\(7xy^5\left(x-1\right)-3x^2y^4\left(1-x\right)+5xy^3\left(x-1\right)\)

\(=7xy^5\left(x-1\right)+3x^2y^4\left(x-1\right)+6xy^3\left(x-1\right)\)

\(=\left(x-1\right)\left(7xy^5+3x^2y^4-6xy^3\right)=xy\left(x-1\right)\left(7y^4+3xy^3-6y^2\right)\)

 Trả lời:

7xy⁵(x - 1) - 3x²y⁴(1 - x) + 5xy³(x - 1)

= 7xy⁵(x - 1) + 3x²y⁴(x - 1) + 5xy³(x - 1)

= (7xy⁵ + 3x²y⁴ + 5xy³)(x - 1)

= xy(7y⁴ + 3xy³ + 5y²)(x - 1)

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

\(\Leftrightarrow\left[\left(x+1\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x+2\right)\right]+4x^2=0\)

\(\Leftrightarrow\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2=0\)

Nếu x = 0 thì PT vô nghiệm

Nếu x khác 0, chia cả 2 vế cho x² ta được:

\(PT\Leftrightarrow\left(x-\frac{8}{x}-7\right)\left(x-\frac{8}{x}-2\right)+4=0\)

Đặt \(x-\frac{8}{x}=b\) khi đó: 

\(\left(b-7\right)\left(b-2\right)+4=0\)

\(\Leftrightarrow b^2-9b+14+4=0\)

\(\Leftrightarrow b^2-9b+18=0\)

\(\Leftrightarrow\left(b-3\right)\left(b-6\right)=0\Leftrightarrow\orbr{\begin{cases}b-3=0\\b-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}b=3\\b=6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{8}{x}=3\\x-\frac{8}{x}=6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-8=3x\\x^2-8=6x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-3x-8=0\\x^2-6x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3\pm\sqrt{41}}{2}\\x=3\pm\sqrt{17}\end{cases}}\)

Vậy ...

3) \(\left(x-2\right)\left(x-4\right)\left(x-5\right)\left(x-10\right)-54x^2=0\)

\(\Leftrightarrow\left[\left(x-2\right)\left(x-10\right)\right]\left[\left(x-4\right)\left(x-5\right)\right]-54x^2=0\)

\(\Leftrightarrow\left(x^2-12x+20\right)\left(x^2-9x+20\right)-54x^2=0\)

Nếu x = 0 thì PT vô nghiệm

Nếu x khác 0 thì chia cả 2 vế cho x² ta được:
\(PT\Leftrightarrow\left(x+\frac{20}{x}-12\right)\left(x+\frac{20}{x}-9\right)-54=0\)

Đặt \(x+\frac{20}{x}=c\) nên khi đó:

\(\left(c-12\right)\left(c-9\right)-54=0\)

\(\Leftrightarrow c^2-21c+108-54=0\)

\(\Leftrightarrow c^2-21c+54=0\)

\(\Leftrightarrow\left(c-3\right)\left(c-18\right)=0\Leftrightarrow\orbr{\begin{cases}c-3=0\\c-18=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}c=3\\c=18\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{20}{x}=3\\x+\frac{20}{x}=18\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+20=3x\\x^2+20=18x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-3x+20=0\\x^2-18x+20=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-\frac{3}{2}\right)^2=-\frac{71}{4}\left(ktm\right)\\x=9\pm\sqrt{61}\end{cases}}\)

Vậy ...

Đề có sai không cậu ? Tớ nghĩ x=.../2bc chứ

Cho \(x=\frac{b^2+c^2-a^2}{2ab},y=\frac{a^2-\left(b-c\right)^2}{\left(b+c\right)^2-a^2}\)

Tính giá trị của:

\(M=\frac{x+y}{1-xy}\)

1, Xét tam giác AEB và tam giác AFC có

Góc AEB= Góc AFC= 90

góc A chung

Suy ra tam giác AEB đồng dạng vs tam giác AFC 

 suy raAE/AF=AB/AC

suy ra AE. AC=AF. AB

 2,nối E-F

Xét tam giác AEF và  tam giác ABC có

A chung 

AF/AC= AE/ AB (T/c)

suy ra tâm giác AEF đồng dạng tam giác ABC

suy ra AEF= ABC

dùng bđt phụ \(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{\left(x+y\right)^2}{a+b}\) với bđt Cô-si nhé

Ta có : \(x+y+z=0\)

\(\Rightarrow\hept{\begin{cases}x=-\left(y+z\right)\\y=-\left(z+x\right)\\z=-\left(x+y\right)\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x^2=\left(y+z\right)^2\\y^2=\left(z+x\right)^2\\z=\left(x+y\right)^2\end{cases}}\)

\(\Rightarrow ax^2+by^2+cz^2=a\left(y+z\right)^2+b\left(z+x\right)^2+c\left(x+y\right)^2\)

                                       \(=ay^2+az^2+bz^2+bx^2+cx^2+cy^2+2\left(ayz+bzx+cxy\right)\) 

                                       \(=x^2\left(b+c\right)+y^2\left(c+a\right)+z^2\left(a+b\right)+2\left(ayz+bzx+cxy\right)\left(1\right)\)

Từ \(a+b+c=0\)                    \(\Rightarrow\hept{\begin{cases}b+c=-a\\c+a=-b\\a+b=-c\end{cases}}\) 

Thay vào \(\left(1\right)\), ta được :

\(ax^2+by^2+cz^2=-ax^2-by^2-cz^2+2\left(ayz+bzx+cxy\right)\)

Ta có : \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)\(\Rightarrow ayz+bzx+cxy=0\)

\(\Rightarrow ax^2+by^2+cz^2=-ax^2-by^2-cz^2\)

\(\Rightarrow2\left(ax^2+by^2+cz^2\right)=0\)

\(\Rightarrow ax^2+by^2+cz^2=0\left(đpcm\right)\)

Ta có : \(x^2+2y+1=0;y^2+2z+1=0;z^2+2x+1=0\)

\(\Rightarrow x^2+2y+1=y^2+2z+1=z^2+2x+1\)

\(\Rightarrow x^2+2y+1-y^2-2z-1-z^2-2x-1=0\)

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

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

\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-1\right)^2=0\\\left(z+1\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-1=0\\y-1=0\\z+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=1\\z=-1\end{cases}}\)

Thay \(x=1;y=1;z=-1\)vào A ta có :

\(A=1^{2015}+1^{2016}+\left(-1\right)^{2017}=1+1-1=1\)

Vậy A = 1

Từ \(\hept{\begin{cases}x^2+2y+1=0\\y^2+2z+1=0\\z^2+2x+1=0\end{cases}}\)

\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0\)

\(\Rightarrow\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+\left(z^2+2z+1\right)=0\)

\(\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\left(1\right)\)

Vì \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\\\left(z+1\right)^2\ge0\forall z\end{cases}\left(2\right)}\)

Từ \(\left(1\right)\)và \(\left(2\right)\):

\(\Rightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y+1\right)^2=0\\\left(z+1\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+1=0\\y+1=0\\z+1=0\end{cases}}\)

\(\Rightarrow x=y=z=-1\)

\(\Rightarrow A=\left(-1\right)^{2015}+\left(-1\right)^{2016}+\left(-1\right)^{2017}=-1+1-1=-1\)

Vậy \(A=-1\)

Áp dụng cái bổ đề nhưu ở lần trước mình CM cho bạn ý

Nếu \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\) thì ta có:

\(3\left(x^2-6x+9\right)\left(1-x^2\right)\left(6x-10\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2\left(1-x\right)\left(1+x\right)\left(6x-10\right)=0\)

\(\Rightarrow x\in\left\{-1;1;3;\frac{5}{3}\right\}\)

bạn Đoạt Mệnh Thương giải rõ hơn đc ko

Ta có : \(x^2+2xy+x+y^2+4y=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+y=-\left(x+4y\right)\left(2\right)\\x+y=x+4y\left(3\right)\end{cases}}\)

• Từ \(\left(2\right)\): \(\Leftrightarrow2x=-5y\)

\(\Leftrightarrow\frac{x}{-5}=\frac{y}{2}\)

Đặt \(\frac{x}{-5}=\frac{y}{2}=k\left(k\inℤ\right)\)

Vì \(\left(x+y\right)^2\ge0\)nên từ \(\left(1\right)\):

\(\Rightarrow-\left(x+4y\right)\ge0\)

\(\Rightarrow x+4y\le0\)

\(\Rightarrow x\le-4y\left(4\right)\)

Khi đó : \(\hept{\begin{cases}x=-5k\\y=2k\end{cases}}\)

\(\Rightarrow\left(4\right)\Leftrightarrow-5k\le-8k\)

\(\Rightarrow3k\le0\)

\(\Rightarrow k\le0\)

• Từ \(\left(3\right)\): \(\Leftrightarrow3y=0\)

\(\Leftrightarrow y=0\)

Khi đó, PT tương đương với : \(x^2+x=0\)

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

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

Vậy tập nghiệm (x;y) của PT là : \(S=\left\{\left(0;0\right);\left(0;-1\right);\left(-5k;2k\right)\right\}\)\(\left(k\inℤ,k\le0\right)\)