4\(\times\)(1+3x)-2\(\times\left(2-x\right)=0\)
Những câu hỏi liên quan
\(3x\times\left(1-x\right)+\left(x+3\right)\times\left(x-2\right)=-2\times\left(x-4\right)^2\)
Lời giải:
$3x(1-x)+(x+3)(x-2)=-2(x-4)^2$
$\Leftrightarrow (3x-3x^2)+(x^2-2x+3x-6)=-2(x^2-8x+16)$
$\Leftrightarrow -2x^2+4x-6=-2x^2+16x-32$
$\Leftrightarrow 12x=26\Rightarrow x=\frac{13}{6}$
Vậy........
Giải phương trình:
\(3x\times\left(1-x\right)+\left(x+3\right)\times\left(x-2\right)=-2\times\left(x-4\right)^2\)
Ta có : \(3x\left(1-x\right)+\left(x+3\right)\left(x-2\right)=-2\left(x-4\right)^2\)
=> \(3x\left(1-x\right)+\left(x+3\right)\left(x-2\right)=-2\left(x^2-8x+16\right)\)
=> \(3x-3x^2+x^2+3x-2x-6=-2x^2+16x-32\)
=> \(3x-3x^2+x^2+3x-2x-6+2x^2-16x+32=0\)
=> \(-12x+26=0\)
=> \(x=\frac{26}{12}=\frac{13}{6}\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{\frac{13}{6}\right\}\)
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 biết:
\(\frac{-1}{2}\times\left(3x-1\right)+\frac{3}{4}\left(3-2x\right)=-3\times\left(\frac{x}{2}-1\right)-\left(\frac{4}{5}\right)^{-1}\)
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
Đọc tiếp
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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Rút gọn \(B=\left(x^4-x+\frac{x-3}{x^3+1}\times\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right)\times\frac{4x^2+6x+1}{\left(x+3\right)\left(4-x\right)}\)
Bài 1 Tìm \(x\)
a)\(\left(\dfrac{1}{2}\times x-3\right)\times\left(-\dfrac{1}{3}+x\right)=0\)
b)\(\dfrac{1}{2}\times x^2-\dfrac{1}{5}\times x=0\)
c)\(\dfrac{1`}{4}\times x=\dfrac{1}{16}\times x^2\)
d)\(9\times x^2=1\)
e)\(\left(x-5\right)^2=4\)
a) \(\left(\dfrac{1}{2}x-3\right)\left(-\dfrac{1}{3}+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=0+3\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{1}{2}\\x=0-\left(-\dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)
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d) \(9x^2=1\)
\(\Leftrightarrow x^2=1:9\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
\(\Leftrightarrow x^2=\left(\dfrac{1}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
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e) \(\left(x-5\right)^2=4\)
\(\Leftrightarrow\left(x-5\right)^2=2^2\)
\(\Leftrightarrow x-5=2\)
\(\Leftrightarrow x=2+5\)
\(\Leftrightarrow x=7\)
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Xem thêm câu trả lời
1.Rút gọn biểu thức:
\(a,\)\(x\times\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(b,\)\(3x\times\left(x-2\right)-5x\times\left(1-x\right)-8\times\left(x^2-3\right)\)
\(c,\)\(\left(2x-6\right)\times\left(x+3\right)-5\times\left(2x^2-x+7\right)\)
thực hiện phép tính
a)\(\dfrac{2x^2-20x+50}{3x+3}\times\dfrac{x^2-1}{4\left(x-5\right)^2}\)
b) \(\dfrac{6x-3}{5x^2+x}\times\dfrac{25x^2+10x+1}{1-8x^3}\)
c) \(\dfrac{3x^2-x}{x^2-1}\times\dfrac{1-x^4}{\left(1-3x\right)^3}\)
a/ \(\dfrac{2x^2-20x+50}{3x+3}\cdot\dfrac{x^2-1}{4\left(x-5\right)^2}=\dfrac{2\left(x^2-10x+25\right)\cdot\left(x^2-1\right)}{3\left(x+1\right)\cdot4\left(x-5\right)^2}=\dfrac{2\left(x-5\right)^2\left(x-1\right)\left(x+1\right)}{12\left(x+1\right)\left(x-5\right)^2}=\dfrac{x+1}{6}\)
b/ \(\dfrac{6x-3}{5x^2+x}\cdot\dfrac{25x^2+10x+1}{1-8x^2}=-\dfrac{3\left(1-2x\right)\cdot\left(5x+1\right)^2}{x\left(5x+1\right)\left(1-2x\right)\left(1+2x+4x^2\right)}=\dfrac{3\left(5x+1\right)}{x\left(4x^2+2x+1\right)}\)
c/ \(\dfrac{3x^2-x}{x^2-1}\cdot\dfrac{1-x^4}{\left(1-3x\right)^3}=\dfrac{x-3x^2}{1-x^2}\cdot\dfrac{\left(1-x^2\right)\left(1+x^2\right)}{\left(1-3x\right)^3}=\dfrac{x\left(1-3x\right)\left(1-x^2\right)\left(1+x^2\right)}{\left(1-x^2\right)\left(1-3x\right)^3}=\dfrac{x\left(x^2+1\right)}{\left(1-3x\right)^3}\)
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Dễ thế mà bạn ( người ko quen) ko làm đc !
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