right)^3+\left(3x-1\right)^3\)

\(\Leftrightarrow8x^3+12x^2+6x+1+27x^3-27x^2+9x-1=125x^3\)

\(\Leftrightarrow35x^3-15x^2+15x=125x^3\)

\(\Leftrightarrow90x^3+15x^2-15x=0\)

\(\Leftrightarrow x\left(90x^2+15x-15\right)=0\)

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

\(\Leftrightarrow x=0;x=-\frac{1}{2};x=\frac{1}{3}\)

Câu c có cách giải rất hay đó nha :) 

\(\left(2x-5\right)^3+27\left(x-1\right)^3+\left(8-5x\right)^3=0\)

\(\Leftrightarrow\left(2x-5\right)^3+\left(3x-3\right)^3+\left(8-5x\right)^3=0\)

Đặt \(2x-5=a;3x-3=b;8-5x=c\Rightarrow a+b+c=0\)

\(\Rightarrow a+b=-c\)

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

\(\Leftrightarrow a^3+3ab\left(a+b\right)+b^3=-c^3\)

\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=-3ab\left(-c\right)=3abc\)

Khi đó:

\(\left(2x-5\right)^3+27\left(x-1\right)^3+\left(8-5x\right)^3=0\)

\(\Leftrightarrow3\left(2x-5\right)\left(3x-3\right)\left(8-5x\right)=0\)

\(\Leftrightarrow x=\frac{5}{2};x=1;x=\frac{8}{5}\)

a. \(\frac{x}{2016}+\frac{x-1}{2015}+\frac{x-2}{2014}+\frac{x-3}{2013}=4\)

\(\rightarrow\left(\frac{x}{2016}-1\right)+\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)+\left(\frac{x-3}{2013}-1\right)=0\)

\(\rightarrow\frac{x-2016}{2016}+\frac{x-2016}{2015}+\frac{x-2016}{2014}+\frac{x-2016}{2013}=0\)

\(\rightarrow\left(x-2016\right).\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{1014}+\frac{1}{2013}\right)=0\)

Vì \(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}\ne0\)

\(\rightarrow x-2016=0\)

\(\rightarrow x=2016\)

Vậy ...

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

<=> \(\orbr{\begin{cases}x^2+2x+2=0\\x^2+2x+3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x+1\right)^2+1=0\left(vl\right)\\\left(x+1\right)^2+2=0\left(vl\right)\end{cases}}\)

=> pt vô nghiệm

b) \(\left(x+3\right)\left(x-3\right)\left(x^2-11\right)+3=2\)

<=> \(\left(x^2-9\right)\left(x^2-11\right)+1=0\)

<=> \(\left(x^2-9\right)^2-2\left(x^2-9\right)+1=0\)

<=> \(\left(x^2-9-1\right)^2=0\)

<=> \(x^2-10=0\)

<=> \(x=\pm\sqrt{10}\)

c) \(\left(x+3\right)^4+\left(x+5\right)^4=2\)

<=> \(\left(x+4-1\right)^4+\left(x+4+1\right)^4=2\)

Đặt x + 4 = a

<=> \(\left(a-1\right)^4+\left(a+1\right)^4=2\)

<=> \(a^4-4a^3+6a^2-4a+1+a^4+4a^3+6a^2+4a+1=2\)

<=> \(a^4+12a^2=0\)

<=> \(a^2\left(a^2+12\right)=0\)

<=> a = 0 (vì a² + 12 > 0)

Vậy S = {0}

Làm nốt c 

Đặt \(x+4=t\)

\(\Leftrightarrow\left(t-1\right)^4+\left(t+1\right)^4=2\)

\(\Leftrightarrow\left(t^2+2t+t\right)\left(t^2-2t+1\right)+\left(t+1\right)^4=2\)

\(\Leftrightarrow2t^4+12t^2+2=2\Leftrightarrow2t^4+12t^2=0\)

\(\Leftrightarrow2t^2\left(t^2+6\right)=0\Leftrightarrow t^2=-6\) ( vô lí )

Phương trình vô nghiệm

Bài 1:

a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)

\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)

\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)

Suy ra: \(12x-45-12x^2+45x=0\)

\(\Leftrightarrow-12x^2+57x-45=0\)

\(\Leftrightarrow-12x^2+12x+45x-45=0\)

\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)

\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)

mà \(-3\ne0\)

nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)

b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

\(\Leftrightarrow\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}+\frac{\left(x-3\right)^2}{4}=0\)

\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)

Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)

\(\Leftrightarrow-x^2+16x-39=0\)

\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)

\(\Leftrightarrow x^2-13x-3x+39=0\)

\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)

\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)

Vậy: Tập nghiệm S={3;13}

c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)

\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)

\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)

\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)

Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)

\(\Leftrightarrow-21x^2+26x+11=0\)

\(\Leftrightarrow-21x^2-7x+33x+11=0\)

\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)

Lời giải :

Đặt \(\hept{\begin{cases}x^2+3x-4=a\\2x^2-5x+3=b\end{cases}}\)

\(\Rightarrow a+b=\left(x^2+3x-4\right)+\left(2x^2-5x+3\right)=3x^2-2x-1\)

Khi đó phương trình đã cho trở thành :

\(a^3+b^3=\left(a+b\right)^3\)

\(\Leftrightarrow a^3+b^3=a^3+b^3+3ab.\left(a+b\right)\)

\(\Leftrightarrow3ab.\left(a+b\right)=0\) \(\Rightarrow\orbr{\begin{cases}a+b=0\\ab=0\end{cases}}\)

+) Với \(a+b=0\Rightarrow3x^2-2x-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}\)

+) Với \(ab=0\Rightarrow\left(x^2+3x-4\right).\left(2x^2-5x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+3x-4=0\left(1\right)\\2x^2-5x+3=0\left(2\right)\end{cases}}\)

Pt (1) \(\Leftrightarrow\left(x-1\right)\left(x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)

Pt (2) \(\Leftrightarrow\left(x-1\right)\left(2x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{2}\end{cases}}\)

Vạy phương trình đã cho có tập nghiệm \(S=\left\{-4,-\frac{1}{3},1,\frac{3}{2}\right\}\)