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`x :3*5 = 3/4 :(-5/6)`
`x :15 =3/4*(-6/5)=-9/10`
`x = -9/10 *15 =-27/2`
`x-1*2/2 = 8/x -1.2`
`x- 1*1 = 8/x -2`
`x-8/x = -2+1`
`x-8/x =-1`
`x^2 -8x =-x`
`x^2 -8x +x=0`
`x^2 -7x =0`
`x(x-7) =0`
`=>[(x=0),(x=7):}`
`a, x \div 15=-9/10`
`x=-9/10*14`
`x=-27/2`
`b, (x-1*2)/2=8/(x-1*2)`
\(\left(x-1\cdot2\right)\cdot\left(x-1\cdot2\right)=8\cdot2\)
`(x-1*2)^2=16`
`(x-1*2)^2=(+-4)^2`
\(\Rightarrow\left[{}\begin{matrix}x-1\cdot2=4\\x-1\cdot2=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4+2\\x=\left(-4\right)+2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
\(a,\left(x+12\right)\left(x-6\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+12>0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+12< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-12\\x>6\end{matrix}\right.\\\left\{{}\begin{matrix}x< -12\\x< 6\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>6\\x< -12\end{matrix}\right.\)
\(b,\left(10-x\right)\left(3-x\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}10-x< 0\\3-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}10-x>0\\3-x< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>10\\x< 3\left(vô.lí\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x< 10\\x>3\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x< 10\\x>3\end{matrix}\right.\)
\(a,\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+12>0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+12< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>6\\x< -12\end{matrix}\right.\\ \Rightarrow x\in\left\{...;-15;-14;-13;7;8;9;...\right\}\\ b,\Rightarrow\left(x-10\right)\left(x-3\right)< 0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10>0\\x-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-10< 0\\x-3>0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>10;x< 3\left(\text{loại}\right)\\3< x< 10\end{matrix}\right.\\ \Rightarrow x\in\left\{4;5;6;7;8;9\right\}\)
a) (Tớ đọc đề không hiểu, không nhớ :v)
b) a. `2(x-5) = 0`
`x-5 = 0:2`
`x-5 = 0`
`x=0+5`
`x=5`
b) `12(x-35) = 0`
`x-35=0:12`
`x-35 =0`
`x=0+35`
`x=35`
c) `(x-10).(x-13) = 0`
`=> x-10=0`
`x-13=0`
`=> x=0+10`
`x=0+13`
`=>x=10`
`x=13`.
a) Tích 2 số bằng không khi 1 trong 2 số bằng 0
b) \(2\left(x-5\right)=0\)
\(\Rightarrow x-5=0\)
\(\Rightarrow x=5\)
\(12\left(x-35\right)=0\)
\(\Rightarrow x-35=0\)
\(\Rightarrow x=35\)
\(\left(x-10\right)\left(x-13\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-10=0\\x-13=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=13\end{matrix}\right.\)
a)
\(x+\left(x+2\right)+\left(x+4\right)+...+\left(x+98\right)=0\)
\(x+x+2+x+4+...+x+98=0\)
\(50x+\left(98+2\right).\left[\left(98-2\right):2+1\right]:2=0\)
\(50x+100.49:2=0\)
\(50x+49.50=0\)
\(50x=0-49.50\)
\(50x=-2450\)
\(x=-2450:50\)
\(x=-49\)
b)
\(\left(x-5\right)+\left(x-4\right)+\left(x-3\right)+...+\left(x+11\right)+\left(x+12\right)=99\)
\(x+x+x+...+x-5-4-3-...+11+12=99\)
\(18x+6+7\text{+ 8 + 9 + 10 + 11 + 12 = 99}\)
\(18x+63=99\)
\(18x=99-63\)
\(18x=36\)
\(x=36:18\)
\(x=2\)
a) \(18-\left(2x+5\right)=9\)
\(2x+5=18-9\)
\(2x+5=9\)
\(2x=9-5\)
\(2x=4\)
\(x=2\)
a) \(18-\left(2x+5\right)=9\)
\(\Rightarrow2x+5=18-9=9\)
\(\Rightarrow2x=9-5=4\Rightarrow x=4:2=2\)
b) \(23x-4=32\Rightarrow23x=32+4=36\Rightarrow x=\dfrac{36}{23}\)
c) \(\left(3x+2\right)^2=64\)
\(\Rightarrow\left[{}\begin{matrix}3x+2=8\\3x+2=-8\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{10}{3}\end{matrix}\right.\)
d) \(x\left(2x-12\right)=0\Rightarrow6x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
a) \(3\sqrt{x-3}=12\left(đk:x\ge3\right)\)
\(\Leftrightarrow\sqrt{x-3}=4\)
\(\Leftrightarrow x-3=16\Leftrightarrow x=19\left(tm\right)\)
b) \(\sqrt{16\left(1-2x\right)}-8=0\left(đk:x\le\dfrac{1}{2}\right)\)
\(\Leftrightarrow4\sqrt{1-2x}=8\Leftrightarrow\sqrt{1-2x}=2\)
\(\Leftrightarrow1-2x=4\Leftrightarrow x=-\dfrac{3}{2}\left(tm\right)\)
c) \(\sqrt{4\left(9-6x+x^2\right)}-12=0\)
\(\Leftrightarrow2\sqrt{\left(x-3\right)^2}=12\)
\(\Leftrightarrow\left|x-3\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=6\\x-3=-6\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)
a: ta có: \(3\sqrt{x-3}=12\)
\(\Leftrightarrow x-3=16\)
hay x=19
b: Ta có: \(\sqrt{16\left(1-2x\right)}-8=0\)
\(\Leftrightarrow1-2x=4\)
\(\Leftrightarrow2x=-3\)
hay \(x=-\dfrac{3}{2}\)
a: \(3\left(x-3\right)-6x=0\)
=>\(3x-9-6x=0\)
=>-3x-9=0
=>3x+9=0
=>3x=-9
=>\(x=-\dfrac{9}{3}=-3\)
b: Đề thiếu vế phải rồi bạn
c: \(2\left(x-3\right)+3x=9\)
=>2x-6+3x=9
=>5x-6=9
=>5x=6+9=15
=>x=15/5=3
d: \(x\left(x-11\right)+2\left(x-11\right)=0\)
=>\(\left(x-11\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-11=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-2\end{matrix}\right.\)
e: \(x\left(x+2\right)+8=x^2\)
=>\(x^2+2x+8=x^2\)
=>2x+8=0
=>2x=-8
=>x=-8/2=-4
f: \(8\left(x+1\right)+2x=-2\)
=>\(8x+8+2x=-2\)
=>10x=-2-8=-10
=>\(x=-\dfrac{10}{10}=-1\)
g: 12-3(x+2)=0
=>3(x+2)=12
=>x+2=12/3=4
=>x=4-2=2
a) x = 15. b) x = 11. c) x = 68. d) x = 310
Tìm xx biết: \left(x^{4}\right)^{3}=\dfrac{x^{19}}{x^{6}}(x4)3=x6x19
Trả lời: x=x=