tìm x :\(\left(5\times x+4\right)\times\left(12\div x-1\right)=0\)
JA
Những câu hỏi liên quan
Tìm x
a, \(\left(2-x\right)\times\left(\dfrac{4}{5}-x\right)< 0\)
b, \(\left(x-\dfrac{3}{2}\right)\times\left(2\times x-1\right)>0\)
c, \(\left(x-\dfrac{4}{7}\right)\div\left(x+\dfrac{1}{2}\right)>0\)
d, \(\left(2\times x-5\right)\div\left(x+1\dfrac{3}{4}\right)< 0\)
Giúp mk với
\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
=>Trong 2 số phải có 1 số âm và 1 số dương
Mà \(2-x>\dfrac{4}{5}-x\)
=>\(\dfrac{4}{5}< x< 2\)
Vậy...
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Tìm x \(\frac{2}{\left(x-1\right)\times\left(x-3\right)}+\frac{5}{\left(x-3\right)\times\left(x-8\right)}+\frac{12}{\left(x-8\right)\times\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)-3/4
Tìm x:
a.\(\frac{1}{3}\times\left(x-1\right)+\frac{2}{5}\times\left(x+1\right)=0\)
b.\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}x-5\right)\)
c.\(\left(x+\frac{1}{2}\right)\times\left(x-\frac{3}{4}\right)=0\)
Các bn ơi giúp mk với chiều mk đi học rồi!!!!!!!!!!!!
Tìm xin Z:a. 2timesleft(x-5right)-3timesleft(x-4right)left(-6right)+15timesleft(-3right)b.left(2x-5right)-722c.left(2timesleft|xright|-6right)times1144d.left|x+3right|+left|x+5right|+left|x+9right|4xe.left|x-1right|+left|x-5right|4g.x+left(x+1right)+left(x+2right)+.......+1000
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Tìm \(x\in Z\):
a. \(2\times\left(x-5\right)-3\times\left(x-4\right)=\left(-6\right)+15\times\left(-3\right)\)
b.\(\left(2x-5\right)-7=22\)
c.\(\left(2\times\left|x\right|-6\right)\times11=44\)
d.\(\left|x+3\right|+\left|x+5\right|+\left|x+9\right|=4x\)
e.\(\left|x-1\right|+\left|x-5\right|=4\)
g.\(x+\left(x+1\right)+\left(x+2\right)+.......+100=0\)
a) x=53
b) x=17
c) x=5;x=-5
d) x=17
e) x=5
g) ???
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Tìm x biết left|2x-4right|+left|3y+12right|+left(2z-10right)^{10}le0left|x-3right|+left|2y-6right|+left(4x-3yright)^2le0left(x-7right)timesleft(x+3right)0left(x-7right)timesleft(x+3right)timesleft(x-5right)0
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Tìm x biết
\(\left|2x-4\right|+\left|3y+12\right|+\left(2z-10\right)^{10}\le0\)
\(\left|x-3\right|+\left|2y-6\right|+\left(4x-3y\right)^2\le0\)
\(\left(x-7\right)\times\left(x+3\right)>0\)
\(\left(x-7\right)\times\left(x+3\right)\times\left(x-5\right)<0\)
Tìm x thuộc Z biết
a. left(x-1right)timesleft(2x-4right)0
b. left(x^2+5right)timesleft(x-5right)0
c. left(x^2+5right)timesleft(x^2-2right)0
d. left(x^2-1right)timesleft(x^2-9right) 0
e. left|x^2-2xright|x
g. left|2x+1right|+left|x+8right|4x
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Tìm x thuộc Z biết
a. \(\left(x-1\right)\times\left(2x-4\right)=0\)
b. \(\left(x^2+5\right)\times\left(x-5\right)=0\)
c. \(\left(x^2+5\right)\times\left(x^2-2\right)=0\)
d. \(\left(x^2-1\right)\times\left(x^2-9\right)< 0\)
e. \(\left|x^2-2x\right|=x\)
g. \(\left|2x+1\right|+\left|x+8\right|=4x\)
a) \(\left(x-1\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\2x-4=0\Rightarrow x=2\end{matrix}\right.\)
b) \(\left(x^2+5\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x-5=0\Rightarrow x=5\end{matrix}\right.\)
mà \(x\in Z\Rightarrow x=5\)
c) \(\left(x^2+5\right)\left(x^2-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x^2-2=0\Rightarrow x=\sqrt{2}\end{matrix}\right.\)
mà \(x\in Z\Rightarrow x\in\varnothing\)
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Tìm x,y biết
\(\left(x-3\right)^2+\left(y+2\right)^2=0\)
\(2\times x+2^{x+3}=136\)
\(\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\)
\(\left(2\times x-5\right)^{2000}+\left(3\times y+4\right)^{2002}\le0\)
\(\left(x-3\right)^2+\left(y+2\right)^2=0\)
\(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-3\right)^2+\left(y+2\right)^2\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\\\left(y+2\right)^2=0\Rightarrow y+2=0\Rightarrow y=-2\end{matrix}\right.\)
đề sai câu b các câu sau áp dụng tương tự
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c/ Vì: \(\left(x-12+y\right)^{200}+\left(x-4-x\right)^{200}=0\)
mà \(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}\ge0\forall x,y\\\left(x-4-y\right)^{200}\ge0\forall x,y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-12+y\right)^{200}=0\\\left(x-4-y\right)^{200}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-12+y=0\\x-4-y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=12\\x-y=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=4\end{matrix}\right.\)
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FD
25 tháng 8 2020 lúc 9:38
\(4\times\left(\frac{1}{4}\right)^2+25\times\left[\left(\frac{3}{4}\right)^3\div\left(\frac{5}{4}\right)^3\right]\div\left(\frac{3}{2}\right)^3\)
\(2^3+3\times\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2\div\frac{1}{2}\right]-8\)
\(4.\left(\frac{1}{4}\right)^2+25\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3=4.\frac{1}{16}+25\left(\frac{27}{64}.\frac{64}{125}\right).\frac{8}{27}\)
\(=\frac{1}{4}+25.\frac{27}{125}.\frac{8}{27}=\frac{1}{4}+\frac{8}{5}=\frac{37}{20}\)
\(2^3+3\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8=8+3-1+4.2-8=10\)
cho: \(f\left(x\right)=a\times x^3+4\times x\times\left(x^2-1\right)+8\)
\(g\left(x\right)=x^3-4\times x\times\left(b\times x+1\right)+c-3\)
tìm a, b, c để \(f\left(x\right)=g\left(x\right)\)