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a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
a,A=|x-7|+12
Vì \(\left|x-7\right|\ge0\forall x\)nên \(\left|x-7\right|+12\ge12\forall x\)
Ta thấy A=12 khi |x-7| = 0 => x-7 = 0 => x = 7
Vậy GTNN của A là 12 khi x = 7
b,B=|x+12|+|y-1|+4
Vì \(\left|x+12\right|\ge0\forall x\)
\(\left|y-1\right|\ge0\forall y\)
nên \(\left|x+12\right|+\left|y-1\right|\ge0\forall x,y\)
\(\Rightarrow\left|x+12\right|+\left|y-1\right|+4\ge4\forall x,y\)
Ta thấy B = 4 khi \(\hept{\begin{cases}\left|x+12\right|=0\\\left|y-1\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x+12=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-12\\y=1\end{cases}}\)
Vậy GTNN của B là 4 khi x = -12 và y = 1
\(1a,A=\left|5-x\right|+\left|y-2\right|-3\)
Vì \(\left|5-x\right|\ge vs\forall x,\left|y-2\right|\ge vs\forall y\Rightarrow A\ge3\)
Dấu \("="\) xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|5-x\right|=0\\\left|y-2\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}5-x=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\y=2\end{cases}}\)
Vậy \(A_{min}=3\Leftrightarrow x=5,y=2\)
\(b,B=\left|4-2x\right|+y^2+\left(2-1\right)^2-6\)
\(=\left|4-2x\right|+y^2-5\)
Vì \(\left|4-2x\right|\ge vs\forall x;y^2\ge0vs\forall y\Rightarrow B\ge-5\)
Dấu \("="\) xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|4-2x\right|=0\\y^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}4-2x=0\\y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=0\end{cases}}\)
Vậy \(B_{min}=-5\Leftrightarrow x=2,y=0\)
\(c,C=\frac{1}{2}-\left|x-2\right|\) ( bn xem lại đề nhé )
Bài 1 d)
\(1-2+3-4+5-6+...+2013\)
\(=1+\left(-2+3\right)+\left(-4+5\right)+...+\left(-2012+2013\right)\)
\(=1+1+1+...+1\left(1006s\right)\)
\(=1006.1=1006\)
1. \(3-|2x+1|=-5\)
\(\Rightarrow|2x+1|=8\)
\(\Rightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{2};-\frac{9}{2}\right\}\)
2.\(12+|3-x|=9\)
\(\Rightarrow|3-x|=-3\)
Mà \(|3-x|\ge0\forall x\)
\(\Rightarrow\)Vô lí
Vậy không có x
3.\(|x+9|=12+\left(-9\right)+2\)
\(\Rightarrow|x+9|=5\)
\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)
Vậy \(x\in\left\{-4;-14\right\}\)
4.\(5x-16=40+x\)
\(\Rightarrow5x-x=40+16\)
\(\Rightarrow4x=56\)
\(\Rightarrow x=14\)
Vậy \(x=14\)
5.\(5x-7=-21-2x\)
\(\Rightarrow5x+2x=-21+7\)
\(\Rightarrow7x=-14\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
6.\(\left(2x-1\right)\left(y-2\right)=12\)
Vì \(x,y\inℤ\)nên \(2x-1;y-2\inℤ\)
\(\Rightarrow2x-1;y-2\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Ta có bảng : (em tự xét bảng nhé)
a) \(\left(\frac{2x}{5}-1\right):\left(-5\right)=\frac{1}{7}\)
\(\frac{2x}{5}-1=\frac{1}{7}.\left(-5\right)\)
\(\frac{2x}{5}-1=\frac{-5}{7}\)
\(\frac{2x}{5}=\frac{-5}{7}+\frac{7}{7}\)
\(\frac{2x}{5}=\frac{2}{7}\)
\(=>2x.7=2.5\)
\(=>14x=10\)
\(=>x=\frac{5}{7}\)
c) \(\left|3,5+2,5x\right|-2,5=3,5\)
\(\left|3,5+2,5x\right|=3,5+2,5\)
\(\left|3,5+2,5x\right|=6\)
\(TH1\) \(3,5+2,5x=6\) \(TH2\) \(3,5+2,5x=-6\)
\(2,5x=6-3,5\) \(2,5x=-6-3,5\)
\(2,5x=2,5\) \(2,5x=-9.5\)
\(x=1\) \(x=-3,8\)
vậy \(x=1\) hoặc \(x=-3,8\)
câu d) làm tương tự như câu c)
ko ghi đề
\(=25,97+\left(6,54+103,46\right)\)
\(=25,97+110\)
\(=135,97\)
Bài 1
a, Có thể lập xy=21 <=> x=3;y=7 hoặc x=-3;y=-7
<=> x=7;y=3 hoặc x=-7;y=-3 ....v..v...
b, \(\left(x+5\right)\left(y-3\right)=15\)
\(\Rightarrow\orbr{\begin{cases}x+5=15\\y-3=15\end{cases}\Rightarrow\orbr{\begin{cases}x=10\\y=18\end{cases}}}\)
c, \(\left(2x-1\right)\left(y-3\right)=12\)
\(\Rightarrow\orbr{\begin{cases}2x-1=12\\y-3=12\end{cases}\Rightarrow\orbr{\begin{cases}2x=13\\y=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{13}{2}\\y=15\end{cases}}}\)
Bài 2
Ư(6)={1;2;3;6} => 1+2+3+6=12
Ư(8)={1;2;4;8} => 1+2+4+8 =15
=> Tổng 2 ước này đều \(⋮3\)
๖²⁴ʱミ★Šїℓεŋէ❄Bʉℓℓ★彡⁀ᶦᵈᵒᶫ mù mắt =)) t làm mẫu câu b thôi, c nhìn vào mà làm
b) \(\left(x+5\right)\left(y-3\right)=15\)
\(\Rightarrow y-3=\frac{15}{x+5}\Rightarrow y=3+\frac{15}{x+5}\)
\(\Rightarrow x+5\inƯ\left(15\right)\)
Ta có: \(Ư\left(15\right)=\left\{-15;-5;-3;-1;0;1;3;5;15\right\}\)
\(x=\left\{0;-10;-8;-6;-20;-4;-2;0;10\right\}\)
Vì \(x\inℕ\Rightarrow x=\left\{0;10\right\}\)
\(\Rightarrow y=\left\{6;4\right\}\)
Vậy: (x,y) = {(0;10); (6;4)}
\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)
Câu a) xem lại đề giùm nhé em
b) \(\left(x-1\right)^3=9^3\)
\(x-1=9\)
\(x=10\)
Vậy \(x=10\)
c) \(\left(x-1\right)^2=25\)
\(x-1=5\) hoặc \(x-1=-5\)
* \(x-1=5\)
\(x=6\)
* \(x-1=-5\)
\(x=-4\)
Vậy \(x=-4\); \(x=6\)
d) \(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
Vậy \(x=2\)
e) Sửa đề: \(\left(2x+4\right)^3=64\)
\(\left(2x+4\right)^3=4^3\)
\(2x+4=4\)
\(2x=0\)
\(x=0\)
Vậy \(x=0\)