giải phương trình x^2-4x-6=căn(2x^2-8x+12)
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NH
0
LN
2
NT
14 tháng 9 2017
a) căn(2x+5) - căn(3-x) = x2 -5x + 8
Điều kiện : \(-\frac{5}{2}\Leftarrow x\Leftarrow3\)
căn(2x+5) - căn(3-x) = x^2-5x+8
\(\Leftrightarrow\)[căn(2x+5)-3]-[căn(3-x)-1]=x2 -5x+6
nhân liên hợp
\(\Leftrightarrow\)(2x+5-9) / [căn(2x+5)+3] -(3-x-1) / [căn (3-x)+1]=(x-2)(x-3)
\(\Leftrightarrow\)(2x-4) / [căn (2x+5)+3] -(2-x) / [ căn (3-x)+1]-(x-2)(x-3)=0
\(\Leftrightarrow\)(x-2).M=0
\(\Leftrightarrow\)x=2 hoặc M=0
M=2 / [căn(2x+5)+3]+1 / [căn(3-x)+1]-x+3
2/[can(2x+5)+3]+1/[can(3-x)+1]>0 voi moi x
voi -5/2<=x<=3 <->3-x thuoc[0;11/2]
nen M>0
vay x=2
b/ 2+ căn(3-8x) = 6x + căn(4x-1)
dk[1/4;8/3]
6x-2+căn(4x-1)-căn(3-8x)=0
<->2(3x-1)+(4x-1-3+8x)/[căn(4x-1)+căn(...
<->2(3x-1)+(12x-4)/[căn(4x-1)+căn(3-8x...
<->2(3x-1)+4(3x-1)/[căn(4x-1)+căn(3-8x...
<->(3x-1){2+4/[căn(4x-1)+căn(3-8x)]}=0
2+4/[căn(4x-1)+căn(3-8x)>0
nen 3x-1=0
x=1/3
14 tháng 9 2017
a) căn(2x+5) - căn(3-x) = x^2-5x+8
dkxd -5/2<=x<=3
căn(2x+5) - căn(3-x) = x^2-5x+8
<->[can(2x+5)-3]-[can(3-x)-1]=x^2-5x+6
nhan lien hop
<->(2x+5-9)/[can(2x+5)+3] -(3-x-1)/[can(3-x)+1]=(x-2)(x-3)
<->(2x-4)/[can(2x+5)+3] -(2-x)/[can(3-x)+1]-(x-2)(x-3)=0
<->(x-2).M=0
<->x=2 hoac M=0
M=2/[can(2x+5)+3]+1/[can(3-x)+1]-x+3
2/[can(2x+5)+3]+1/[can(3-x)+1]>0 voi moi x
voi -5/2<=x<=3 <->3-x thuoc[0;11/2]
nen M>0
vay x=2
b/ 2+ căn(3-8x) = 6x + căn(4x-1)
dk[1/4;8/3]
6x-2+căn(4x-1)-căn(3-8x)=0
<->2(3x-1)+(4x-1-3+8x)/[căn(4x-1)+căn(...
<->2(3x-1)+(12x-4)/[căn(4x-1)+căn(3-8x...
<->2(3x-1)+4(3x-1)/[căn(4x-1)+căn(3-8x...
<->(3x-1){2+4/[căn(4x-1)+căn(3-8x)]}=0
2+4/[căn(4x-1)+căn(3-8x)>0
nen 3x-1=0
x=1/3
VT
1
NC
4 tháng 3 2021
\(\frac{x^2-2x+2}{x-1}+\frac{x^2-8x+20}{x-4}=\frac{x^2-4x+6}{x-2}+\frac{x^2-6x+12}{x-3}\)\(ĐKXĐ:x\ne1;2;3;4\)
\(\Leftrightarrow\frac{\left(x-1\right)^2+1}{x-1}+\frac{\left(x-4\right)^2+4}{x-4}=\frac{\left(x-2\right)^2+2}{x-2}+\frac{\left(x-3\right)^2+3}{x-3}\)
\(\Leftrightarrow\left(\frac{\left(x-1\right)^2}{x-1}+\frac{1}{x-1}\right)+\left(\frac{\left(x-4\right)^2}{x-4}+\frac{4}{x-4}\right)=\left(\frac{\left(x-2\right)^2}{x-2}+\frac{2}{x-2}\right)+\left(\frac{\left(x-3\right)^2}{x-3}+\frac{3}{x-3}\right)\)
\(\Leftrightarrow x-1+\frac{1}{x-1}+x-4+\frac{1}{x-4}=x-2+\frac{1}{x-2}+x-3+\frac{1}{x-3}\)
\(\Leftrightarrow\frac{1}{x-1}+\frac{4}{x-4}=\frac{2}{x-2}+\frac{3}{x-3}\)
\(\Leftrightarrow\frac{x-4+4x-4}{\left(x-1\right)\left(x-4\right)}=\frac{2x-6+3x-6}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{5x-8}{x^2-5x+4}=\frac{5x-12}{x^2-5x+6}\)
\(\Leftrightarrow\left(5x-8\right)\left(x^2-5x+6\right)=\left(5x-12\right)\left(x^2-5x+4\right)\)
Tự giải ra rồi tìm x nhé
13 tháng 3 2022
dấu suy ra số 4 là 1/(x+1) + 1/(x+4) mà.
1 tháng 3 2021
`d,(10x+3)/12=1+(6+8x)/9`
`<=>(10x+3)/12=(8x+15)/9`
`<=>30x+9=32x+60`
`<=>2x=-51`
`<=>x=-51/2`
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 4 2022
a.
\(3\sqrt{-x^2+x+6}\ge2\left(1-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x^2+x+6\ge0\\1-2x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-2x\ge0\\9\left(-x^2+x+6\right)\ge4\left(1-2x\right)^2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\le x\le3\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\25\left(x^2-x-2\right)\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x\le3\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\-1\le x\le2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-1\le x\le3\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 4 2022
b.
ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\dfrac{2x^2+8x+5-16x}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-4x+5-4x}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
\(\Leftrightarrow x=\dfrac{4\pm\sqrt{6}}{2}\)
26 tháng 7 2017
ĐK \(x\ne\left\{1;2;3;4\right\}\)
Ta có \(\frac{x^2-2x+2}{x-1}+\frac{x^2-8x+20}{x-4}=\frac{x^2-4x+6}{x-2}+\frac{x^2-6x+12}{x-3}\)
\(\Leftrightarrow\frac{\left(x-1\right)^2+1}{x-1}+\frac{\left(x-4\right)^2+4}{x-4}=\frac{\left(x-2\right)^2+2}{x-2}+\frac{\left(x-3\right)^2+3}{x-3}\)
\(\Leftrightarrow x-1+\frac{1}{x-1}+x-4+\frac{4}{x-4}=x-2+\frac{2}{x-2}+x-3+\frac{3}{x-3}\)
\(\Leftrightarrow\frac{1}{x-1}+\frac{4}{x-4}=\frac{2}{x-2}+\frac{3}{x-3}\)
\(\Leftrightarrow\frac{5x-8}{x^2-5x+4}=\frac{5x-12}{x^2-5x+6}\)\(\Leftrightarrow\left(5x-8\right)\left(x^2-5x+6\right)=\left(5x-12\right)\left(x^2-5x+4\right)\)
\(\Leftrightarrow5x^3-25x^2+30x-8x^2+40x-48=5x^3-25x^2+20x-12x^2+60x-48\)
\(\Leftrightarrow4x^2-10x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{2}\end{cases}\left(tm\right)}\)
Vậy x=0 hoặc x=5/2
ND
1
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 1 2022
ĐKXĐ: \(x\ge0\)
\(\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
\(\Leftrightarrow...\)
Đặt \(\sqrt{2x^2-8x+12}=a>0\)thì được
\(2\left(x^2-4x-6\right)=2\sqrt{2x^2-8x+12}\)
\(\Leftrightarrow2x^2-8x-12=2\sqrt{2x^2-8x+12}\)
\(\Rightarrow a^2-2a-24=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=6\\a=-4\left(loai\right)\end{cases}}\)
\(\Rightarrow\sqrt{2x^2-8x+12}=6\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=6\end{cases}}\)