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\(\Leftrightarrow2x\left(x+5\right)-3\left(x-2\right)=7x+1\)
\(\Leftrightarrow2x^2+10x-3x+6-7x-1=0\)
\(\Leftrightarrow2x^2+5=0\)(vô lý)
ĐKXĐ:\(\left\{{}\begin{matrix}x\ne2\\x\ne-5\end{matrix}\right.\)
\(\dfrac{2x}{x-2}-\dfrac{3}{x+5}=\dfrac{7x+1}{x^2+3x-10}\\ \Leftrightarrow\dfrac{2x\left(x+5\right)}{\left(x+5\right)\left(x-2\right)}-\dfrac{3\left(x-2\right)}{\left(x+5\right)\left(x-2\right)}=\dfrac{7x+1}{x^2-2x+5x-10}\\ \Leftrightarrow\dfrac{2x^2+10x}{\left(x+5\right)\left(x-2\right)}-\dfrac{3x-6}{\left(x+5\right)\left(x-2\right)}=\dfrac{7x+1}{x\left(x-2\right)+5\left(x-2\right)}\\ \Leftrightarrow\dfrac{2x^2+10x}{\left(x+5\right)\left(x-2\right)}-\dfrac{3x-6}{\left(x+5\right)\left(x-2\right)}-\dfrac{7x+1}{\left(x+5\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2+10x-3x+6-7x-1}{\left(x+5\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{2x^2+5}{\left(x+5\right)\left(x-2\right)}=0\\ \Rightarrow2x^2+5=0\left(vô.lí\right)\)
Vậy pt vô nghiệm
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\(=\left(x^2+2x+1\right)+\left(y^2-8y+16\right)=\left(x+1\right)^2+\left(y-4\right)^2\ge0\forall x,y\)
dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\)
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Bài 1:
a) \(2x\left(x^2-5x+6\right)=2x^3-10x^2+12x\)
b) \(\left(7x^5+14x^2y^3-28x^3y^2\right):7x^2=x^3+2y^3-4xy^2\)
Bài 2:
\(x^2+y^2+2x-8y+17=\left(x^2+2x+1\right)+\left(y^2-8y+16\right)=\left(x+1\right)^2+\left(y-4\right)^2\ge0\forall x,y\)
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Lời giải:
a.
\(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)
\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)
\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)
b.
\(\frac{1}{x+2}=\frac{x(2-x)}{x(x+2)(2-x)}=\frac{x(2-x)}{x(4-x^2)}\)
\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)
c.
\(\frac{4x^2-3x+5}{x^3-1}\)
\(\frac{1-2x}{x^2+x+1}=\frac{(1-2x)(x-1)}{(x-1)(x^2+x+1)}=\frac{-2x^2+3x-1}{x^3-1}\)
\(-2=\frac{-2(x^3-1)}{x^3-1}\)
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k cho mk nha
x^4-2x^3+3x^2-2x+1
=(x^4-2x^3+x^2)+(x^2-2x+1)
=x^2(x^2-2x+1)+(x^2-2x+1)
=(x^2+1)(x^2-2x+1)
=(x^2+1)(x-1)^2
\(\left(2x-3\right)^2-\left(x-5\right)\left(4x^2-1\right)=7x+6\)
=>\(4x^2-12x+9-\left(4x^3-x-20x^2+5\right)=7x+6\)
=>\(4x^2-12x+9-4x^3+20x^2+x-5-7x-6=0\)
=>\(-4x^3+24x^2-18x-2=0\)
=>\(-4x^3+4x^2+20x^2-20x+2x-2=0\)
=>\(-4x^2\left(x-1\right)+20x\left(x-1\right)+2\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(-4x^2+20x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\-4x^2+20x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5\pm3\sqrt{3}}{2}\end{matrix}\right.\)
mong các bạn giúp đỡ
ngày 12/6 là mình đi học rồi vào buổi sáng