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1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
\(a,\left\{{}\begin{matrix}\left|x-3y\right|\ge0\\\left|y+4\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-3y=0\\y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3y=-12\\y=-4\end{matrix}\right.\)
\(b,Sửa:\left|x-y-5\right|+\left(y+3\right)^2=0\\ \left\{{}\begin{matrix}\left|x-y-5\right|\ge0\\\left(y+3\right)^2\ge0\end{matrix}\right.\Rightarrow VT\ge0\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-y-5=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+5=2\\y=-3\end{matrix}\right.\)
\(c,\left\{{}\begin{matrix}\left|x+y-1\right|\ge0\\\left(y-2\right)^4\ge0\end{matrix}\right.\Rightarrow VT\ge0\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+y-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-y=-1\\y=2\end{matrix}\right.\)
\(d,\left\{{}\begin{matrix}\left|x+3y-1\right|\ge0\\3\left|y+2\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+3y-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-3y=7\\y=-2\end{matrix}\right.\)
\(e,Sửa:\left|2021-x\right|+\left|2y-2022\right|=0\\ \left\{{}\begin{matrix}\left|2021-x\right|\ge0\\\left|2y-2022\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}2021-x=0\\2y-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\y=1011\end{matrix}\right.\)
\(\Rightarrow-3< x< 2\\ \Rightarrow x\in\left\{-2;-1;0;1\right\}\\ \Rightarrow B\)
a) x - 8 - (12 - 2x) = -20
=> x - 8 - 12 + 2x = -20
=> (x + 2x) + (-8 - 12) = -20
=> 3x - 20 = -20
=> 3x = 0 => x = 0
b) -27 + (x + 8) - ( +11) = 2
=> -27 + x + 8 - 11 = 2
=> -27 + x = 2 + 11 - 8
=> -27 + x = 5
=> x = 5 - (-27) = 32
c) -2x - 16 = -2 - (3x + 9)
=> -2x - 16 = -2 - 3x - 9
=> -2x - 16 + 2 + 3x + 9 = 0
=> (-2x + 3x) + (-16 + 2 + 9) = 0
=> x - 5 = 0
=> x = 5
\(a,x-8-\left(12-2x\right)=-20\)
\(x-8-12+2x=-20\)
\(x+2x-8-12=-20\)
\(3x-20=-20\)
\(3x=-20+20\)
\(3x=0\)
\(x=0\)
\(b,-27+\left(x+8\right)-\left(+11\right)=2\)
\(-27+x+8-11=2\)
\(x-27+8-11=2\)
\(x-30=2\)
\(x=2+30\)
\(x=32\)
\(c,-2x-16=2-\left(3x+9\right)\)
\(-2x-16=2-3x-9\)
\(-2x+3x=2-9+16\)
\(x=9\)
Học tốt
a) Liệt kê
x = {-7;-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6;7}
Tính tổng là: -7+-6+-5+-4+.....+4+5+6+7
= (-7+7)+(-6+6)+(-5+5)+....+(-1+1)+0
= 0+0+0....+0
= 0
b) Liệt kê
x = {-5;-4;-3;-2;-1;0;1;2;3}
Tính tổng: -5+-4+-3+-2+-2+0+1+2+3
= (-3+3)+(-2+2)+(-1+1)+0+-5+-4
= 0+0+0+0+ -9
= -9
c) Liệt kê:
x = { -19;-18;-17;-16;....;18;19;20}
Tính tổng: -19+-18+-17+-16+....+15+16+17+18+19+20
= (-19+19)+(-18+18)+...+(-1+1)+0+20
= 0 + 0+...+0+20
= 20
*TÌM X:
a) 2x -35 = 15
2x = 15 + 35
2x = 50
x = 50 :2
x = 25
b) 3x + 17 = 2
3x = 17+2
3x = 19
x = 19 : 3
x = 6,33
c) /x-1/ = 0
\(\hept{\begin{cases}x-1=0\\x-1=-0\left(loai\right)\end{cases}}\)
Vậy x-1 = 0
x = 0 +1 = 1