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câu 1
a)\(ĐKXĐ:x^3-8\ne0=>x\ne2\)
b)\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2-2x+4\right)}{\left(x-2\right)\left(x^2-2x+4\right)}=\frac{3}{x-2}\left(#\right)\)
Thay \(x=\frac{4001}{2000}\)zô \(\left(#\right)\)ta được
\(\frac{3}{\frac{4001}{2000}-2}=\frac{3}{\frac{4001}{2000}-\frac{4000}{2000}}=\frac{3}{\frac{1}{2000}}=6000\)
\(D=\frac{x^{2}-2x+2018}{x^{2}}\)
\(D=\frac{x^{2}-2*x*1+1+2017}{x^{2}}\)
\(D= \frac{(x-1)^{2}+2017}{x^{2}}\)
Nhận xét: Để D Đặt GTNN thì \((x-1)^{2} + 2017\) Đạt GTNN
Mà \((x-1)^{2} \geq 0\) . Nên:
\((x-1)^{2}+2017\)\(\geq 2017\). GTNN của \((x-1)^{2}+2017=2017 \) Khi x-1=0 => x=1
Thay x=1 vào D
GTNN D=2017
\(P=2x\left(x+y\right)=2x^2+2xy\) Với x khác y, x khác -y
\(3x^2+y^2+2x-2y=1\)\(\Leftrightarrow2x^2+2xy+y^2+x^2+1-2xy+2x-2y=2\)
\(\Leftrightarrow P+\left(x-y+1\right)^2=2\)\(\Leftrightarrow P=2-\left(x-y+1\right)^2\le2\)vì \(\left(x-y+1\right)^2\ge0\)với mọi x, y là số thực
Vì P nguyên dương => P=1
Khi đó \(\left(x-y+1\right)^2=1\Leftrightarrow\orbr{\begin{cases}x-y+1=-1\\x-y+1=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y=-2\\x-y=0\left(loai\right)\end{cases}}\)
vì x khác y
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
a) \(P=\frac{2}{2x+3}+\frac{3}{2x+1}-\frac{6x+5}{\left(2x+3\right)\left(2x-3\right)}\)
\(=\frac{2\left(2x+1\right)\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}+\frac{3\left(2x+3\right)\left(2x-3\right)}{\left(2x+1\right)\left(2x+3\right)\left(2x-3\right)}-\frac{\left(6x+5\right)\left(2x+1\right)}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)
\(=\frac{\left(4x+2\right)\left(2x-3\right)+3\left(4x^2-9\right)-12x^2-16x-5}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)
\(=\frac{8x^2-8x-6+12x^2-27-12x^2-16x-5}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)
\(=\frac{8x^2-24x-38}{\left(2x+3\right)\left(2x-3\right)\left(2x+1\right)}\)
Check hộ mình xem nghi nghi sai sai
b) \(Q=\left(\frac{x+1}{2x-1}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left(\frac{x+1}{2x-1}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right).\frac{4x^2-4}{5}\)
\(=\left(\frac{2\left(x+1\right)\left(x-1\right)\left(x+1\right)}{2\left(2x-1\right)\left(x-1\right)\left(x+1\right)}+\frac{2.3\left(2x-1\right)}{2\left(x-1\right)\left(x+1\right)\left(2x-1\right)}-\frac{\left(x+3\right)\left(2x-1\right)\left(x-1\right)}{2\left(x+1\right)\left(2x-1\right)\left(x-1\right)}\right).\frac{4x^2-4}{5}\)
\(=\frac{2\left(x+1\right)\left(x^2-1\right)+12x-6-\left(2x^2+5x-3\right)\left(x-1\right)}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)
\(=\frac{2\left(x^3+x^2-x-1\right)+12x-6-2x^3-5x^2+3x+2x^2+5x-3}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)
\(=\frac{2x^3+2x^2-2x-2+20x-2x^3-3x^2-9}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4x^2-4}{5}\)
\(=\frac{-x^2+18x-11}{2\left(2x-1\right)\left(x+1\right)\left(x-1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{-x^2+18x-11}{\left(2x-1\right)}.\frac{2}{5}\)
\(=\frac{-2x^2+36x-22}{5\left(2x-1\right)}\)
\(A=\frac{-3}{2+\left|2x-5\right|}\)
Ta có: \(\left|2x-5\right|\ge0\forall x\)
\(\Rightarrow2+\left|2x-5\right|\ge2\forall x\)
\(\Rightarrow\frac{1}{2+\left|2x-5\right|}\le\frac{1}{2}\forall x\)
\(\Rightarrow\frac{-3}{2+\left|2x-5\right|}\ge\frac{-3}{2}\forall x\)
hay \(A\ge\frac{-3}{2}\forall x\)
GTNN của \(A=\frac{-3}{2}\)
Dấu "=" xảy ra khi:
\(\left|2x-5\right|=0\)
\(\Rightarrow2x-5=0\)
\(\Rightarrow x=\frac{5}{2}\)