Giải phương trình: (x+1)2(1+2/x)2+(1+1/x)2 = 8(1+2/x)2
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DP
7 tháng 6 2017
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(x+x+x+x+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(4x+\frac{15}{16}=1\)
\(4x=1-\frac{15}{16}\)
\(4x=\frac{1}{16}\)
\(x=\frac{1}{16}\div4\)
\(x=\frac{1}{64}\)
6 tháng 6 2017
ta coi x để ra mội bên ta có 1/2 + 1/4 + 1/8 + 1/6 = 25/24
x + 25/24 = 1
x = 1 - 25/24
x = -1/24
mk nha
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Cô Hoàng Huyền
Admin
VIP
8 tháng 8 2017
1) \(\Delta'=1^2-\left(m-1\right)=2-m\)
Để pt có 2 nghiệm thì \(\Delta'\ge0\Leftrightarrow2-m\ge0\Leftrightarrow m\le2\)
Khi đó \(x_1=1+\sqrt{2-m};x_2=1-\sqrt{2-m}\)
TH1: \(2\left(1+\sqrt{2-m}\right)-\left(1-\sqrt{2-m}\right)=7\Leftrightarrow1+3\sqrt{2-m}=7\)
\(\Leftrightarrow\sqrt{2-m}=2\Leftrightarrow2-m=4\Rightarrow m=-2\left(tm\right)\)
TH2: \(2\left(1-\sqrt{2-m}\right)-\left(1+\sqrt{2-m}\right)=7\Leftrightarrow1-3\sqrt{2-m}=7\) (VÔ LÝ)
Vậy m = - 2.
2) \(P=\frac{x^4+3x^2+1}{x^2+1}=\frac{\left(x^4+2x^2+1\right)+\left(x^2+1\right)+2}{x^2+1}=\left(x^2+1\right)+\frac{2}{x^2+1}+1\)
Vì \(x^2+1\ge1\), áp dụng bđt Cô si ta có:
\(\left(x^2+1\right)+\frac{2}{x^2+1}\ge2\sqrt{\left(x^2+1\right).\frac{2}{x^2+1}}=2\sqrt{2}\)
Vậy \(P\ge2\sqrt{2}+1\)
Dấu bằng xảy ra khi
\(x^2+1=\frac{2}{x^2+1}\Leftrightarrow x^2+1=\sqrt{2}\Rightarrow x^2=\sqrt{2}-1\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\sqrt{2}-1}\\x=-\sqrt{\sqrt{2}-1}\end{cases}}\)
22 tháng 6 2018
Ta có: \(\hept{\begin{cases}\left(\frac{1}{x}+y\right)+\left(\frac{1}{x}-y\right)=\frac{5}{8}\\\left(\frac{1}{x}+y\right)-\left(\frac{1}{x}-y\right)=-\frac{3}{8}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{2}{x}=\frac{5}{8}\\2y=-\frac{3}{8}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{16}{5}\\y=-\frac{3}{16}\end{cases}}}\)
PA
1
TN
24 tháng 9 2016
Đk:\(x\ge1\)
\(pt\Leftrightarrow3\left(x-2\right)\sqrt{x-1}\sqrt{x^2+x+1}+18\left(x-1\right)=x\left(x^2+x+1\right)\)
Chia 2 vế của pt cho \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)ta đc:
\(3\left(x-2\right)\frac{\sqrt{x-1}}{\sqrt{x^2+x+1}}+\frac{18\left(x-1\right)}{x^2+x+1}=x\)
Đặt \(y=\frac{\sqrt{x-1}}{\sqrt{x^2+x+1}}\left(y\ge0\right)\) pt trở thành
\(3\left(x-2\right)y+18y^2-x=0\)
\(\Leftrightarrow\left(3y-1\right)\left(6y+x\right)=0\)
\(\Leftrightarrow3y-1=0\left(y\ge0;x\ge1\Rightarrow6y+x\ge1\right)\)
\(\Leftrightarrow y=\frac{1}{3}\)\(\Leftrightarrow\frac{\sqrt{x-1}}{\sqrt{x^2+x+1}}=\frac{1}{3}\)
\(\Leftrightarrow9\left(x-1\right)=x^2+x+1\)
\(\Leftrightarrow x^2-8x+10=0\)
\(\Leftrightarrow x=4\pm\sqrt{6}\)
Vậy...
\(\left(x+1\right)^2\left(1+\frac{2}{x}\right)^2+\left(1+\frac{1}{x}\right)^2=8\left(1+\frac{2}{x}\right)^2\left(ĐK:x\ne0\right)\)
\(\Leftrightarrow\left[\left(x+1\right)\left(1+\frac{2}{x}\right)\right]^2+\left(\frac{x+1}{x}\right)^2=8\left(\frac{x+2}{x}\right)^2\)
\(\Leftrightarrow\left[\left(x+1\right)\cdot\frac{x+2}{x}\right]^2+\frac{\left(x+1\right)^2}{x^2}=8\cdot\frac{\left(x+2\right)^2}{x^2}\)
\(\Leftrightarrow\left[\frac{\left(x+1\right)\left(x+2\right)}{x}\right]^2+\frac{x^2+2x+1}{x^2}=\frac{8\left(x+2\right)^2}{x^2}\)
\(\Leftrightarrow\left(\frac{x^2+3x+2}{x}\right)^2+\frac{x^2+2x+1}{x^2}=\frac{8x^2+32x+32}{x^2}\)
\(\Leftrightarrow\frac{\left(x^2+3x+2\right)^2}{x^2}+\frac{x^2+2x+1}{x^2}=\frac{8x^2+32x+32}{x^2}\)
\(\Leftrightarrow\frac{x^4+13x^2+4+6x^3+12x}{x^2}+\frac{x^2+2x+1}{x^2}-\frac{8x^2+32x+32}{x^2}=0\)
\(\Leftrightarrow\frac{x^4+6x^2-27+6x^3-18x}{x^2}=0\)
=> \(x^4+6x^3+6x^2-18x-27=0\)
<=> \(x^4+3x^3+3x^3+9x^2-3x^2-9x-9x-27=0\)
<=> \(x^3\left(x+3\right)+3x^2\left(x+3\right)-3x\left(x+3\right)-9\left(x+3\right)=0\)
<=> \(\left(x+3\right)\left(x^3+3x^2-3x-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x^3+3x^2-3x-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\pm\sqrt{3}\end{cases}\left(tmđk\right)}}\)