Hỏi đáp môn Toán

d) Điều kiện: x > 0.

Đặt \(t=\ln x\) khi đó phương trình trở thành:

\(t^3-3t^2-4t+12=0\)

\(\Leftrightarrow t^2\left(t-3\right)-4\left(t-3\right)=0\)

\(\Leftrightarrow\left(t-3\right)\left(t-2\right)\left(t+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=3\\t=2\\t=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\ln x=3\\\ln x=2\\\ln x=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=e^3\\x=e^2\\x=\dfrac{1}{e^2}\end{matrix}\right.\)

Cả 3 nghiệm đều thỏa mãn.

c) Điều kiện: x > 0.

Khi đó biến đổi phương trình như sau:

\(2^{\log_3x^2}.5^{\log_3x}=400\)

\(\Leftrightarrow2^{2\log_3x}.5^{\log_3x}=400\)

\(\Leftrightarrow\left(2^2.5\right)^{\log_3x}=400\)

\(\Leftrightarrow20^{\log_3x}=20^2\)

\(\Leftrightarrow\log_3x=2\)

\(\Leftrightarrow x=3^2=9\) (thỏa mãn)

ĐK : \(x\ge0\) và \(x\ne1\)

\(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\left(15\sqrt{x}-11\right)-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2-3\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

\(P\le\dfrac{2}{3}\Leftrightarrow\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\le\dfrac{2}{3}\)

\(\Leftrightarrow-15\sqrt{x}+6\le2\sqrt{x}+6\)

\(\Leftrightarrow-17\sqrt{x}\le0\)( Luôn đúng với mọi \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\) )

GIÚP MÌNH VỚI MÌNH ĐANG VỘI

Vì: \(x^2\ge0\forall x\Rightarrow x^2+2\ge2>0\)

=> Đa thức vô nghiệm (đpcm)

c)\(\left\{{}\begin{matrix}x=\dfrac{2m+1}{m+1}\in Z\\y=\dfrac{m}{m+1}\in Z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2-\dfrac{1}{m+1}\in Z\\1-\dfrac{1}{m+1}\in Z\end{matrix}\right.\Rightarrow}m+1\in U\left(1\right)=\left\{\pm1\right\}}\)

Tìm được m=0 và m=-2

b)\(\left\{{}\begin{matrix}x=\dfrac{2m+1}{m+1}\left(1\right)\\y=\dfrac{m}{m+1}\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Rightarrow x\left(m+1\right)=2m+1\Leftrightarrow mx+x=2m+1\Leftrightarrow m=\dfrac{1-x}{x-2}\left(3\right)\)

Thay \(\left(3\right)\) vào \(\left(2\right):y=\dfrac{\dfrac{1-x}{x-2}}{\dfrac{1-x}{x-2}+1}=x-1\)

Lời giải:

Ta có:

\(x-y-1=0\Rightarrow x-y=1\)

\(x-2xy-y=-2018\)

\(\Rightarrow 2xy=x-y+2018=1+2018=2019\)

Do đó:

\((x+y)^2-2021=x^2+2xy+y^2-2021\)

\(=x^2-2xy+y^2+4xy-2021\)

\(=(x-y)^2+4xy-2021=1^2+4.2019-2021=6056\)

c)

\(C=x^4+4x-1\)

\(=x^4-2x^2+1+2x^2+4x-2\)

\(=(x^2-1)^2+2(x^2+2x+1)-4\)

\(=(x^2-1)^2+2(x+1)^2-4\)

\(=(x-1)^2(x+1)^2+2(x+1)^2-4=(x+1)^2[(x-1)^2+2]-4\)

Thấy rằng:

\((x+1)^2\geq 0; (x-1)^2+2>0\Rightarrow (x+1)^2[(x-1)^2+2]\geq 0\)

\(\Rightarrow C\geq 0-4=-4\)

Vậy $C_{\min}=-4$ khi \((x+1)^2=0\Leftrightarrow x=-1\)

d)

\(D=4x^2+\frac{9}{x^2}=(2x)^2+(\frac{3}{x})^2-2.2x.\frac{3}{x}+12\)

\(=(2x-\frac{3}{x})^2+12\geq 0+12=12\)

Vậy $D_{\min}=12$ khi \(2x-\frac{3}{x}=0\Leftrightarrow x=\pm \sqrt{\frac{3}{2}}\)

Lời giải:

a)

Ta có: \(A=4x^2-x-2=(2x)^2-2.2x.\frac{1}{4}x+(\frac{1}{4})^2-\frac{33}{16}\)

\(=(2x-\frac{1}{4})^2-\frac{33}{16}\)

Vì \((2x-\frac{1}{4})^2\geq 0, \forall x\in\mathbb{R}\Rightarrow A\ge 0-\frac{33}{16}=-\frac{33}{16}\)

Vậy GTNN của $A$ là $\frac{-33}{16}$ khi $x=\frac{1}{8}$

b)

\(B=\frac{2x^2+6x-3}{5}=\frac{2(x^2+3x+\frac{9}{4})-\frac{15}{2}}{5}\)

\(=\frac{2(x+\frac{3}{2})^2-\frac{15}{2}}{5}\geq \frac{2.0-\frac{15}{2}}{5}=\frac{-3}{2}\)

Vậy \(B_{\min}=\frac{-3}{2}\Leftrightarrow (x+\frac{3}{2})^2=0\Leftrightarrow x=\frac{-3}{2}\)