Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta thấy: a = 2 x b = 3 x c nên a gấp 2 lần c và gấp 3 lần c.
Mà a, b, c< 10 nên a chia hết cho 2 và 3.
Vậy, a = 0, 6.
+ Nếu a = 0 thì b = 0 và c = 0
+ Nếu a = 6 thì b = 3 và c = 2
a) \(6x^2-2x=2x\left(3x-1\right)\)
\(2x\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}2x=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(S=\left\{0;\dfrac{1}{3}\right\}\)
b) \(x^2+5x+6=x^2+2x+3x+6=x\left(x+2\right)+3\left(x+2\right)=\left(x+3\right)\left(x+2\right)\)
\(\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{-3;-2\right\}\)
a) choA(x) = 0
\(=>-18+2x=0\)
\(=>2x=18=>x=9\)
b) cho B(x) = 0
\(=>\left(x+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
a) \(-3x^3+5x^2-2x=0\\ \Leftrightarrow3x^3-5x^2+2x=0\\ \Leftrightarrow x\left(3x^2-5x+2\right)=0\\ \Leftrightarrow x\left(3x-2\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\\x=1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{0;\dfrac{2}{3};1\right\}\)
b) \(\dfrac{-1}{2}x^4+\dfrac{1}{8}x^2=0\\ \Leftrightarrow\dfrac{-1}{2}x^2\left(x^2-\dfrac{1}{4}\right)=0\\ \Leftrightarrow\dfrac{-1}{2}x^2\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{0;\dfrac{1}{2};\dfrac{-1}{2}\right\}\)
= > a 3 phần b 2 phần và c 1 phần
a là : 180 : ( 3 + 2 + 1 ) x 3 = 90
b là : 90 : 2 = 45
c là : 90 : 3 = 30
= > a = 90 ; b = 45 và c = 30
+1 L-I-K-E - 0 L-I-K-E
= > a 3 phần b 2 phần và c 1 phần
a là : 180 : ( 3 + 2 + 1 ) x 3 = 90
b là : 90 : 2 = 45
c là : 90 : 3 = 30
= > a = 90 ; b = 45 và c = 30
\(a,\Rightarrow\left[{}\begin{matrix}x-1=2x\\1-x=2x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Rightarrow\left[{}\begin{matrix}x+x-2=2\left(x\ge2\right)\\x+2-x=2\left(0\le x< 2\right)\\-x+2-x=2\left(x< 0\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\left(x\ge2\right)\left(tm\right)\\x=0\left(0\le x< 2\right)\left(tm\right)\\x=0\left(x< 0\right)\left(ktm\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
a: Ta có: \(\left|x-1\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2x\left(x\ge1\right)\\x-1=-2x\left(x< 1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=\dfrac{1}{3}\left(nhận\right)\end{matrix}\right.\)
a: \(\left|7-2x\right|+7=2x\)
=>\(\left|2x-7\right|+7=2x\)
=>\(\left|2x-7\right|=2x-7\)
=>2x-7>=0
=>\(x>=\dfrac{7}{2}\)
b: \(\left|1-x\right|=4x+1\)
=>\(\left|x-1\right|=4x+1\)
=>\(\left\{{}\begin{matrix}4x+1>=0\\\left(4x+1\right)^2=\left(x-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(4x+1\right)^2-\left(x-1\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(4x+1-x+1\right)\left(4x+1+x-1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\5x\left(3x+2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
c: \(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|3,2+\dfrac{2}{5}\right|\)
=>\(\left|x-\dfrac{1}{3}\right|=\dfrac{16}{5}+\dfrac{2}{5}-\dfrac{4}{5}=\dfrac{14}{5}\)
=>\(\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{14}{5}\\x-\dfrac{1}{3}=-\dfrac{14}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}+\dfrac{1}{3}=\dfrac{42+5}{15}=\dfrac{47}{15}\\x=-\dfrac{14}{5}+\dfrac{1}{3}=\dfrac{-42+5}{15}=-\dfrac{37}{15}\end{matrix}\right.\)
d: \(\left|x-7\right|+2x+5=6\)
=>\(\left|x-7\right|=6-2x-5=-2x+1\)
=>\(\left\{{}\begin{matrix}-2x+1>=0\\\left(-2x+1\right)^2=\left(x-7\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x-1\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x-1+x-7\right)\left(2x-1-x+7\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(3x-8\right)\left(x+6\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{8}{3}\left(loại\right)\\x=-6\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)
e: 3x-|2x-1|=2
=>|2x-1|=3x-2
=>\(\left\{{}\begin{matrix}3x-2>=0\\\left(3x-2\right)^2=\left(2x-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2\right)^2-\left(2x-1\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2-2x+1\right)\left(3x-2+2x-1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(x-1\right)\left(5x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x-1=0\\5x-3=0\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x=1\left(nhận\right)\\x=\dfrac{3}{5}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)