Tìm x biết:
a) |2x-1|=|x+2|
b) 2.3x.32=18 (x thuộc N)
c) \(\dfrac{x+2}{5}=\dfrac{2-3x}{3}\)
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a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
a: =>2x-1=-2
=>2x=-1
hay x=-1/2
b: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\-\dfrac{2}{5}x-7=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};-\dfrac{35}{2}\right\}\)
c: x/8=9/4
nên x/8=18/8
hay x=18
d: \(\Leftrightarrow\left(x-3\right)^2=36\)
=>x-3=6 hoặc x-3=-6
=>x=9 hoặc x=-3
e: =>-1,7x=6,12
hay x=-3,6
h: =>x-3,4=27,6
hay x=31
a) \(\dfrac{1}{3}\div\left(2x-1\right)=\dfrac{-1}{6}\)
\(\left(2x-1\right).\dfrac{1}{3}\div\left(2x-1\right)=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=-1\left(2x-1\right)\div6\)
\(\dfrac{1}{3}=-2x+1\div6\)
\(x=-\dfrac{1}{2}\)
b) \(\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0\)
\(TH1:3x+2=0\)
\(3x=0-2\)
\(3x=-2\)
\(x=\dfrac{-2}{3}\)
\(TH2:\left(-\dfrac{2}{5}x-7\right)=0\)
\(\left(\dfrac{-2}{5}x-7\right)=0\)
\(\left(\dfrac{-2x}{5}+\dfrac{5\left(-7\right)}{5}\right)=0\)
\(\left(\dfrac{-2x-35}{5}\right)=0\)
\(-2x-35=0\)
\(-2x=0+35\)
\(x=-\dfrac{35}{2}\)
c) \(\dfrac{x}{8}=\dfrac{9}{4}\)
\(\Leftrightarrow x=\dfrac{9.8}{4}=\dfrac{72}{4}=18\)
\(x=18\)
d) \(\dfrac{x-3}{2}=\dfrac{18}{x-3}\)
\(x-3=18+2\)
\(x=20-3\)
\(x=17\)
e) \(4,5x-6,2x=6,12\)
\(\dfrac{9x}{2}-6,2.x=6,12\)
\(\dfrac{9x}{2}+\dfrac{-31x}{5}=6,12\)
\(\dfrac{5.9x}{10}+\dfrac{2\left(-31\right)x}{10}=6.12\)
\(\dfrac{45x-62x}{10}=6.12\)
\(=-17x\div10=6.12\)
\(-17x=10.6.12\)
\(x=-3,6\)
h) \(11,4-\left(x-3,4\right)=-16,2\)
\(x-3,4=-16,2+11,4\)
\(x-3,4=-4,8\)
\(x=-1,4\)
a. \(A=\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\left(ĐKXĐ:x\ne1;x\ne-3\right)\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{x+3}{x-1}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{\left(x+3\right)^2}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{2-3x+x^2+6x+9-x^2+1}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}.\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{3x+12}=\dfrac{x^2+x+1}{x+3}\)
\(M=A.B=\dfrac{x^2+x+1}{x+3}.\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x-2}{x+3}\)
b. -Để M thuộc Z thì:
\(\left(x^2+x-2\right)⋮\left(x+3\right)\)
\(\Rightarrow\left(x^2+3x-2x-6+4\right)⋮\left(x+3\right)\)
\(\Rightarrow\left[x\left(x+3\right)-2\left(x+3\right)+4\right]⋮\left(x+3\right)\)
\(\Rightarrow4⋮\left(x+3\right)\)
\(\Rightarrow x+3\in\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow x\in\left\{-2;-1;1;-4;-5;-7\right\}\)
c. \(A^{-1}-B=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{x^3-1}\)
\(=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2-x+3x-3-x^2-x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
\(=\dfrac{1}{x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
\(Max=\dfrac{4}{3}\Leftrightarrow x=\dfrac{-1}{2}\)
\(a,=\dfrac{4x+8}{x^2+2x}=\dfrac{4\left(x+2\right)}{x\left(x+2\right)}=\dfrac{4}{x}\\ b,=\dfrac{\left(2x-3\right)-\left(2x-4\right)}{x-2}=\dfrac{2x-3-2x+4}{x-2}=\dfrac{1}{x-2}\\ c,=\dfrac{2x-1-3x-2}{x+3}=\dfrac{-x-3}{x+3}=\dfrac{-\left(x+3\right)}{x+3}=-1\\ d,=\dfrac{11x-18+x}{2x-3}=\dfrac{12x-18}{2x-3}=\dfrac{6\left(2x-3\right)}{2x-3}=6\)
\(e,=\dfrac{3x-6-9x+3}{2x+1}=\dfrac{-6x-3}{2x+1}=\dfrac{-3\left(2x+1\right)}{2x+1}=-3\)
a: \(A\left(x\right)+B\left(x\right)\)
\(=-2x^3+11x^2-5x-\dfrac{1}{5}+2x^3-3x^2-7x+\dfrac{1}{5}\)
\(=8x^2-12x\)
b: C(x)=A(x)-B(x)
\(=-2x^3+11x^2-5x-\dfrac{1}{5}-2x^3+3x^2+7x-\dfrac{1}{5}\)
\(=-4x^3+14x^2+2x-\dfrac{2}{5}\)
a: \(=\dfrac{x^3-x^2+x-1}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x+2\right)\left(x+1\right)}-\dfrac{3x}{\left(x-2\right)\left(x+1\right)}+\dfrac{2x+5}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x-1\right)\left(x^2+1\right)\left(x+1\right)-x^2+4x-4-3x^2-6x+2x+5}{\left(x+2\right)\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{x^4-1-4x^2+1}{\left(x+2\right)\left(x-2\right)\left(x+1\right)}=\dfrac{x^2\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)\left(x+1\right)}\)
=x^2/x+1
b: Sửa đề: \(\dfrac{19x^2-30x+9}{2x^3+54}-\dfrac{x-3}{2x^2+6x}-\dfrac{3x^2}{2x^2-6x+18}\) \(=\dfrac{19x^2-30x+9}{2\left(x+3\right)\left(x^2-3x+9\right)}-\dfrac{x-3}{2x\left(x+3\right)}-\dfrac{3x^2}{2\left(x^2-3x+9\right)}\)
\(=\dfrac{19x^3-30x^2+9x-\left(x-3\right)\left(x^2-3x+9\right)-3x^3\left(x+3\right)}{2x\left(x+3\right)\left(x^2-3x+9\right)}\)
\(=\dfrac{19x^3-30x^2+9x-3x^4-9x^3-\left(x^3-3x^2+9x-3x^2+9x-27\right)}{2x\left(x+3\right)\left(x^2-3x+9\right)}\)
\(=\dfrac{-3x^4+10x^3-30x^2+9x-x^3+6x^2-18x+27}{2x\left(x+3\right)\left(x^2-3x+9\right)}\)
\(=\dfrac{-3x^4+10x^3-24x^2-9x+27}{2x\left(x+3\right)\left(x^2-3x+9\right)}\)
a) \(\dfrac{3x^2y}{2xy^5}=\dfrac{3x}{2y^4}\)
b) \(\dfrac{3x^2-3x}{x-1}=\dfrac{3x\left(x-1\right)}{x-1}=3x\)
c) \(\dfrac{ab^2-a^2b}{2a^2+a}=\dfrac{ab\left(b-a\right)}{a\left(2a+1\right)}=\dfrac{b\left(b-a\right)}{2a+1}=\dfrac{b^2-ab}{2a+1}\)
d) \(\dfrac{12\left(x^4-1\right)}{18\left(x^2-1\right)}=\dfrac{2\left(x^2-1\right)\left(x^2+1\right)}{3\left(x^2-1\right)}=\dfrac{2\left(x^2+1\right)}{3}\)
`a, (3x^2y)/(2xy^5)`
`= (3x)/(2y^4)`
`b, (3x^2-3x)/(x-1)`
`= (3x(x-1))/(x-1)`
`= 3x`
`c, (ab^2-a^2b)/(2a^2+a)`
`= (b(a-b))/((2a+1))`
`d, (12(x^4-1))/(18(x^2-1)) = (2(x^2+1))/3`.
a: ĐKXĐ: x>=2/3
\(\dfrac{x-2}{\sqrt{3x-2}+2}=9\)
=>\(x-2=9\sqrt{3x-2}+18\)
=>\(9\sqrt{3x-2}=x-2-18=x-20\)
=>\(\Leftrightarrow\left\{{}\begin{matrix}x>=20\\81\left(3x-2\right)=x^2-40x+400\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=20\\x^2-40x+400-243x+162=0\end{matrix}\right.\)
=>x>=20 và x^2-283x+562=0
=>x=281(nhận) hoặc x=2(loại)
b: ĐKXĐ: x>=2/5
\(\sqrt{5x-2}=9\)
=>5x-2=81
=>5x=83
=>x=83/5
c: ĐKXĐ: x>=-1; x<>8
\(\dfrac{2x-16}{\sqrt{x+1}-3}=5\)
=>\(2x-16=5\sqrt{x+1}-15\)
=>\(\sqrt{25x+25}=2x-16+15=2x-1\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\4x^2-4x+1=25x+25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\4x^2-29x-24=0\end{matrix}\right.\)
=>x=8(nhận) hoặc x=-3/4(loại)
a)|2x-1|=|x+2|
*)Nếu x\(\ge\dfrac{1}{2}\)
=>2x-1=x+2
<=>x=3(TM)
*)Nếu -2\(\le\)x<\(\dfrac{1}{2}\)
=>1-2x=x+2
<=>3x=-1
<=>x=\(-\dfrac{1}{3}\)(TM)
*)Nếu x<-2
=>1-2x=-x-2
<=>x=3(L)
Vậy S={3;\(-\dfrac{1}{3}\)}
b)2.3x.32=18
<=>3x+2=9=32
<=>x+2=2
<=>x=0
Vậy S={0}
c)\(\dfrac{x+2}{5}=\dfrac{2-3x}{3}\)
<=>3x+6=10-15x
<=>18x=16
<=>x=\(\dfrac{8}{9}\)
Vậy S={\(\dfrac{8}{9}\)}
ý c sai thì phải![lolang lolang](https://hoc24.vn/media/olmeditor/plugins/smiley/images/lolang.png)