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a) 2x + 6 = 12
2x = 6
x = 3
b) ( 3x + 3 ) - 5 = 13
3x + 3 = 18
3x = 15
x = 5
Vậy,...........
\(\left(2x\right)+6=12\) \(\left(3x+3\right)-5=13\)
\(2x=12-6\) \(3x+3=13+5\)
\(2x=6\) \(3x+3=18\)
\(x=6:2\) \(3x=18-3\)
\(x=3\) \(3x=15\Rightarrow x=5\)
\(P=1+\frac{x+3}{x^2+5x+6}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)
\(P=1+\frac{1}{x+2}:\left(\frac{4x^2.2}{4x^2\left(x-2\right)}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{1}{x+2}\right)\)
\(P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\)
\(P=1+\frac{1}{x+2}:\left(\frac{2x+4-x-x+2}{\left(x+2\right)\left(x-2\right)}\right)\)
\(P=1+\frac{1}{x+2}:\frac{6}{\left(x+2\right)\left(x-2\right)}=1+\frac{\left(x+2\right)\left(x-2\right)}{6\left(x+2\right)}=1+\frac{x-2}{6}\)
\(=\frac{x+4}{6}.P=0\Leftrightarrow x=-4\)
\(P>0\Leftrightarrow x>-4\)
x . 2 + x . 3 = 15
x. ( 2 + 3 ) = 15
x . 5 = 15
x = 15 : 5
x = 3
\(\frac{9}{15}x\frac{12}{1}x\frac{5}{6}\)=\(\frac{9x12x5}{15x1x6}\)=\(\frac{9x6x2x5}{5x3x1x6}\)= \(\frac{9x2}{3x1}\)=\(\frac{18}{3}\)= 6
Toán này đâu có phải toán lớp 1? Mà toán này là toán lớp 5.
7,
\(\Leftrightarrow x=\sqrt{x+2}\left(\frac{\sqrt{x}}{1+\sqrt{1-\sqrt{x}}}\right)^2\)
\(\Leftrightarrow x=\frac{\sqrt{x+2}.x}{2-\sqrt{x}+2\sqrt{1-\sqrt{x}}}\Leftrightarrow\frac{\sqrt{x+2}}{2-\sqrt{x}+2\sqrt{1-\sqrt{x}}}=1\)
đến đây tự làm
7 đề như tớ
8. (x-1)^2 +\(x\sqrt{x-\frac{1}{x}}\)
9. \(\sqrt{1+x}+\sqrt{3-3x}=\sqrt{4x^2+1}\)
\(6+4=210\)
\(210=\overline{\left(6-4\right)\left(6+4\right)}\)
Tương tự :
\(\overline{\left(9-2\right)\left(9+2\right)}=711\)
\(\overline{\left(8-5\right)\left(8+5\right)}=313\)
...
\(\overline{\left(15-3\right)\left(15+3\right)}=1218\)
6 + 4 = 210
9 + 2 = 711
8 + 5 = 313
5 + 7 \(\ne\)37
7 + 6 = 113
9 + 8 = 117
10 + 6 = 416
15 + 3 = 1218
6 - 4 = 210
9 - 2 = 711
8 - 5 = 313
5 - 3 \(\ne\)37
7 - 6 = 113
9 - 8 = 117
10 - 6 = 416
15 - 3 = 1218
- Đề sai :>
- Quy luật kép nào đó mà giờ tui 'ếu' nhìn thấy được :))
\(\left(\frac{x^2+3x}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)
\(=\left(\frac{x\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right)\)
\(=\left(\frac{x}{x^2+9}+\frac{3}{x^2+9}\right):\left(\frac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\right)=\frac{x+3}{x^2+9}:\frac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\)
\(=\frac{\left(x+3\right)\left(x-3\right)\left(x^2+9\right)}{\left(x^2+9\right)\left(x^2-6x+9\right)}=\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-3\right)}=\frac{x+3}{x-3}\)
b) \(Voix>0\Rightarrow P\ne\varnothing\)(mk ko chac)
c) \(P\inℤ\Leftrightarrow x+3⋮x-3\Leftrightarrow x-3\in\left\{-1;-2;-3;-6;1;2;3;6\right\}\)
sau do tinh
cau nay la toan lp 8 nha
| 3x | + 6 = 15
| 3x | = 9
+) 3x = 9
=> x = 3
+) 3x = -9
=>x = -3
Vậy,..........
\(/3x/+6=15\)
\(/3x/=15-6\)
\(/3x/=9\)
\(\Rightarrow3x=9\) hoặc \(3x=-9\)
\(TH1:3x=9\) \(TH2:3x=-9\)
\(x=9:3\) \(x=-9:3\)
\(x=3\) \(x=-3\)
Vậy x = 3 hoặc x = -3