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b)\(\frac{1}{x+\sqrt{x^2+x}}+\frac{1}{x-\sqrt{x^2+x}}=x\)
\(\Leftrightarrow\frac{x-\sqrt{x^2+x}}{\left(x+\sqrt{x^2+x}\right)\left(x-\sqrt{x^2+x}\right)}+\frac{x+\sqrt{x^2+x}}{\left(x-\sqrt{x^2+x}\right)\left(x+\sqrt{x^2+x}\right)}-\frac{x\left(x+\sqrt{x^2+x}\right)\left(x-\sqrt{x^2+x}\right)}{\left(x+\sqrt{x^2+x}\right)\left(x-\sqrt{x^2+x}\right)}=0\)
\(\Leftrightarrow\frac{x-\sqrt{x^2+x}+x+\sqrt{x^2+x}-x^2}{\left(x+\sqrt{x^2+x}\right)\left(x-\sqrt{x^2+x}\right)}=0\)
\(\Leftrightarrow\frac{-x^2+2x}{\left(x+\sqrt{x^2+x}\right)\left(x-\sqrt{x^2+x}\right)}=0\)
\(\Leftrightarrow\frac{-x\left(x+2\right)}{\left(x+\sqrt{x^2+x}\right)\left(x-\sqrt{x^2+x}\right)}=0\)
Dễ thấy: x=0 ko là nghiệm nên \(x+2=0\Rightarrow x=-2\)
c)\(\sqrt{2x+4}-2\sqrt{2-x}=\frac{12x-8}{\sqrt{9x^2+16}}\)
\(\Leftrightarrow\frac{\left(2x+4\right)-4\left(2-x\right)}{\sqrt{2x+4}+2\sqrt{2-x}}=\frac{4\left(3x-2\right)}{\sqrt{9x^2+16}}\)
\(\Leftrightarrow\frac{2\left(3x-2\right)}{\sqrt{2x+4}+2\sqrt{2-x}}=\frac{4\left(3x-2\right)}{\sqrt{9x^2+16}}\)
\(\Leftrightarrow\frac{2\left(3x-2\right)}{\sqrt{2x+4}+2\sqrt{2-x}}-\frac{4\left(3x-2\right)}{\sqrt{9x^2+16}}=0\)
\(\Leftrightarrow\left(3x-2\right)\left(\frac{2}{\sqrt{2x+4}+2\sqrt{2-x}}-\frac{4}{\sqrt{9x^2+16}}\right)=0\)
\(\Leftrightarrow x=\frac{2}{3}\)
d)\(2x^2+4x=\sqrt{\frac{x+3}{2}}\)
ĐK:\(x\ge-3\)
\(\Leftrightarrow4x^4+16x^3+16x^2=\frac{x+3}{2}\)
\(\Leftrightarrow\frac{8x^4+32x^3+32x^2-x-3}{2}=0\)
\(\Leftrightarrow8x^4+32x^3+32x^2-x-3=0\)
\(\Leftrightarrow\left(2x^2+3x-1\right)\left(4x^2+10x+3\right)=0\)
d)\(2x^2+4x=\sqrt{\frac{x+3}{2}}\)
ĐK:\(x\ge-3\)
\(\Leftrightarrow4x^4+16x^3+16x^2=\frac{x+3}{2}\)
\(\Leftrightarrow\frac{8x^4+32x^3+32x^2-x-3}{2}=0\)
\(\Leftrightarrow8x^4+32x^3+32x^2-x-3=0\)
\(\Leftrightarrow\left(2x^2+3x-1\right)\left(4x^2+10x+3\right)=0\)
a, \(x^2+\frac{9x^2}{\left(x-3\right)^2}=16\) (đk: \(x\ne3\))
<=> \(x^2+\frac{6x^2}{x-3}+\frac{9x^2}{\left(x-3\right)^2}-\frac{6x^2}{x-3}-16=0\)
<=>\(\left(x+\frac{3x}{x-3}\right)^2-\frac{6x^2}{x-3}-16=0\)
<=>\(\left(\frac{x^2}{x-3}\right)^2-\frac{6x^2}{x-3}-16=0\)
Đặt a=\(\frac{x^2}{x-3}\)
Có: \(a^2-6a-16=0\)
<=> (a+2)(a-8)=0
<=> \(\left[{}\begin{matrix}a=-2\\a=8\end{matrix}\right.\)
=> \(\frac{x^2}{x-3}=-2\) hoặc \(\frac{x^2}{x-3}=8\)
Tại \(\frac{x^2}{x-3}=-2\) <=> \(x^2+2x-6=0\)
\(\Delta=2^2-4\left(-6\right)=28>0\)
=> \(\sqrt{\Delta}=\sqrt{28}\)
=> \(\left[{}\begin{matrix}x_1=\frac{-2+2\sqrt{7}}{2}=-1+\sqrt{7}\\x_2=\frac{-2-2\sqrt{7}}{2}=-1-\sqrt{7}\end{matrix}\right.\)(tm)
Tại \(\frac{x^2}{x-3}=8\) <=> \(x^2-8x+24=0\) <=> (x-4)2+8=0( vô nghiệm)
Vậy....
Phạm Minh Quang,tth,HISINOMA KINIMADON,guyễn Huy Thắng,Lê Thị Thục Hiền,Nguyễn Thanh Hằng,Nguyễn Thị Diễm Quỳnh,.... Akai Haruma,Nguyễn Việt Lâm
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