1. Tìm x trong các trường đẳng thức sau:
a) |2x -3|= 5
b) |2x- 1| = | 2x+ 3|
c) | x-1| + 3x= 1
d) |5x -3| -x= 7
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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 \(x=-\dfrac{1}{2}\)
\(|5x-3|-x=7\)
\(|5x-3|=7+x\)
\(\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}}\)
\(\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}\)
\(\orbr{\begin{cases}4x=10\\6x=-4\end{cases}}\)
\(\orbr{\begin{cases}x=2,5\\x=\frac{-2}{3}\end{cases}}\)
Vậy x = 2,5 hoặc x = -2/3
Hi Hi!
a) \(\left|3x-1\right|=5\)
\(\Rightarrow\orbr{\begin{cases}3x-1=5\\3x-1=-5\end{cases}}\Rightarrow\orbr{\begin{cases}3x=6\\3x=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{-4}{3}\end{cases}}\)
b) \(\left|x-1\right|+11=45\)
\(\Rightarrow\left|x-1\right|=35\)
\(\Rightarrow\orbr{\begin{cases}x-1=35\\x-1=-35\end{cases}\Rightarrow\orbr{\begin{cases}x=36\\x=-34\end{cases}}}\)
c)\(\left|2x+1\right|=\left|2x-3\right|\)
\(\Rightarrow\orbr{\begin{cases}2x+1=2x-3\\2x+1=-2x+3\end{cases}\Rightarrow\orbr{\begin{cases}2x-2x=-3-1\\2x+2x=3-1\end{cases}\Rightarrow}\orbr{\begin{cases}0=-4\\4x=2\end{cases}\Rightarrow}\orbr{\begin{cases}vôlis\\x=\frac{1}{2}\end{cases}}}\)
d)\(\left|x+1\right|-5x=7\)
\(\Rightarrow\left|x+1\right|=7+5x\)
\(\Rightarrow\orbr{\begin{cases}x+1=7+5x\\x+1=-7-5x\end{cases}\Rightarrow\orbr{\begin{cases}x-5x=7-1\\x+5x=-7-1\end{cases}\Rightarrow}\orbr{\begin{cases}-4x=6\\6x=-8\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{4}{3}\end{cases}}}\)
hok tốt!!!
Bài 1
A= (x-2)(2x-1)-2x(x+3)=2x2-x-4x+2-2x2-6x=-11x+2
Bài 1:
a) \(A=\left(x-2\right)\left(2x-1\right)-2x\left(x+3\right)\)
\(A=2x^2-x-4x+2-2x^2-6x\)
\(A=-11x+2\)
b) \(B=\left(3x-2\right)\left(2x+1\right)-\left(6x-1\right)\left(x+2\right)\)
\(B=6x^2+3x-4x-2-6x^2-12x+x+2\)
\(B=-12x\)
c) \(C=6x\left(2x+3\right)-\left(4x-1\right)\left(3x-2\right)\)
\(C=12x^2+18x-12x^2+8x+3x-2\)
\(C=29x-2\)
d) \(D=\left(2x+3\right)\left(5x-2\right)+\left(x+4\right)\left(2x-1\right)-6x\left(2x-3\right)\)
\(D=10x^2-4x+15x-6+2x^2-x+8x-4-12x^2+18x\)
\(D=36x-10\)
\(a,\Rightarrow\left[{}\begin{matrix}2x-3=5\\3-2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\\ b,\Rightarrow\left|x-1\right|=1-3x\\ \Rightarrow\left[{}\begin{matrix}x-1=1-3x\\x-1=3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
a) \(\Rightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=8\\2x=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
b) \(\left|x-1\right|+3x=1\left(đk:x\le\dfrac{1}{3}\right)\)
\(\Rightarrow x-1=3x-1\)
\(\Rightarrow2x=0\Rightarrow x=0\left(tm\right)\)
Bài 1:
\(a,6x^2-15x^3y\\ b,=-\dfrac{2}{3}x^2y^3+\dfrac{2}{3}x^4y-\dfrac{8}{3}xy\)
Bài 2:
\(a,=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\\ b,=3x^2-6x-5x+5x^2-8x^2+24=24-11x\\ c,=x^5+x^3-2x^3-2x=x^5-x^3-2x\)
a. 3(x-2)-10=5(2x + 1)
<=> 3x - 6 - 10 = 10x + 5
<=> 3x - 10x = 5 + 6 + 10
<=> -7x = 21
<=> x = -3
b. 3x + 2=8 -2(x-7)
<=> 3x + 2 = 8 - 2x + 14
<=> 3x + 2x = 8 + 14 - 2
<=> 5x = 20
<=> x = 4
c. 2x-(2+5x)= 4(x + 3)
<=> 2x - 2 - 5x = 4x + 12
<=> 2x - 5x - 4x = 12 + 2
<=> -7x = 14
<=> x = -2
d. 5-(x +8)=3x + 3(x-9)
<=> 5 - x - 8 = 3x + 3x - 27
<=> -x - 3x - 3x = -27 + 8 - 5
<=> -7x = -24
<=> x = 24/7
e. 3x - 18 + x= 12-(5x + 3)
<=> 3x - 18 + x = 12 - 5x - 3
<=> 3x + x - 5x = 12 - 3 + 18
<=> -x = 27
<=> x = - 27
a. 3(x-2)-10=5(2x + 1)
<=> 3x - 6 - 10 = 10x + 5
<=> 3x - 10x = 5 + 6 + 10
<=> -7x = 21
<=> x = -3
b. 3x + 2=8 -2(x-7)
<=> 3x + 2 = 8 - 2x + 14
<=> 3x + 2x = 8 + 14 - 2
<=> 5x = 20
<=> x = 4
c. 2x-(2+5x)= 4(x + 3)
<=> 2x - 2 - 5x = 4x + 12
<=> 2x - 5x - 4x = 12 + 2
<=> -7x = 14
<=> x = -2
d. 5-(x +8)=3x + 3(x-9)
<=> 5 - x - 8 = 3x + 3x - 27
<=> -x - 3x - 3x = -27 + 8 - 5
<=> -7x = -24
<=> x = 24/7
e. 3x - 18 + x= 12-(5x + 3)
<=> 3x - 18 + x = 12 - 5x - 3
<=> 3x + x - 5x = 12 - 3 + 18
<=> -x = 27
<=> x = - 27
|2x-3|=5 suy ra:th1:2x-3=5 2x=5+3 2x=8 x=8:2 x=4 th2:2x-3=-5 2x=-5+3 x=-2 x=-2:2 x=-1
1.
a) | 2x+3 |= 5
=>2x+3=\(\pm\) 5
=>\(\left[\begin{array}{nghiempt}2x+3=5\\2x+3=-5\end{array}\right.\) => \(\left[\begin{array}{nghiempt}2x=2\\2x=-8\end{array}\right.\) => \(\left[\begin{array}{nghiempt}x=1\\x=-4\end{array}\right.\)
Vậy x\(\in\)\(\left\{1;-4\right\}\)