đề bài sai nhé, 6x phảy là 6y
\(\Leftrightarrow\left(4x^2+y^2+z^2-4xy-4xz+2yz\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)
\(\Leftrightarrow\left(-2x+y+z\right)^2+\left(y-3\right)^2+\left(z-5\right)^2=0\)
Vì \(\left(-2x+y+z\right)^2\ge0\)
\(\left(y-3\right)^2\ge0\)
\(\left(z-5\right)^2\ge0\)
\(\Rightarrow\left(-2x+y+z\right)^2+\left(y-3\right)^2+\left(z-5\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow y=3;z=5;x=4\)
\(\left(x-4\right)^{2015}+\left(y-4\right)^{2015}+\left(z-4\right)^{2015}=\left(4-4\right)^{2015}+\left(3-4\right)^{2015}+\left(5-4\right)^{2015}=0\)

Phân tích cái trên thành hằng đẳng thức bậc 2 là đc, tìm ra x;y;z rồi thay vào M

làm rõ ra cho tớ được không? Không hiểu sao tớ phân tích không ra :((

Ta có : \(4x^2+2y^2+2z^2-4xy-4zx+2yz-6y-10z+34=0\)

\(\Rightarrow\left(4x^2+y^2+z^2-4xy-4zx+2yz\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)

\(\Rightarrow\left(2x-y-z\right)^2+\left(y-3\right)^2+\left(z-5\right)^2=0\)

Vì \(\hept{\begin{cases}\left(2x-y-z\right)^2\ge0\forall x,y,z\\\left(y-3\right)^2\ge0\forall y\\\left(z-5\right)^2\ge0\forall z\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\left(2x-y-z\right)^2=0\\\left(y-3\right)^2=0\\\left(z-5\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x-3-5=0\\y=3\\z=5\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x=8\\y=3\\z=5\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}\left(1\right)\)

Lại có : \(S=\left(x-4\right)^{2017}+\left(y-4\right)^{2017}+\left(z-4\right)^{2017}\)

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

\(S=0^{2017}+\left(-1\right)^{2017}+1^{2017}\)

    \(=0-1+1=0\)

Vậy \(S=0\)

1/a) 12 - x= 1-(-5)

      12 - x = 6

             x= 12-6

             x=6

 b)| x+4|= 12

x+4 = \(\pm\)12

*x+4=12

     x=8

*x+4= -12

    x=-16

2/Tìm n

\(n-5⋮n+2\)

=> \(n+2-7⋮n+2\)

mà \(n+2⋮n+2\)

=> 7\(⋮\)n+2

=> n+2 \(\varepsilon\)Ư(7)= {1;-1;7;-7}

3/a)4.(-5)² + 2.(-12)

= 2.2.(-5)² + 2.(-12)

=2[2.25.(-12)]

=2.(-600)

=-1200

a.

\(\left\{{}\begin{matrix}x^4+y^4=34\\y=2-x\end{matrix}\right.\)

\(\Rightarrow x^4+\left(x-2\right)^4=34\)

Đặt \(x-1=t\)

\(\Rightarrow\left(t+1\right)^4+\left(t-1\right)^4=34\)

\(\Leftrightarrow t^4+6t^2-16=0\Rightarrow\left[{}\begin{matrix}t^2=2\\t^2=-8\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}t=\sqrt{2}\Rightarrow x=\sqrt{2}+1\Rightarrow y=1-\sqrt{2}\\t=-\sqrt{2}\Rightarrow x=1-\sqrt{2}\Rightarrow y=1+\sqrt{2}\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}xy^2-x^2y+6x-y^2-y-6=0\\x^2y-xy^2+6y-x^2-x-6=0\end{matrix}\right.\) (1)

Lần lượt cộng 2 vế và trừ 2 vế ta được:

\(\left\{{}\begin{matrix}-x^2-y^2+5x+5y-12=0\\2xy\left(y-x\right)+7\left(x-y\right)+\left(x-y\right)\left(x+y\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-5\left(x+y\right)+12=0\\\left(y-x\right)\left(2xy-x-y-7\right)=0\end{matrix}\right.\)

Th1: \(\left\{{}\begin{matrix}x=y\\x^2+y^2-5\left(x+y\right)+12=0\end{matrix}\right.\)

\(\Rightarrow2x^2-10x+12=0\Rightarrow...\)

TH2: \(\left\{{}\begin{matrix}2xy-\left(x+y\right)-7=0\\x^2+y^2-5\left(x+y\right)+12=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2xy-\left(x+y\right)-7=0\\\left(x+y\right)^2-2xy-5\left(x+y\right)+12=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2v-u-7=0\\u^2-2v-5u+12=0\end{matrix}\right.\)

\(\Rightarrow u^2-6u+5=0\)

\(\Leftrightarrow...\)

tìm x a

e xin lỗi em ghi nhầm là lớp 7 chứ ko pk lớp 8 đâu ạ

\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)+\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\frac{31}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow x=31\)

\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Leftrightarrow\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Leftrightarrow\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Leftrightarrow\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Leftrightarrow x+34-x-3=x\)

\(\Leftrightarrow x=31\)

\(ĐKXĐ\): \(x\ne-3\); \(x\ne-10\); \(x\ne-21\); \(x\ne-34\)

\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Leftrightarrow\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Leftrightarrow\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Leftrightarrow\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow\left(x+34\right)-\left(x+3\right)=x\)

\(\Leftrightarrow x+34-x-3=x\)

\(\Leftrightarrow x=31\)( thỏa mãn )

Vậy \(x=31\)