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a) \(\left|2x-3\right|=5\)
⇒ \(\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=5+3=8\\2x=\left(-5\right)+3=-2\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=8:2\\x=\left(-2\right):2\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{4;-1\right\}.\)
b) Đề sai rồi, bạn xem lại nhé.
c) \(\left|x-1\right|+3x=11\)
⇒ \(\left|x-1\right|=11-3x\)
+) Với \(x\ge1\)
⇒ \(x-1=11-3x\)
⇒ \(x+3x=11+1\)
⇒ \(4x=12\)
⇒ \(x=12:4\)
⇒ \(x=3.\)
+) Với \(x< 1\)
⇒ \(1-x=11-3x\)
⇒ \(1-11=\left(-3x\right)+x\)
⇒ \(-10=-2x\)
⇒ \(x=\left(-10\right):\left(-2\right)\)
⇒ \(x=5.\)
Vậy \(x\in\left\{3;5\right\}.\)
Chúc bạn học tốt!
a) |2x - 3| = 5
2x - 3 = 5 Hoặc 2x - 3 = -5
* 2x - 3 = 5
2x = 5 + 3
2x = 8
x = 8 : 2
x = 4
2x - 3 = -5
2x = -5 + 3
2x = -2
x = -2 : 2
x = -1
Vậy x ϵ {-2; -1}
a) \(\left|2x-3\right|=5\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=5\\2x-3=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=8\\2x=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)
Vậy : \(x\in\left\{4,-1\right\}\)
b) \(\left|2x-1\right|=\left|2x-3\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=2x-3\\2x-1=3-2x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-2x=-3+1\\2x+2x=3+1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}0=-2\\4x=4\end{cases}}\)
\(\Rightarrow x=1\)
Vậy : \(x=1\)
c) \(\left|x-1\right|+3x=11\)
\(\Leftrightarrow\left|x-1\right|=11-3x\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=11-3x\\x-1=3x-11\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+3x=11+1\\3x-x=-1+11\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=12\\2x=10\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)
Vậy : \(x\in\left\{3,5\right\}\)
d) \(\left|5x-3\right|-x=7\)
\(\Leftrightarrow\left|5x-3\right|=7+x\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=10\\6x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{4}=\frac{5}{2}\\x=-\frac{4}{6}=-\frac{2}{3}\end{cases}}\)
Vậy : \(x\in\left\{\frac{5}{2},-\frac{2}{3}\right\}\)
a,\(|2x-3|=5\)
\(\Rightarrow2x-3=5\)hoặc\(2x-3=-5\)
\(2x=5+3\)hoặc\(2x=-5+3\)
\(2x=8\)hoặc\(2x=-2\)
\(\Rightarrow x=\frac{8}{2}=4\)hoặc\(x=-\frac{2}{2}=-1\)
Vậy.......
b,Sai vế phải r bn ơi
c,\(|x-1|+3x=11\)
\(|x-1|=11-3x\)
\(\Rightarrow x-1=+\left(11-3x\right)\)hoặc\(\Rightarrow x-1=-\left(11-3x\right)\)
\(x-1=11-3x\)hoặc\(x-1=-11+3x\)
\(x+3x=11+1\)hoặc\(x-3x=-11+1\)
\(4x=12\)hoặc\(-2x=-10\)
\(\Rightarrow x=\frac{12}{4}=3\)hoặc\(x=\frac{-10}{-2}=5\)
Vậy.....
d,\(|5x-3|-x=7\)
\(|5x-3|=7+x\)
\(\Rightarrow5x-3=+\left(7+x\right)\)hoặc\(5x-3=-\left(7+x\right)\)
\(5x-3=7+x\)hoặc\(5x-3=-7-x\)
\(5x-x=7+3\)hoặc\(5x+x=-7+3\)
\(4x=10\)hoặc\(6x=-4\)
\(\Rightarrow x=\frac{10}{4}=2,5\)hoặc\(x=-\frac{4}{6}=-\frac{2}{3}\)
Vậy......
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
a: \(=x^2-2x-3x^2+5x-4+2x^2-3x+7=3\)
b: \(=2x^3-4x^2+x-1-5+x^2-2x^3+3x^2-x=4\)
c: \(=1-x-\dfrac{3}{5}x^2-x^4+2x+6+0.6x^2+x^4-x=7\)
a) Ta có: \(5x^2-3x\left(x+2\right)\)
\(=5x^2-3x^2-6x\)
\(=2x^2-6x\)
b) Ta có: \(3x\left(x-5\right)-5x\left(x+7\right)\)
\(=3x^2-15x-5x^2-35x\)
\(=-2x^2-50x\)
c) Ta có: \(3x^2y\left(2x^2-y\right)-2x^2\left(2x^2y-y^2\right)\)
\(=3x^2y\left(2x^2-y\right)-2x^2y\left(2x^2-y\right)\)
\(=x^2y\left(2x^2-y\right)=2x^4y-x^2y^2\)
d) Ta có: \(3x^2\left(2y-1\right)-\left[2x^2\cdot\left(5y-3\right)-2x\left(x-1\right)\right]\)
\(=6x^2y-3x^2-\left[10x^2y-6x^2-2x^2+2x\right]\)
\(=6x^2y-3x^2-10x^2y+6x^2+2x^2-2x\)
\(=-4x^2y+5x^2-2x\)
e) Ta có: \(4x\left(x^3-4x^2\right)+2x\left(2x^3-x^2+7x\right)\)
\(=4x^4-16x^3+4x^4-2x^3+14x^2\)
\(=8x^4-18x^3+14x^2\)
f) Ta có: \(25x-4\left(3x-1\right)+7x\left(5-2x^2\right)\)
\(=25x-12x+4+35x-14x^3\)
\(=-14x^3+48x+4\)
a) Ta có: \(\left|2x-1\right|=\left|2x+3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2x+3\left(loại\right)\\2x-1=-2x-3\end{matrix}\right.\Leftrightarrow2x+2x=-3+1\)
\(\Leftrightarrow4x=-2\)
hay \(x=-\dfrac{1}{2}\)