1-2+3-4-(5)-6-(-7)-8-(-9)-10-..........-(-2009)-2010-(-2011))
[3.(-2)-(-8)].(-7)-(-2).(-5)
2,
(2x - 3)2+5x-6 với x=-3
x2-2.x.y+y2 với x=-2, y=3
giá trị tuyệt đói của x+ 2.x.y-y2 với x=-5 y=3
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( Mik làm mấy phần mà bạn dưới chưa làm)
11) xy+x+y=9
\(\Leftrightarrow\) xy+x+y+1=9+1
\(\Leftrightarrow\left(xy+x\right)+\left(y+1\right)\)=10
\(\Leftrightarrow x\left(y+1\right)+\left(y+1\right)=10\)
\(\Leftrightarrow\) (x+1)(y+1)=10=1.10=10.1=-1.-10=-10.-1=2.5=5.2=-2.-5=-5.-2
\(\Rightarrow\) TH1: x+1=1 ; y+1=10
\(\Leftrightarrow x=0;y=9\)
TH2: x+1=10;y+1=1
\(\Leftrightarrow\)x=9;y=0
TH3: x+1=-1;y+1=-10
\(\Leftrightarrow\) x=-2;y=-11
...........
Vậy:........
( Bạn tự làm nốt chứ dài quá, mik chỉ hướng dẫn cách làm bài thôi)
1) -x = -7
=> x = 7
2) - x = 17
=> x = - 17
3) |x| = 17
=> x = ±17
4) -(-x) = |-17|
=> x = 17
5) - 19 - x = 17
=> - x = 17 + 19
=> x = - 36
6) - 19 - x = - 17
=> - x = - 17 + 19
=> -x = 2
=> x = - 2
7) - 5 - (10 - x) = 7
=> - 5 - 10 + x = 7
=> - 15 + x = 7
=> x = 7 + 15
=> x = 22
8) |x + 3| + 7 = 12
=> |x + 3| = 12 - 7
=> |x + 3| = 5
=> x + 3 = 5 hoặc x + 3 =- 5
=> x = 2 hoặc x = - 8
9) 2 - |x - 2| = x
=> - |x - 2| - x = - 2
TH1: x >= 2
- (x - 2) - x = - 2
=> - x + 2 - x =- 2
=> - 2x = - 4
=> x = 2 (nhận)
TH2: x < 2
-[-(x - 2)] - x = - 2
=> x - 2 - x = - 2
=> 0x = 0 (vô số nghiệm)
\(x^2=1\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
\(x^2=3\Rightarrow\left[{}\begin{matrix}x=-\sqrt{3}\\x=\sqrt{3}\end{matrix}\right.\)
\(x^2=5\Rightarrow\left[{}\begin{matrix}x=-\sqrt{5}\\x=\sqrt{5}\end{matrix}\right.\Rightarrow x=-\sqrt{5}\left(vì.x< 0\right)\)
\(x^2=7\Rightarrow\left[{}\begin{matrix}x=-\sqrt{7}\\x=\sqrt{7}\end{matrix}\right.\Rightarrow x=-\sqrt{7}\left(vì.x< 0\right)\)
\(x^2=9\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
\(\left(x-2\right)^2=2\Rightarrow\left[{}\begin{matrix}x-2=-\sqrt{2}\\x-2=\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2-\sqrt{2}\\x=2+\sqrt{2}\end{matrix}\right.\)
\(\left(x-4\right)^2=4\Rightarrow\left[{}\begin{matrix}x-2=-2\\x-2=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\left(x-6\right)^2=6\Rightarrow\left[{}\begin{matrix}x-6=-\sqrt{6}\\x-6=\sqrt{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6-\sqrt{6}\\x=6+\sqrt{6}\end{matrix}\right.\)
\(\left(x-8\right)^2=8\Rightarrow\left[{}\begin{matrix}x-8=-2\sqrt{2}\\x-8=2\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8-2\sqrt{2}\\x=2+2\sqrt{2}\end{matrix}\right.\)
\(\left(x-10\right)^2=10\Rightarrow\left[{}\begin{matrix}x-10=-\sqrt{10}\\x-10=\sqrt{10}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-\sqrt{10}\\x=10+\sqrt{10}\end{matrix}\right.\)
\(\left(x-\sqrt{3}\right)^2=3\Rightarrow\left[{}\begin{matrix}x-\sqrt{3}=-\sqrt{3}\\x-\sqrt{3}=\sqrt{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{3}\end{matrix}\right.\)
\(\left(x-\sqrt{5}\right)^2=5\Rightarrow\left[{}\begin{matrix}x-\sqrt{5}=-\sqrt{5}\\x-\sqrt{5}=\sqrt{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{5}\end{matrix}\right.\)
9: \(\left(-2x\right)\left(3x^2-2x+4\right)=-6x^3+4x^2-8x\)
a) \(2xy+2x-y=8\)
\(\Rightarrow\ 2x\left(y+1\right)-\left(y+1\right)=7\)
\(\Leftrightarrow\left(2x-1\right)\left(y+1\right)=7\)
\(\Rightarrow\left[\begin{matrix}\begin{cases}2x-1=-7\\y+1=-1\end{cases}\\\begin{cases}2x-1=-1\\y+1=-7\end{cases}\end{matrix}\right.\left[\begin{matrix}\begin{cases}2x-1=7\\y+1=1\end{cases}\\\begin{cases}2x-1=1\\y+1=7\end{cases}\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}\left[\begin{matrix}\begin{cases}x=4\\y=0\end{cases}\end{matrix}\right.\\\left[\begin{matrix}\begin{cases}x=1\\y=6\end{cases}\\\left[\begin{matrix}\begin{cases}x=-3\\y=-2\end{cases}\\\begin{cases}x=0\\y=-8\end{cases}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
c)\(x^2+xy+x+y=2\)
\(\Leftrightarrow x\left(x+1\right)+y\left(x+1\right)=2\)
\(\Leftrightarrow\left(x+y\right)\left(x+1\right)=2\)
\(\Rightarrow\left[\begin{matrix}\left[\begin{matrix}\begin{cases}x+y=2\\x+1=1\end{cases}\\\begin{cases}x+y=1\\x+1=2\end{cases}\end{matrix}\right.\\\left[\begin{matrix}\begin{cases}x+y=-2\\x+1=-1\end{cases}\\\begin{cases}x+y=-1\\x+1=-2\end{cases}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}\left[\begin{matrix}\begin{cases}x=0\\y=2\end{cases}\\\begin{cases}x=1\\y=0\end{cases}\end{matrix}\right.\\\left[\begin{matrix}\begin{cases}x=-2\\y=0\end{cases}\\\begin{cases}x=-3\\y=2\end{cases}\end{matrix}\right.\end{matrix}\right.\)