1.tìm x biết:
a. 3/4 - 2 . |3x+1| = 5/8
b.|3x+2| - |x-3| = 7/2
c |2x+1| + |3x-2| -1/1/3(hỗn sỗ) =0
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29 tháng 7 2021
a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)
\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
hay \(x=-\dfrac{1}{4}\)
c) Ta có: \(8x^3-50x=0\)
\(\Leftrightarrow2x\left(4x^2-25\right)=0\)
\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
29 tháng 7 2021
e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
2
10 tháng 1 2023
Bài `1:`
`a)3x^3+6x^2=3x^2(x+2)`
`b)x^2-y^2-2x+2y=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)`
Bài `2:`
`a)(2x-1)^2-25=0`
`<=>(2x-1-5)(2x-1+5)=0`
`<=>(2x-6)(2x+4)=0`
`<=>[(x=3),(x=-2):}`
`b)Q.(x^2+3x+1)=x^3+2x^2-2x-1`
`<=>Q=[x^3+2x^2-2x-1]/[x^2+3x+1]`
`<=>Q=[x^3-x^2+3x^2-3x+x-1]/[x^2+3x+1]`
`<=>Q=[(x-1)(x^2+3x+1)]/[x^2+3x+1]=x-1`
TL
1
3 tháng 12 2021
\(a,\Rightarrow5x+3x^2-3x^2-x+2=6\\ \Rightarrow4x=4\Rightarrow x=1\\ b,\Rightarrow\left(2x+\dfrac{1}{2}-1+2x\right)\left(2x+\dfrac{1}{2}+1-2x\right)=2\\ \Rightarrow\dfrac{3}{2}\left(4x-\dfrac{1}{2}\right)=2\\ \Rightarrow6x-\dfrac{3}{4}=2\\ \Rightarrow6x=\dfrac{11}{4}\\ \Rightarrow x=\dfrac{11}{24}\\ c,\Rightarrow\left(x+3\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
18 tháng 5 2021
a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)
\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)
\(\Leftrightarrow-9x=18\)
hay x=-2
Vậy: S={-2}
b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)
\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
\(\Leftrightarrow14x=7\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)
\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)
\(\Leftrightarrow5.2x=-6.5\)
hay \(x=-\dfrac{5}{4}\)
Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)
18 tháng 5 2021
d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x+16=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
Vậy: S={-5}
e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
Vậy: S={0}
NT
1
8 tháng 9 2021
a: Ta có: \(\left(8x^2-4x\right):\left(-4x\right)-\left(x+2\right)=8\)
\(\Leftrightarrow-2x+1-x-2=8\)
\(\Leftrightarrow-3x=9\)
hay x=-3
b: Ta có: \(\left(2x^4-3x^3+x^2\right):\left(-\dfrac{1}{2}x^2\right)+4\left(x-1\right)^2=0\)
\(\Leftrightarrow-4x^2+6x-2+4x^2-8x+4=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1
19 tháng 4 2023
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
8 tháng 9 2021
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
8 tháng 9 2021
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
8 tháng 9 2023
2:
a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8
=>x^2-x-12-x^2+4x+5=8
=>3x-7=8
=>3x=15
=>x=5
b: =>3x^2+3x-2x-2-3x^2-21x=13
=>-20x=15
=>x=-3/4
c: =>x^2-25-x^2-2x=9
=>-2x=25+9=34
=>x=-17
d: =>x^3-1-x^3+3x=1
=>3x-1=1
=>3x=2
=>x=2/3
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
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......
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