tìm x trong các đẳng thức:
a) \(\left|2x-3\right|=5\) b)\(\left|2x-1\right|=\left|2x=3\right|\)
c) \(\left|x-1\right|+3x=11\) d)\(\left|5x-3\right|-x=7\)
Tìm x trong các đẳng thức :
\(a.\left|2x-1\right|=\left|2x+3\right|\)
\(b.\left|x-1\right|+3x=1\)
\(c.\left|5x-3\right|-x=7\)
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 \(x=-\dfrac{1}{2}\)
Tìm x trong các đẳng thức:
a)\(\left|2x-1\right|=\left|2x+3\right|\)
b)\(\left|x-1\right|+3x=5\)
c)\(\left|5x-3\right|-x=7\)
tìm x trong các đẳng thức:
a) \(\left|2x-3\right|=5\) b)\(\left|2x-1\right|=\left|2x=3\right|\)
c) \(\left|x-1\right|+3x=11\) d)\(\left|5x-3\right|-x=7\)
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......
Tìm x trong các đẳng thức:
a) \(\left|2x-3\right|=5\)
b) \(\left|2x-1\right|=\left|2x+3\right|\)
c) \(\left|x-1\right|+3x=1\)
d) \(\left|5x-3\right|-x=7\)
a) \(\left|2x-3\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=8\\2x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
Bài 11 : Tìm GTNN của của các biểu thức sau :
a ) \(A=\left|x+3\right|+\left|2x-5\right|+\left|x-7\right|.\)
b ) \(B=\left|x+2\right|+\left|3x-4\right|+\left|x-2\right|+5\)
c ) \(M=\left|x+2\right|+\left|x-3\right|\)
d ) \(C=\left|2x+5\right|+\left|2x+1\right|+\left|2x-7\right|+\left|2x-4\right|+4\)
e ) \(D=\left|3x-6\right|+\left|3x-9\right|+\left|3x-12\right|+\left|3x-15\right|+2018\)
Bài tập. Giải các phương trình sau:
a) \(\left|7-x\right|+2x=3\)
b) \(\left|2x-3\right|-4x-9=0\)
c) \(\left|3x+5\right|=\left|2-5x\right|\)
d) \(x\left|x-3\right|-\left|x^2+x+1\right|=1\)
a: =>|x-7|=3-2x
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)
b: =>|2x-3|=4x+9
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)
c: =>3x+5=2-5x hoặc 3x+5=5x-2
=>8x=-3 hoặc -2x=-7
=>x=-3/8 hoặc x=7/2
chứng minh rằng các biểu thức sau không phụ thuộc vào x:
a. \(A=\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)
b. \(B=\left(x^2-2\right)\left(x^2+x-1\right)-x\left(x^3+x^2-3x-2\right)\)
c. \(C=x\left(x^3+x^2-3x-2\right)-\left(x^2-2\right)\left(x^2+x-1\right)\)
BÀI 6 tìm x
1,\(2x\left(x-5\right)-\left(3x+2x^2\right)=0\) 2,\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
3,\(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\) 4,\(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
5,\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\) 6,\(2x\left(1-x\right)+5=9-2x^2\)
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
Tìm x, biết:
a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
\(\Leftrightarrow21x-15x^2-35+25x-4x+15x^2-4=4\)
\(\Leftrightarrow42x-39=4\)
\(\Leftrightarrow42x=4+39\)
\(\Leftrightarrow42x=43\)
\(\Rightarrow x=\dfrac{43}{42}\)
Tick cho mk nha