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Tớ làm lôn nheé , không chép lại đề đâu
a) 15x3 - 6x2 - 3x
b) ĐKXĐ: x # 1
( x - 1)2 . \(\dfrac{1}{x-1}\)
= x - 1
c) ĐKXĐ: x # 1\(\dfrac{x^2-x-x+1}{x-1}=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\)
d)ĐKXĐ : x # 0 ; x # 5
\(\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}.\dfrac{x-5}{x}=\dfrac{\left(x-5\right)^2}{x^2}\)
a) \(B=\frac{x}{x+1}+\frac{2x-3}{x-1}-\frac{2x^2-x-3}{x^2-1}\)
\(B=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{\left(2x-3\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x^2-x-3}{\left(x-1\right)\left(x+1\right)}\)
\(B=\frac{\left(x^2-x\right)+\left(2x^2+2x-3x-3\right)-\left(2x^2-x-3\right)}{\left(x+1\right)\left(x-1\right)}\)
\(B=\frac{x^2-x+2x^2-x-3-2x^2+x+3}{\left(x+1\right)\left(x-1\right)}\)
\(B=\frac{x^2-x}{\left(x+1\right)\left(x-1\right)}\)
\(B=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(B=\frac{x}{x+1}\)
MÌnh nghĩ đề câu b là với x>-4 mới đúng chứ
\(B=\frac{x}{x+1}+\frac{2x-3}{x-1}-\frac{2x^2-x-3}{\left(x^2-1\right)}.\)
\(=\frac{x\left(x-1\right)+\left(2x-3\right)\left(x+1\right)-2x^2+x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2-x+2x^2-x-3-2x^2+x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2-x}{\left(x-1\right)\left(x+1\right)}=\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x}{x+1}\)
\(\Rightarrow A.B=\frac{x}{\left(x+1\right)}.\frac{x\left(x+1\right)}{\left(x-2\right)}=\frac{x^2}{\left(x-2\right)}=\frac{x^2-4+4}{\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x+2\right)+4}{\left(x-2\right)}=x+2+\frac{4}{x-2}=x-2+\frac{4}{x-2}+4\)
Áp dụng BĐT Cô - Si cho 2 số dương \(x-2;\frac{4}{x-2}\)ta có :
\(x-2+\frac{4}{x-2}\ge2\sqrt{\frac{\left(x-2\right).4}{x-2}}=2\sqrt{4}=4\)
\(\Rightarrow x-2+\frac{4}{x-2}\ge4\Rightarrow x-2+\frac{4}{x-2}+4\ge8\)
Hay \(S_{min}=4\Leftrightarrow x-2=\frac{4}{x-2}\)
\(\Rightarrow\frac{\left(x-2\right)^2}{\left(x-2\right)}=\frac{4}{x-2}\Rightarrow x^2+4x+4=4\)
\(\Rightarrow x^2+4x=0\Rightarrow x\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=-4\left(ktm\right)\end{cases}}\)\(\Rightarrow...\)
1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)
b, \(ax^3+bx^2+5x-50⋮\left(x^2+3x-10\right)\)
\(\Rightarrow f\left(x\right)=ax^3+bx^2+5x-50⋮\left(x-2\right)\left(x+5\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=0\\f\left(-5\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(2\right)=8a+4b+10-50=0\\f\left(-5\right)=-125a+25b-25-50=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=4\left(2a+b\right)=40\\f\left(-5\right)=-25\left(5a-b\right)=75\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=2a+b=1\\f\left(-5\right)=5a-b=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{2}{7}\\b=\dfrac{11}{7}\end{matrix}\right.\)
a) \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+b^2x^2+a^2y^2+b^2y^2=a^2x^2+b^2y^2+2abxy\)
\(\Leftrightarrow b^2x^2-2abxy+a^2y^2=0\)
\(\Leftrightarrow\left(bx\right)^2-2\cdot bx\cdot ay+\left(ay\right)^2=0\)
\(\Leftrightarrow\left(bx-ay\right)^2=0\Rightarrow bx=ay\Rightarrow\left(\frac{a}{x}=\frac{b}{y}\right)\)
b) \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)
\(\Leftrightarrow a^2x^2+b^2x^2+c^2x^2+a^2y^2+b^2y^2+c^2y^2+a^2z^2+b^2z^2+c^2z^2\)
\(=a^2x^2+b^2y^2+c^2z^2+2abxy+2bcyz+2acxz\)
\(\Leftrightarrow b^2x^2-2bxay+a^2y^2+b^2z^2-2bzcy+c^2y^2+a^2z^2-2azcx+c^2x^2=0\)
\(\Leftrightarrow\left(bx-ay\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)
\(\hept{\begin{cases}bx=ay\\bz=cy\\az=cx\end{cases}\Rightarrow\hept{\begin{cases}\frac{a}{x}=\frac{b}{y}\\\frac{b}{y}=\frac{c}{z}\\\frac{a}{x}=\frac{c}{z}\end{cases}}\Rightarrow\left(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\right)}\)
c) \(\left(a+b\right)^2=2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2+b^2+2ab=2a^2+2b^2\)
\(\Leftrightarrow a^2-2ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)^2=0\Leftrightarrow a=b\)