Tìm x biết:
3(x-4)+x(x-4)=0
Mik cảm ơn ahjhj =)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
<=> \(\frac{x-2}{7}.\frac{x+3}{5}.\frac{x+4}{3}=0\)
<=> \(\frac{x-2}{7}=0\)hoặc \(\frac{x+3}{5}=0\); \(\frac{x+4}{3}=0\)
Nếu \(\frac{x-2}{7}=0\)<=> \(x-2=0\)<=> \(x=2\)
Nếu \(\frac{x+3}{5}=0\)<=> \(x+3=0\) <=> \(x=3\)
Nếu \(\frac{x+4}{3}=0\)<=> \(x+4=0\)<=> \(x=4\)
Vây x= 2 hoặc 3; 4
0,25 x 3 + 1 : 4 x 7
= 0,25 (3+7)
= 0,25 x 10
= 2,5
X x 1.2 + X x 1.8 = 45
X [1(2+8)] = 45
X [1 x 10] = 45
X x 10 = 45
X = 45 : 10
X = 4,5
\(3\left(2x-3\right)\left(3x+2\right)-2\left(x+4\right)\left(4x-3\right)+9x\left(4-x\right)=0\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow\left(x^2-3x\right)+\left(-2x+6\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
a) \(PT\Leftrightarrow x^2-4x+1=3x-5\)
\(\Leftrightarrow x^2-7x+6=0\Leftrightarrow\left(x-1\right)\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=6\end{cases}}\)
b) \(PT\Leftrightarrow x^2\left(2x-3\right)-\left(2x-3\right)=0\Leftrightarrow\left(x^2-1\right)\left(2x-3\right)=0\Leftrightarrow x\in\left\{\pm1;\frac{3}{2}\right\}\)
a) ( x + 1 ) , ( 3x + 3 ) = 0
\(\Rightarrow\orbr{\begin{cases}x+1=0\\3x+3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0-1\\3x=0-3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\3x=-3\end{cases}}\)
\(\Rightarrow\) \(x=-3:3=-1\)
Vậy x = -1
b) ( x2 + 2 ) . ( x - 3 ) = 0
\(\Rightarrow\orbr{\begin{cases}x^2+2=0\\x-3=0\end{cases}}\)
\(\Rightarrow\) \(\orbr{\begin{cases}x^2=0-2\\x=0+3\end{cases}}\)
\(\Rightarrow\) \(\orbr{\begin{cases}x^2=-2\\x=3\end{cases}}\)
\(\Rightarrow\) \(x\in\varnothing\)
Vậy x = 3
c) 4|x-4| = 4
\(\Rightarrow\)|x-4| = 4 : 4
\(\Rightarrow\)|x-4| = 1
\(\Rightarrow\orbr{\begin{cases}x-4=1\\x-4=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1+4\\x=\left(-1\right)+4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}\)
Vậy x \(\in\) { 3;5 }
a)\(\orbr{\begin{cases}x+1=0\\3x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\3x=-3\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=-1\end{cases}\Rightarrow}x=-1}\)
b)\(\orbr{\begin{cases}x^2+2=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-2\\x=3\end{cases}}}\)Vì : \(x^2\ge0\)\(\Rightarrow\orbr{\begin{cases}x\in\varnothing\\x=3\end{cases}\Rightarrow x=3}\)
c)\(4.\left|x-4\right|=4\)
\(\Rightarrow\left|x-4\right|=4:4=1\)
\(\Rightarrow\orbr{\begin{cases}x-4=1\\x-4=-1\end{cases}\Rightarrow\hept{\begin{cases}x=5\\x=3\end{cases}}}\)
A. 2.\(|3x+1|\)=\(\frac{3}{4}\)-\(\frac{5}{8}\)
2.\(|3x+1|\)=1/8
\(|3x+1|\)=1/8:2
\(|3x+1|\)=1/16
TH1 : 3x+1=1/16
3x=1/16-1
3x=-15/16
x=-15/16:3
x=-5/16
a,\(\frac{3}{4}-2.\left|3x+1\right|=\frac{5}{8}\)
\(\Rightarrow2.\left|3x+1\right|=\frac{3}{4}-\frac{5}{8}=\frac{6}{8}-\frac{5}{8}=\frac{1}{8}\)
\(\Rightarrow\left|3x+1\right|=\frac{1}{8}.\frac{1}{2}=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}3x+1=\frac{1}{16}\\3x+1=\frac{-1}{16}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3x=\frac{1}{16}-1=\frac{-15}{16}\\3x=\frac{-1}{16}-1=\frac{-17}{16}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-15}{16}.\frac{1}{3}=\frac{-5}{16}\\x=\frac{-17}{16}.\frac{1}{3}=\frac{-17}{48}\end{cases}}\)
Vậy....
b,\(\left|3x+2\right|-\left|x-3\right|=\frac{7}{2}\left(1\right)\)
Ta có bảng xét dấu
x | \(\frac{-2}{3}\) 3 |
3x+2 | - 0 + | + |
x-3 | - | - 0 + |
Nếu x<\(\frac{-2}{3}\) thì \(\left|3x+2\right|-\left|x-3\right|\) \(=-3x-2-3+x\)
\(=-2x-5\)
Từ (1) \(\Rightarrow-2x-5=\frac{7}{2}\)
\(\Rightarrow-2x=\frac{7}{2}+5=\frac{17}{2}\)
\(\Rightarrow x=\frac{17}{2}\cdot\frac{-1}{2}=\frac{-17}{4}\)(thỏa mãn x<\(\frac{-2}{3}\)
Nếu \(\frac{-2}{3}\le x\le3\)thì \(\left|3x+2\right|-\left|x-3\right|=3x+2-\left(3-x\right)\)
\(=3x+2-3+x\)
\(=2x-1\)
Từ (1)\(\Rightarrow\)\(2x-1=\frac{7}{2}\)
\(\Rightarrow2x=\frac{9}{2}\)
\(\Rightarrow x=\frac{9}{4}\)(thỏa mãn......
Còn trưonwfg hợp cuối bạn tự làm nốt nhé
a
\(x+x^2-x^3-x^4=0\\ \Leftrightarrow x\left(1+x\right)-x^3\left(1+x\right)=0\\ \Leftrightarrow\left(1+x\right)\left(x-x^3\right)=0\\ \Leftrightarrow\left(1+x\right).x.\left(1-x^2\right)=0\\ \Leftrightarrow\left(1+x\right).x.\left(1-x\right)\left(1+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
b
x^3 chứ: )
\(x^3+27+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow x^3+3^3+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow\left(x+3\right).x.\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
Bạn tham khảo tại đây:
https://hoc24.vn/cau-hoi/giup-minh-voiiiii-minh-cam-on-tim-xy-biet-dfracx4-dfrac2y13-dfracx-2y-1y-voi-y-0.4107067269450
3(x-4)+x(x-4)=0
<=> (x-4)(3+x)=0
=> x - 4 = 0 hoặc x + 3 = 0
<=> x = 4 hoặc x = -3
3(x - 4) + x(x - 4) = 0
(x - 4)(3 + x) = 0
x - 4 = 0 hoặc 3 + x = 0
*) x - 4 = 0
x = 0 + 4
x = 4
*) 3 + x = 0
x = 0 - 3
x = -3
Vậy x = -3; x = 4