Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
Thêm nữa câu a) Tính: M(x) + N(x)+ P(x)
B) Tính M(x) - N (x) - P(x)
ok rồi giúp mình với nha
c: \(P\left(-1\right)=-3-5-4+2+6+4=0\)
Vậy: x=-1 là nghiệm của P(x)
\(Q\left(-1\right)=4+1+3+2-7+1=4< >0\)
=>x=-1 không là nghiệm của Q(x)
\(\dfrac{1}{3}P\left(x\right)=\dfrac{5}{3}x^5-\dfrac{4}{3}x^4-\dfrac{2}{3}x^3+\dfrac{4}{3}x^2+x+2\)
\(\dfrac{3}{4}Q\left(x\right)=-\dfrac{3}{4}x^5+\dfrac{3}{2}x^4-\dfrac{3}{2}x^3+\dfrac{9}{4}x^2-\dfrac{3}{4}x+\dfrac{3}{16}\)
Do đó: \(\dfrac{1}{3}P\left(x\right)-\dfrac{3}{4}Q\left(x\right)=\dfrac{29}{12}x^5-\dfrac{17}{6}x^4+\dfrac{5}{6}x^3-\dfrac{11}{12}x^2+\dfrac{7}{4}x+\dfrac{29}{16}\)
|3x+4|=x+2
=>\(\left\{{}\begin{matrix}x+2>=0\\\left(3x+4\right)^2=\left(x+2\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-2\\\left(3x+4-x-2\right)\left(3x+4+x+2\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-2\\\left(2x+2\right)\left(4x+6\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-2\\x\in\left\{-1;-\dfrac{3}{2}\right\}\end{matrix}\right.\Leftrightarrow x\in\left\{-1;-\dfrac{3}{2}\right\}\)
|5x-6|=4-x
=>\(\left\{{}\begin{matrix}4-x>=0\\\left(5x-6\right)^2=\left(4-x\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =4\\\left(5x-6-4+x\right)\left(5x-6+4-x\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =4\\\left(6x-10\right)\left(4x-2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{\dfrac{5}{3};\dfrac{1}{2}\right\}\)
|5-2x|=x-3
=>|2x-5|=x-3
=>\(\left\{{}\begin{matrix}x-3>=0\\\left(2x-5\right)^2=\left(x-3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=3\\\left(2x-5\right)^2-\left(x-3\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=3\\\left(2x-5-x+3\right)\left(2x-5+x-3\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=3\\\left(x-2\right)\left(3x-8\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
|3-2x|=6+4x
=>|2x-3|=4x+6
=>\(\left\{{}\begin{matrix}4x+6>=0\\\left(4x+6\right)^2=\left(2x-3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{3}{2}\\\left(4x+6-2x+3\right)\left(4x+6+2x-3\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{3}{2}\\\left(2x+9\right)\left(6x+3\right)=0\end{matrix}\right.\Leftrightarrow x=-\dfrac{1}{2}\)
|6-3x|=3x
=>|3x-6|=3x
=>|x-2|=x
=>\(\left\{{}\begin{matrix}x>=0\\\left(x-2\right)^2=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=0\\-4x+4=0\end{matrix}\right.\Leftrightarrow x=1\)