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suy nghi nha
a)\(123-5:\left(x+4\right)=38\)
\(5:\left(x+4\right)=123-38\)
\(5:\left(x+4\right)=85\)
\(x+4=5:85\)
\(x=\dfrac{1}{17}-4\)
\(x=-\dfrac{67}{17}\)
b)\(70-5.\left(x-3\right)=45\)
\(5.\left(x-3\right)=70-45\)
\(5.\left(x-3\right)=35\)
\(x-3=35:5\)
\(x-3=7\)
\(x=7+3\)
\(x=10\)
\(\left|1-3x\right|=\dfrac{1}{2}\)
\(\Rightarrow1-3x=\pm\dfrac{1}{2}\)
*)Xét \(1-3x=\dfrac{1}{2}\)
\(\Rightarrow3x=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{6}\)
*)Xét \(1-3x=-\dfrac{1}{2}\)
\(\Rightarrow3x=\dfrac{3}{2}\Rightarrow x=\dfrac{1}{2}\)
\(\left(x-2\right)\left(x-4\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2< 0\\x-4>0\end{matrix}\right.=>4< x< 2\left(1\right)\\\left\{{}\begin{matrix}x-2>0\\x-4< 0\end{matrix}\right.=>2< x< 4\left(2\right)}\end{matrix}\right.\)(1 ) vô lý=> loại
=> (x-2).(x-4)<0 <=> 2<x<4
b. ta có\(x^2+1>0\forall x\)
=>(x2 -1).(x2+1)<0 <=> (x2 -1)<0 <=> x2<1
<=> -1<x<1
câu c bạn làm tương tự
x4 - 2x3 - x + 7 chia hết cho x - 3
= x.(x3 - 2.x2 - 1 ) + 7 chia hết cho x - 3
= x.[x2.(x - 2 - 1 )] + 7 chia hết cho x - 3
= x.x2.(x - 3) + 7 chia hết cho x - 3
Vì x.x2.(x - 3) chia hết cho x - 3 nên 7 chia hết cho x - 3 .
=> x - 3 \(\in\) Ư(7)
Ư(7) = {1;-1;7;-7}
=> x - 3 \(\in\) {1;-1;7;-7}
=> x \(\in\) {4;2;10;-4}.
\(\dfrac{1}{3}x.\dfrac{2}{5}\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x=0\\\dfrac{2}{5}\left(x-1\right)=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy x = 0 hoặc x = 1
\(\dfrac{1}{3}x.\dfrac{2}{5}\left(x-1\right)=0\\ =>\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(\left(20.2^4-12.2^3-48.2^2\right)^2:\left(-8\right)^3\)
\(=\left(20.16-12.9-48.4\right)^2:\left(-8\right)^3\)
\(=32^2:-512\)
\(=1024:-512=-2\)
\(\left(-2\right)\left(-3\right):\left(-1\right)-\left(-3\right)\left(-2\right):\left(-6\right)+\left(-2\right)\)
\(=-6-\left(-1\right)+\left(-2\right)\)
\(=-7\)
\(1.\left(-2\right)-\left(-3\right).\left(-4\right)-\left(-2\right).\left(-3\right)\)
\(=\left(-2\right)-12-6\)
\(=-20\)
a) \(\left|-5x+3\right|-x+5=4\)
th1: \(-5x+3\ge0\Leftrightarrow5x\le3\Leftrightarrow x\le\dfrac{3}{5}\)
\(\Rightarrow\left|-5x+3\right|-x+5=4\Leftrightarrow-5x+3-x+5=4\)
\(\Leftrightarrow-5x-x=4-3-5\Leftrightarrow-6x=-4\Leftrightarrow x=\dfrac{-4}{-6}=\dfrac{2}{3}\left(loại\right)\)
th2: \(-5x+3< 0\Leftrightarrow5x>3\Leftrightarrow x>\dfrac{3}{5}\)
\(\Rightarrow\left|-5x+3\right|-x+5=4\Leftrightarrow5x-3-x+5=4\)
\(\Leftrightarrow5x-x=4+3-5\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{2}{4}=\dfrac{1}{2}\left(loại\right)\)
vậy phương trình vô ngiệm
a)x=3,-3 vì nếu x=3 thì 3+3=6:3
nếu x=-3 thì -3+3=-6;-3
b)UC(10,6)
= 2
\(5.x+x=15:3+13\\ \Rightarrow\left(5+1\right).x=5+13=18\\ \Rightarrow6x=18\\ \Rightarrow x=\dfrac{18}{6}=3\)
\(\dfrac{2}{3}x+\dfrac{1}{3}x+\dfrac{1}{2}=\dfrac{2}{3}\\ \Rightarrow x\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=\dfrac{2}{3}-\dfrac{1}{2}\\ \Rightarrow x=\dfrac{4}{6}-\dfrac{3}{6}=\dfrac{1}{6}\)
\(\dfrac{2}{5}x+x=\dfrac{3}{2}\\ \Rightarrow x\left(\dfrac{2}{5}+1\right)=\dfrac{3}{2}\\ \Rightarrow x\left(\dfrac{2}{5}+\dfrac{5}{5}\right)=\dfrac{3}{2}\\ \Rightarrow x.\dfrac{7}{5}=\dfrac{3}{2}\\ \Rightarrow x=\dfrac{3}{2}:\dfrac{7}{5}\\ \Rightarrow x=\dfrac{3.5}{7.2}=\dfrac{15}{14}\)
\(5\cdot x+x=15:3+13\)
\(\Rightarrow x\cdot\left(5+1\right)=5+13\)
\(\Rightarrow x\cdot6=18\)
\(\Rightarrow x=\dfrac{18}{6}=3\)
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\(\dfrac{2}{3}\cdot x+\dfrac{1}{3}\cdot x+\dfrac{1}{2}=\dfrac{2}{3}\)
\(\Rightarrow x\cdot\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=\dfrac{2}{3}-\dfrac{1}{2}\)
\(\Rightarrow x\cdot1=\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{1}{6}\)
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\(\dfrac{2}{5}\cdot x+x=\dfrac{3}{2}\)
\(\Rightarrow x\cdot\left(\dfrac{2}{5}+1\right)=\dfrac{3}{2}\)
\(\Rightarrow x\cdot\dfrac{7}{5}=\dfrac{3}{2}\)
\(\Rightarrow x=\dfrac{3}{2}:\dfrac{7}{5}\)
\(\Rightarrow x=\dfrac{15}{14}\)